Given that the weights are normally distributed with mean \(\mu = 0.75\) kg and standard deviation \(\sigma\) kg, and 68% of the weights are less than 0.9 kg, we can use the standard normal distribution to find \(\sigma\).

The probability statement is:

\(P(X < 0.9) = P\left( Z < \frac{0.9 - 0.75}{\sigma} \right) = 0.68\)

From standard normal distribution tables, \(P(Z < 0.468) \approx 0.68\).

Thus, \(\frac{0.9 - 0.75}{\sigma} = 0.468\).

Solving for \(\sigma\):

\(\sigma = \frac{0.15}{0.468} \approx 0.3205\)

Therefore, the value of \(\sigma\) is approximately between 0.3205 and 0.3215.

(b) We use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\). Given:

\(z_1 = \frac{3 - \mu}{\sigma} = -1.329\)

\(z_2 = \frac{8 - \mu}{\sigma} = 0.878\)

Solving these equations simultaneously, we find:

\(\sigma = 2.27, \mu = 6.01\)

(c) To find the number of leaves more than 1 standard deviation from the mean, we calculate:

\(P(Z < -1) + P(Z > 1) = 2 \times \Phi(1) - \Phi(-1)\)

\(= 2 - 2 \times 0.8413 = 0.3174\)

Number of leaves: \(2000 \times 0.3174 = 634.8\), so approximately 634 or 635 leaves.

(i) Given that 5% of the fish are longer than 50 cm, we use the z-score for 95% which is 1.645. The equation is:

\(1.645 = \frac{50 - 38}{\sigma}\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{50 - 38}{1.645} = 7.29\)

(ii) To find the proportion of fish shorter than 30 cm, we standardize:

\(z = \frac{30 - 38}{7.29} = -1.097\)

Using the standard normal distribution table, \(P(z < -1.097) = 1 - \Phi(1.097)\)

\(= 1 - 0.8637 = 0.136\)

(iii) The probability that a fish is longer than 50 cm is 0.05. For 9 fish, the probability that none are longer than 50 cm is \((0.95)^9\).

The probability that at least one is longer than 50 cm is:

\(1 - (0.95)^9 = 0.370\)

Given that the tyre pressures follow a normal distribution with mean \(\mu = 1.9\) bars and standard deviation \(\sigma = 0.15\) bars, we need to find the value of \(b\) such that 80% of the data lies within \(1.9 - b\) and \(1.9 + b\).

Since 80% of the data is within these limits, we have \(P(1.9 - b < X < 1.9 + b) = 0.8\).

This corresponds to a cumulative probability of 0.8, which means the z-scores are \(\pm 1.282\) (or approximately \(\pm 1.28\)).

Using the z-score formula: \(z = \frac{b}{0.15}\), we have:

\(\pm 1.282 = \frac{b}{0.15}\)

Solving for \(b\), we get:

\(b = 1.282 \times 0.15 = 0.1923\)

Thus, the safety limits are:

\(1.9 - 0.1923 = 1.7077 \approx 1.71\)

\(1.9 + 0.1923 = 2.0923 \approx 2.09\)

Therefore, the limits are between 1.71 and 2.09 bars.

(i) We know that the probability of the lunch break lasting more than 52 minutes is \(\frac{1}{4} = 0.25\). This corresponds to a standard normal value \(z\) such that \(P(Z > z) = 0.25\). From standard normal distribution tables, \(z \approx 0.674\).

Using the formula for the standard normal variable, \(z = \frac{52 - \mu}{5}\), we have:

\(\frac{52 - \mu}{5} = 0.674\)

Solving for \(\mu\):

\(52 - \mu = 5 \times 0.674\)

\(52 - \mu = 3.37\)

\(\mu = 52 - 3.37 = 48.63\)

Rounding to one decimal place, \(\mu = 48.6\).

(ii) We need to find the probability that the lunch break lasts between 40 and 46 minutes. First, standardize these values:

\(z_1 = \frac{40 - 48.63}{5} = -1.726\)

\(z_2 = \frac{46 - 48.63}{5} = -0.526\)

Using the standard normal distribution table, find the probabilities:

\(P(Z < -1.726) \approx 0.0422\)

\(P(Z < -0.526) \approx 0.2995\)

The probability that the lunch break lasts between 40 and 46 minutes is:

\(P(-1.726 < Z < -0.526) = 0.2995 - 0.0422 = 0.2573\)

The probability that this happens on every one of the next four days is:

\((0.2573)^4 = 0.00438\)

Given that 69% of the distribution is less than 28, we find the corresponding z-score from the standard normal distribution table, which is approximately 0.496. Thus, we have:

\(28 - \mu = 0.496\sigma\)

Similarly, for 90% less than 35, the z-score is approximately 1.282. Thus, we have:

\(35 - \mu = 1.282\sigma\)

We now have two equations:

1. \(28 - \mu = 0.496\sigma\)
2. \(35 - \mu = 1.282\sigma\)

Subtract the first equation from the second:

\((35 - \mu) - (28 - \mu) = 1.282\sigma - 0.496\sigma\)

\(7 = 0.786\sigma\)

Solving for \(\sigma\):

\(\sigma = \frac{7}{0.786} \approx 8.91\)

Substitute \(\sigma = 8.91\) back into the first equation:

\(28 - \mu = 0.496 \times 8.91\)

\(28 - \mu = 4.42\)

\(\mu = 28 - 4.42 = 23.6\)

Thus, the mean \(\mu = 23.6\) and the standard deviation \(\sigma = 8.91\).

Let the height of the sunflowers be a normally distributed random variable \(X\) with mean \(\mu = 115\) cm and standard deviation \(\sigma\).

We know that 80% of the heights are greater than 103 cm, which means 20% are less than 103 cm. This corresponds to a cumulative probability of 0.20.

Using the standard normal distribution table, we find that the z-score corresponding to a cumulative probability of 0.20 is approximately \(z = -0.842\).

We use the z-score formula:

\(z = \frac{X - \mu}{\sigma}\)

Substituting the known values:

\(-0.842 = \frac{103 - 115}{\sigma}\)

\(-0.842 = \frac{-12}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{-12}{-0.842}\)

\(\sigma \approx 14.3\)

To find the least distance that must be thrown to qualify for a certificate, we need to find the 90th percentile of the normal distribution, as the top 10% of pupils qualify.

The mean \\(\mu \\\) is 35.0 m and the standard deviation \\(\sigma \\\) is 11.6 m. We need to find the value of \\(x \\\) such that the cumulative probability is 0.90.

Using the standard normal distribution table, the z-score corresponding to the 90th percentile is approximately \\(z = 1.282 \\\).

We use the z-score formula:

\\(z = \frac{x - \mu}{\sigma} \\\)

Substitute the known values:

\\(1.282 = \frac{x - 35.0}{11.6} \\\)

Solve for \\(x \\\):

\\(x - 35.0 = 1.282 \times 11.6 \\\)

\\(x - 35.0 = 14.8592 \\\)

\\(x = 35.0 + 14.8592 \\\)

\\(x \approx 49.9 \\\)

Therefore, the least distance that must be thrown to qualify for a certificate is 49.9 m.

(i) Since \(P(X > 3.6) = 0.5\), the mean \(\mu\) is 3.6 because the probability of being greater than the mean in a normal distribution is 0.5.

For \(P(X > 2.8) = 0.6554\), we standardize: \(\frac{2.8 - \mu}{\sigma} = -0.4\).

Substitute \(\mu = 3.6\): \(\frac{2.8 - 3.6}{\sigma} = -0.4\).

Solve for \(\sigma\): \(\sigma = \frac{-0.8}{-0.4} = 2\).

(ii) Let \(p = 0.6554\) be the probability that an observation is greater than 2.8. We use the binomial distribution with \(n = 4\) and \(p = 0.6554\).

The probability that at least two observations are greater than 2.8 is:

\(P(X \geq 2) = \binom{4}{2} (0.6554)^2 (0.3446)^2 + \binom{4}{3} (0.6554)^3 (0.3446) + \binom{4}{4} (0.6554)^4\).

Calculate each term:

\(\binom{4}{2} (0.6554)^2 (0.3446)^2 = 0.3061\)

\(\binom{4}{3} (0.6554)^3 (0.3446) = 0.3881\)

\(\binom{4}{4} (0.6554)^4 = 0.1845\)

Sum these probabilities: \(0.3061 + 0.3881 + 0.1845 = 0.879\).

(a) To find the probability that a randomly chosen male leopard weighs between 46 and 62 kg, we standardize the values using the formula:

\(P(46 < X < 62) = P\left( \frac{46 - 55}{6} < Z < \frac{62 - 55}{6} \right)\)

This simplifies to:

\(P(-1.5 < Z < 1.1667)\)

Using the standard normal distribution table, we find:

\(\Phi(1.1667) - \Phi(-1.5) = 0.8784 + (0.9332 - 1) = 0.812\)

(b) Given that 25% of female leopards weigh less than 36 kg, we find the z-value corresponding to 0.25, which is approximately -0.674. Using the standardization formula:

\(\frac{36 - 42}{\sigma} = -0.674\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{36 - 42}{-0.674} = 8.9\)

(c) To find the probability that both leopards weigh less than 46 kg, we calculate:

For male leopards: \(P(X < 46) = 1 - \Phi\left(\frac{46 - 55}{6}\right) = 0.0668\)

For female leopards: \(P(Y < 46) = \Phi\left(\frac{46 - 42}{8.9}\right) = 0.6732\)

The probability that both events occur is:

\(P(\text{both}) = 0.0668 \times 0.6732 = 0.0450\)

Given that the running times are normally distributed with mean \(\mu = 41.2\) and standard deviation \(\sigma = 3.6\), we need to find the value of \(t\) such that 95% of the times are greater than \(t\).

This implies that 5% of the times are less than \(t\). The corresponding z-score for the 5th percentile is \(z = -1.645\).

Using the standardization formula:

\(\frac{t - 41.2}{3.6} = -1.645\)

Solving for \(t\):

\(t - 41.2 = -1.645 \times 3.6\)

\(t - 41.2 = -5.922\)

\(t = 41.2 - 5.922\)

\(t = 35.278\)

Rounding to one decimal place, \(t = 35.3\).

Given that the times are normally distributed with mean \(\mu = 32.2\) and standard deviation \(\sigma = 9.6\), we need to find the value of \(t\) such that 20% of employees take longer than \(t\) minutes.

This implies that 80% of employees take less than \(t\) minutes. We need to find the corresponding \(z\)-value for 0.8 in the standard normal distribution table.

The \(z\)-value for 0.8 is approximately 0.842.

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(0.842 = \frac{t - 32.2}{9.6}\)

Solve for \(t\):

\(t - 32.2 = 0.842 \times 9.6\)

\(t - 32.2 = 8.0832\)

\(t = 32.2 + 8.0832\)

\(t = 40.2832\)

Rounding to one decimal place, \(t = 40.3\).

We are given that the times are normally distributed with mean \(\mu = 125\) and standard deviation \(\sigma = 24\). We need to find the value of \(t\) such that 90% of the time, Karli spends more than \(t\) minutes on social media.

This implies that 10% of the time, Karli spends less than \(t\) minutes. We need to find the z-score corresponding to the 10th percentile of the standard normal distribution, which is \(z = -1.282\).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(-1.282 = \frac{t - 125}{24}\)

Solving for \(t\):

\(t - 125 = -1.282 \times 24\)

\(t - 125 = -30.168\)

\(t = 125 - 30.168\)

\(t = 94.832\)

Rounding to one decimal place, \(t \approx 94.2\).

(a) Using the standard normal distribution, we have:

\(z_1 = \frac{4 - \mu}{\sigma} = -1.378\)

\(z_2 = \frac{10 - \mu}{\sigma} = 0.842\)

Solving these equations simultaneously:

\(4 - \mu = -1.378\sigma\)

\(10 - \mu = 0.842\sigma\)

By solving, we find \(\sigma = 2.70\) and \(\mu = 7.72\).

(b) The probability of a leaf length being between \(\mu - 2\sigma\) and \(\mu + 2\sigma\) is given by:

\(\Phi(2) - \Phi(-2) = 2\Phi(2) - 1\)

Using \(\Phi(2) \approx 0.9772\), we have:

\(2 \times 0.9772 - 1 = 0.9544\)

Therefore, the expected number of leaves is \(0.9544 \times 800 = 763.52\), which rounds to 763 or 764.

The probability that a bag weighs more than 1.10 kg is given by:

\(P(X > 1.1) = \frac{72}{2000} = 0.036\)

Using the standard normal distribution, this corresponds to a \(z\)-value of approximately \(z = 1.798\).

We use the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substituting the known values:

\(1.798 = \frac{1.1 - 1.04}{\sigma}\)

\(1.798 = \frac{0.06}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{0.06}{1.798} \approx 0.0334\)

Given that 88% of shoppers spend more than t minutes, this implies that 12% spend less than t minutes. We need to find the z-score corresponding to the 12th percentile of a standard normal distribution.

The z-score for the 12th percentile is approximately -1.175.

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute \(z = -1.175\), \(\mu = 96\), and \(\sigma = 18\):

\(-1.175 = \frac{t - 96}{18}\)

Solving for \(t\):

\(t - 96 = -1.175 \times 18\)

\(t - 96 = -21.15\)

\(t = 96 - 21.15\)

\(t = 74.85\)

Therefore, the value of \(t\) is approximately between 74.85 and 74.9.

Given that the times are normally distributed with a mean of 62 seconds and a standard deviation of 5 seconds, we need to find the time t such that 13% of the members take more than t minutes to swim 100 metres.

\(First, convert t minutes to seconds: t minutes = 60t seconds.\)

We know that 13% corresponds to a z-score of approximately 1.127 (from standard normal distribution tables).

Using the z-score formula:

\(\frac{60t - 62}{5} = 1.127\)

Solving for t:

\(60t - 62 = 5 \times 1.127 = 5.635\)

\(60t = 5.635 + 62 = 67.635\)

\(t = \frac{67.635}{60} \approx 1.13\)

We are given that 75% of the time, Pia takes longer than t minutes. This implies that 25% of the time, she takes less than t minutes. We need to find the z-score corresponding to the 25th percentile of a standard normal distribution.

The z-score for the 25th percentile is approximately (-0.674).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(-0.674 = \frac{t - 10.1}{1.3}\)

Solve for \(t\):

\(t - 10.1 = -0.674 \times 1.3\)

\(t - 10.1 = -0.8762\)

\(t = 10.1 - 0.8762\)

\(t = 9.2238\)

Rounding to two decimal places, \(t = 9.22\).

(a)(i) To find the probability that a randomly chosen member has a height less than 170 cm, we use the standard normal distribution formula:

\(P(X < 170) = P\left(Z < \frac{170 - 166}{10}\right) = P(Z < 0.4)\)

Using standard normal distribution tables, \(P(Z < 0.4) = 0.655\).

(a)(ii) We know that 40% of the members have heights greater than \(h\) cm, so:

\(P\left(Z > \frac{h - 166}{10}\right) = 0.4\)

This implies:

\(\frac{h - 166}{10} = 0.253\)

Solving for \(h\):

\(h = 166 + 10 \times 0.253 = 168.53\)

(b) Given \(\sigma = \frac{2}{3}\mu\), we need to find \(P(X > 0)\):

\(P(X > 0) = P\left(Z > \frac{0 - \mu}{\sigma}\right) = P\left(Z > \frac{0 - \mu}{\frac{2}{3}\mu}\right)\)

\(= P\left(Z > -\frac{3}{2}\right)\)

Using standard normal distribution tables, \(P(Z > -1.5) = 0.933\).

We are given that the time is normally distributed with mean \(\mu = 3.5\) and standard deviation \(\sigma = 0.9\).

We need to find the value of \(t\) such that 90% of the time is greater than \(t\). This implies that 10% of the time is less than \(t\).

Using the standard normal distribution, we find the critical value for 10%, which corresponds to \(z = -1.282\).

We use the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(-1.282 = \frac{t - 3.5}{0.9}\)

Solving for \(t\):

\(t - 3.5 = -1.282 \times 0.9\)

\(t - 3.5 = -1.1538\)

\(t = 3.5 - 1.1538\)

\(t = 2.3462\)

Rounding to two decimal places, \(t = 2.35\).

Given that the distribution is normal with mean \\(\mu = 15.8 \\\) and standard deviation \\(\sigma = 4.2 \\\), we need to find the value of \\(k \\\) such that \\(P(X < k) = 0.75 \\\).

First, find the z-score corresponding to the cumulative probability of 0.75. From standard normal distribution tables, \\(z = 0.674 \\\).

Use the z-score formula:

\\(z = \frac{k - \mu}{\sigma} \\\)

Substitute the known values:

\\(0.674 = \frac{k - 15.8}{4.2} \\\)

Solve for \\(k \\\):

\\(k - 15.8 = 0.674 \times 4.2 \\\)

\\(k - 15.8 = 2.8308 \\\)

\\(k = 15.8 + 2.8308 \\\)

\\(k = 18.6308 \\\)

Rounding to one decimal place, \\(k = 18.6 \\\).

(a) To find the probability that a tree is classified as short, we calculate:

\(P(X < 25) = P\left( Z < \frac{25 - 40}{12} \right) = P(Z < -1.25)\)

Using the standard normal distribution table, \(P(Z < -1.25) = 0.106\).

Thus, the probability is 0.106.

(b) Given that one third of the trees classified as tall or medium are tall, we have:

\(P(\text{tall}) = \frac{1}{3} \times (1 - 0.106) = \frac{0.8944}{3} = 0.298\).

Therefore, the probability that a randomly chosen tree is classified as tall is 0.298.

(c) To find the height above which trees are classified as tall, we use the probability found in part (b):

\(0.2981\) gives \(z = 0.53\).

Using the standard normal distribution formula:

\(\frac{h - 40}{12} = 0.53\)

Solving for \(h\):

\(h = 0.53 \times 12 + 40 = 46.4\).

Thus, trees are classified as tall if their height is above 46.4 m.

Given that the lengths of male snakes are normally distributed, we can use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).

From the problem, 32 out of 200 snakes have lengths less than 45 cm. This corresponds to a probability of \(\frac{32}{200} = 0.16\). Using the standard normal distribution table, this probability corresponds to a z-score of approximately \(-0.994\).

Similarly, 17 out of 200 snakes have lengths more than 56 cm, corresponding to a probability of \(\frac{17}{200} = 0.085\). This corresponds to a z-score of approximately \(1.372\).

We have the following standardization equations:

\(\frac{45 - \mu}{\sigma} = -0.994\)

\(\frac{56 - \mu}{\sigma} = 1.372\)

Solving these two equations simultaneously, we first eliminate \(\mu\) by subtracting the first equation from the second:

\(\frac{56 - \mu}{\sigma} - \frac{45 - \mu}{\sigma} = 1.372 + 0.994\)

\(\frac{11}{\sigma} = 2.366\)

\(\sigma = \frac{11}{2.366} \approx 4.65\)

Substituting \(\sigma = 4.65\) back into one of the original equations to find \(\mu\):

\(\frac{45 - \mu}{4.65} = -0.994\)

\(45 - \mu = -0.994 \times 4.65\)

\(\mu = 45 + 4.6251 \approx 49.6\)

Thus, the estimates for the mean and standard deviation are \(\mu = 49.6\) and \(\sigma = 4.65\).

(a) We know that \(P(X > 87) = 0.22\). Using the standard normal distribution, we have:

\(P\left( Z > \frac{87 - 82}{\sigma} \right) = 0.22\)

This implies:

\(P\left( Z < \frac{5}{\sigma} \right) = 0.78\)

From standard normal distribution tables, \(\frac{5}{\sigma} \approx 0.772\).

Thus, \(\sigma = \frac{5}{0.772} \approx 6.48\).

(b) We need to find \(P(-4 < X - 82 < 4)\), which is equivalent to:

\(P\left( -\frac{4}{\sigma} < Z < \frac{4}{\sigma} \right)\)

Using \(\sigma = 6.48\), we have:

\(P(-0.6176 < Z < 0.6176)\)

From standard normal distribution tables, \(\Phi(0.6176) = 0.7317\).

Therefore, the probability is:

\(2 \times 0.7317 - 1 = 0.463\)

Given that the heights are normally distributed with mean \(\mu = 148\) cm and standard deviation \(\sigma = 8\) cm, we need to find the height \(h\) such that \(P(X < h) = 0.67\).

Using the standard normal distribution, we find the z-score corresponding to a probability of 0.67. From standard normal distribution tables, \(P(Z < 0.44) = 0.67\), so \(z = 0.44\).

The z-score formula is \(z = \frac{h - \mu}{\sigma}\).

Substitute the known values: \(0.44 = \frac{h - 148}{8}\).

Solve for \(h\):

\(h - 148 = 0.44 \times 8\)

\(h - 148 = 3.52\)

\(h = 148 + 3.52\)

\(h = 151.52\)

Rounding to the nearest whole number, \(h = 152\).

Given that the heights are normally distributed, we have:

\(P(X < 10) = \frac{48}{500} = 0.096\)

Using the standard normal distribution table, \(z = -1.305\) for \(P(Z < z) = 0.096\).

Similarly, \(P(X > 24) = \frac{76}{500} = 0.152\)

Using the standard normal distribution table, \(z = 1.028\) for \(P(Z > z) = 0.152\).

We form the equations:

\(10 - \mu = -1.305\sigma\)

\(24 - \mu = 1.028\sigma\)

Subtracting these equations gives:

\(14 = 2.333\sigma\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{14}{2.333} = 6.00\)

Substituting \(\sigma = 6.00\) back into one of the equations:

\(10 - \mu = -1.305 \times 6.00\)

\(10 - \mu = -7.83\)

\(\mu = 17.83\)

Thus, the mean \(\mu\) is approximately 17.8 and the standard deviation \(\sigma\) is 6.00.

Given that 92% of athletes have PBs of more than t seconds, this implies that 8% have PBs less than t seconds. In a standard normal distribution, this corresponds to a z-score of approximately -1.406.

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

where \(\mu = 49.2\) and \(\sigma = 2.8\), we have:

\(\frac{t - 49.2}{2.8} = -1.406\)

Solving for \(t\):

\(t - 49.2 = -1.406 \times 2.8\)

\(t - 49.2 = -3.9368\)

\(t = 49.2 - 3.9368\)

\(t = 45.2632\)

Rounding to one decimal place, \(t = 45.3\).

Given that 96% of ferry crossings take longer than \(t\), this implies that 4% take less time than \(t\). We need to find the corresponding \(z\)-value for the 4th percentile of the standard normal distribution.

The \(z\)-value for 4% in the standard normal distribution is approximately \(z = -1.751\).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute \(z = -1.751\), \(\mu = 85\), and \(\sigma = 6.8\):

\(-1.751 = \frac{t - 85}{6.8}\)

Solving for \(t\):

\(t - 85 = -1.751 \times 6.8\)

\(t - 85 = -11.9088\)

\(t = 85 - 11.9088\)

\(t = 73.0912\)

Rounding to one decimal place, \(t = 73.1\).

Given that the weights are normally distributed, we can use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).

For 20% of pandas weighing more than 121 kg, the corresponding z-score is \(z = 0.842\). Using the standardization formula:

\(z = \frac{121 - \mu}{\sigma}\)

Substituting the values, we get:

\(0.842 = \frac{121 - \mu}{\sigma}\)

So, \(0.842\sigma = 121 - \mu\).

For 71.9% of pandas weighing more than 102 kg, the corresponding z-score is \(z = -0.58\). Using the standardization formula:

\(z = \frac{102 - \mu}{\sigma}\)

Substituting the values, we get:

\(-0.58 = \frac{102 - \mu}{\sigma}\)

So, \(-0.58\sigma = 102 - \mu\).

We now have two equations:

1. \(0.842\sigma = 121 - \mu\)
2. \(-0.58\sigma = 102 - \mu\)

Solving these simultaneously, we eliminate \(\mu\) and solve for \(\sigma\):

\(0.842\sigma + \mu = 121\)

\(-0.58\sigma + \mu = 102\)

Subtract the second equation from the first:

\((0.842 + 0.58)\sigma = 121 - 102\)

\(1.422\sigma = 19\)

\(\sigma = \frac{19}{1.422} \approx 13.4\)

Substitute \(\sigma = 13.4\) back into one of the original equations to find \(\mu\):

\(0.842 \times 13.4 + \mu = 121\)

\(11.2908 + \mu = 121\)

\(\mu = 121 - 11.2908\)

\(\mu \approx 110\)

Thus, the mean \(\mu\) is 110 and the standard deviation \(\sigma\) is 13.4.

(a) We know that \(P(X < 16) = 0.1\). Using the standard normal distribution, we find the z-value corresponding to 0.1, which is approximately \(-1.282\).

Using the standardization formula:

\(\frac{16 - 28}{\sigma} = -1.282\)

Solving for \(\sigma\):

\(\sigma = \frac{12}{1.282} \approx 9.36\)

(c) We need to find \(P(-1.3 < Z < 1.3)\).

Using the standard normal distribution table, \(\Phi(1.3) \approx 0.9032\).

Thus, \(P(-1.3 < Z < 1.3) = 2 \times 0.9032 - 1 = 0.8064\).

In 365 days, the expected number of days is:

\(0.8064 \times 365 \approx 294.336\)

Therefore, the expected number of days is 294 or 295.

We are given that 83.4% of adult male giraffes weigh more than 950 kg. This implies that the probability \(P(X > 950) = 0.834\).

Using the standard normal distribution table, we find that a probability of 0.834 corresponds to a \(z\)-score of approximately -0.97 (since we are looking for the left tail, we use the negative \(z\)-score).

The standardization formula is:

\(z = \frac{X - \mu}{\sigma}\)

Substituting the known values:

\(-0.97 = \frac{950 - 1190}{\sigma}\)

Solving for \(\sigma\):

\(-0.97\sigma = 950 - 1190\)

\(-0.97\sigma = -240\)

\(\sigma = \frac{-240}{-0.97}\)

\(\sigma \approx 247\)

We are given that 90% of the weights lie between (830 - w) kg and (830 + w) kg. This implies that 5% of the weights are below (830 - w) kg and 5% are above (830 + w) kg.

Thus, the probability that a giraffe weighs less than (830 + w) kg is 95%, which corresponds to a z-score of 1.645 for a standard normal distribution.

Using the standardization formula:

\(\frac{(830 + w) - 830}{120} = \frac{w}{120} = 1.645\)

Solving for \(w\):

\(w = 1.645 \times 120 = 197.4\)

Since \(w\) must be an integer, we round to \(w = 197\).

Given that the probability that the time is greater than k is 0.675, we have:

\(P(X > k) = 0.675\)

Using the standard normal distribution, we find the corresponding z-value for the cumulative probability of \(1 - 0.675 = 0.325\).

The z-value for 0.325 is approximately \(z = -0.454\).

Using the standardization formula:

\(z = \frac{k - 140}{12}\)

Substitute \(z = -0.454\):

\(\frac{k - 140}{12} = -0.454\)

Solving for k:

\(k - 140 = -0.454 \times 12\)

\(k - 140 = -5.448\)

\(k = 140 - 5.448\)

\(k = 134.552\)

Rounding to two decimal places, \(k = 134.55\).

Let \(X\) be the lifetime of the light bulb. We know \(X \sim N(2000, \sigma^2)\).

We are given \(P(X > 1800) = 0.96\). This can be rewritten in terms of the standard normal variable \(Z\) as:

\(P\left(Z > \frac{1800 - 2000}{\sigma}\right) = 0.96\).

Using the standard normal distribution table, we find that \(P(Z > -1.751) = 0.96\), so:

\(\frac{200}{\sigma} = 1.751\).

Solving for \(\sigma\), we get:

\(\sigma = \frac{200}{1.751} \approx 114\).

(i) To find the number of days a student uses the machine for less than 4 hours, we calculate \(P(X < 4)\). Standardize using \(Z = \frac{X - \mu}{\sigma}\), where \(\mu = 3.24\) and \(\sigma = 0.96\):

\(Z = \frac{4 - 3.24}{0.96} = 0.7917\)

Using the standard normal distribution table, \(P(Z < 0.7917) = 0.7858\).

Expected days: \(0.7858 \times 365 = 287\) days.

(ii) We need \(P(X > k) = 0.2\), which implies \(P(X < k) = 0.8\). Standardize:

\(P\left(Z < \frac{k - 3.24}{0.96}\right) = 0.8\)

From the standard normal table, \(Z = 0.842\).

\(\frac{k - 3.24}{0.96} = 0.842\)

\(k = 0.842 \times 0.96 + 3.24 = 4.05\)

(iii) We find \(P(-1.5 < Z < 1.5)\):

\(\Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1\)

\(\Phi(1.5) = 0.9332\), so \(2 \times 0.9332 - 1 = 0.866\).

Given that the lengths of fish are normally distributed, we have:

\(P(X < 12) = \frac{42}{400} = 0.105\)

\(P(X > 19) = \frac{58}{400} = 0.145\)

Using the standard normal distribution table, we find the corresponding z-values:

\(\frac{12 - \mu}{\sigma} = -1.253\)

\(\frac{19 - \mu}{\sigma} = 1.058\)

We now have two equations:

\(12 - \mu = -1.253\sigma\)

\(19 - \mu = 1.058\sigma\)

Subtract the first equation from the second:

\(19 - 12 = 1.058\sigma + 1.253\sigma\)

\(7 = 2.311\sigma\)

Solving for \(\sigma\):

\(\sigma = \frac{7}{2.311} \approx 3.03\)

Substitute \(\sigma\) back into one of the equations to find \(\mu\):

\(12 - \mu = -1.253 \times 3.03\)

\(12 - \mu = -3.797\)

\(\mu = 12 + 3.797 \approx 15.8\)

Thus, the estimates for the mean and standard deviation are \(\mu = 15.8\) and \(\sigma = 3.03\).

(i) To find \(\sigma\), we use the standard normal distribution. The probability that \(X > 0\) is 0.25, which corresponds to a \(z\)-value of 0.674 (since the probability to the right of the mean is 0.25).

Using the standardization formula:

\(z = \frac{0 - (-3)}{\sigma} = 0.674\)

Solving for \(\sigma\):

\(\sigma = \frac{3}{0.674} \approx 4.45\)

(ii) The probability that a single value of \(X\) is positive is 0.25. We need to find the probability that fewer than 2 out of 8 values are positive. This is a binomial distribution problem with \(n = 8\) and \(p = 0.25\).

We calculate \(P(0) + P(1)\):

\(P(0) = (0.75)^8\)

\(P(1) = \binom{8}{1} (0.25)(0.75)^7\)

\(P(0) + P(1) = 0.1001 + 0.2670 = 0.367\)

(i) We know that 10% of cartons contain less than 440 millilitres of soup. This corresponds to a z-score of \(z = -1.282\) from the standard normal distribution table.

Using the standardization formula:

\(z = \frac{440 - \mu}{9}\)

Substitute \(z = -1.282\):

\(-1.282 = \frac{440 - \mu}{9}\)

Solving for \(\mu\):

\(440 - \mu = -1.282 \times 9\)

\(440 - \mu = -11.538\)

\(\mu = 440 + 11.538\)

\(\mu = 451.538\)

Rounding to the nearest whole number, \(\mu = 452\).

(ii) We need to find the probability that a carton has a volume more than 1.8 standard deviations above the mean. This corresponds to \(P(z > 1.8)\).

From the standard normal distribution table, \(P(z > 1.8) = 1 - 0.9641 = 0.0359\).

The expected number of cartons is:

\(0.0359 \times 150 = 5.385\)

Rounding to the nearest whole number, the number of cartons is 5.

Let the mean be \(\mu\) and the standard deviation be \(\sigma\).

For 36,800 km, the probability is 0.0082, which corresponds to a \(z\)-score of 2.4 (since \(P(Z > 2.4) = 0.0082\)).

For 31,000 km, the probability is 0.6915, which corresponds to a \(z\)-score of -0.5 (since \(P(Z > -0.5) = 0.6915\)).

Using the standardization formula:

\(z_1 = \frac{36800 - \mu}{\sigma} = 2.4\)

\(z_2 = \frac{31000 - \mu}{\sigma} = -0.5\)

We have two equations:

\(36800 - \mu = 2.4\sigma\)

\(31000 - \mu = -0.5\sigma\)

Subtract the second equation from the first:

\((36800 - \mu) - (31000 - \mu) = 2.4\sigma + 0.5\sigma\)

\(5800 = 2.9\sigma\)

\(\sigma = \frac{5800}{2.9} = 2000\)

Substitute \(\sigma = 2000\) back into one of the equations:

\(36800 - \mu = 2.4 \times 2000\)

\(36800 - \mu = 4800\)

\(\mu = 36800 - 4800 = 32000\)

Thus, the mean is 32,000 km and the standard deviation is 2,000 km.

(i) We know that \(P(X > 410) = \frac{225}{6000} = 0.0375\). Therefore, \(P\left( Z > \frac{410 - 400}{\sigma} \right) = 0.0375\), which implies \(P(Z > z) = 0.0375\). Thus, \(P(Z < z) = 0.9625\).

The corresponding \(z\) value for 0.9625 is approximately 1.78. Therefore, \(\frac{10}{\sigma} = 1.78\), giving \(\sigma = \frac{10}{1.78} = 5.62\) grams.

(ii) We need \(P(Z < -1.5)\) and \(P(Z > 1.5)\). This is \(\Phi(-1.5) + 1 - \Phi(1.5) = 2 - 2\Phi(1.5)\).

Using \(\Phi(1.5) \approx 0.9332\), we have \(2 - 2 \times 0.9332 = 0.1336\).

The expected number of packets is \(500 \times 0.1336 = 66.8\), so approximately 66 or 67 packets.

Let \(z_1\) and \(z_2\) be the standard normal variables corresponding to the given percentages.

For 20% of students having times less than 14.5 minutes, \(z_1 = \frac{14.5 - \mu}{\sigma} = -0.842\).

For 67% of students having times greater than 18.5 minutes, 33% have times less than 18.5 minutes, so \(z_2 = \frac{18.5 - \mu}{\sigma} = -0.44\).

We have the equations:

1. \(\frac{14.5 - \mu}{\sigma} = -0.842\)
2. \(\frac{18.5 - \mu}{\sigma} = -0.44\)

Solving these equations simultaneously, we find:

\(\mu = 22.9\)

\(\sigma = 9.95\)

(i) To find the probability that Josie has to wait longer than 6 minutes, we standardize the normal variable:

\(P(T > 6) = P\left( Z > \frac{6 - 5.3}{2.1} \right) = P(Z > 0.333)\)

Using the standard normal distribution table, \(P(Z > 0.333) = 1 - \Phi(0.333) = 1 - 0.6304 = 0.3696\). Therefore, the probability is approximately 0.370 or 0.369.

(ii) For 5% of days, Josie waits longer than \(x\) minutes, so \(P(T > x) = 0.05\). This corresponds to a \(z\)-value of 1.645. Standardizing gives:

\(1.645 = \frac{x - 5.3}{2.1}\)

Solving for \(x\), we get \(x = 1.645 \times 2.1 + 5.3 = 8.7545\). Thus, \(x \approx 8.75\) or \(8.755\).

(iii) The probability that Josie waits longer than \(x\) minutes on fewer than 3 days in 10 days is a binomial probability problem with \(n = 10\) and \(p = 0.05\). We calculate:

\(P(0, 1, 2) = (0.95)^{10} + \binom{10}{1}(0.05)(0.95)^9 + \binom{10}{2}(0.05)^2(0.95)^8\)

\(= 0.988\) (0.9885 to 4 significant figures).

(iv) The probability that Josie misses the bus is \(P(T < 0)\). Standardizing gives:

\(P\left( Z < \frac{0 - 5.3}{2.1} \right) = P(Z < -2.524)\)

Using the standard normal distribution table, \(P(Z < -2.524) = 1 - \Phi(2.524) = 1 - 0.9942 = 0.0058\).

(i) To find the probability that a boy weighs more than 65 kg, we standardize the value:

\(P(X > 65) = P\left( Z > \frac{65 - 61.4}{12.3} \right) = P(Z > 0.2927)\)

Using the standard normal distribution table, \(P(Z > 0.2927) = 1 - 0.6153 = 0.385\).

(ii) We know \(P(X < 65) = 0.6153\), so \(P(65 < X < k) = 0.25\).

Thus, \(P(X < k) = 0.25 + 0.6153 = 0.8653\).

Find the z-value for 0.8653: \(z = 1.105\).

Standardize: \(1.105 = \frac{k - 61.4}{12.3}\).

Solve for \(k\): \(k = 1.105 \times 12.3 + 61.4 = 75.0\).

(iii) For Brigville, use the z-values for the given probabilities:

\(2.326 = \frac{97.2 - \mu}{\sigma}\) and \(-0.44 = \frac{55.2 - \mu}{\sigma}\).

Solve these simultaneous equations to find \(\mu\) and \(\sigma\):

From \(2.326\sigma = 97.2 - \mu\) and \(-0.44\sigma = 55.2 - \mu\),

Eliminate \(\mu\) to find \(\sigma = 15.2\).

Substitute back to find \(\mu = 61.9\).

(i) To find the proportion of large pineapples, calculate \(P(X > 570)\):

Standardize: \(z = \frac{570 - 500}{91.5} = 0.7650\)

\(P(Z > 0.7650) = 1 - P(Z < 0.7650) = 1 - 0.7779 = 0.222\)

To find the proportion of small pineapples, calculate \(P(X < 390)\):

Standardize: \(z = \frac{390 - 500}{91.5} = -1.202\)

\(P(Z < -1.202) = 0.1147\)

The proportion of medium pineapples is \(1 - (0.222 + 0.115) = 0.663\).

(ii) To find the weight exceeded by the heaviest 5% of pineapples, find \(x\) such that \(P(X > x) = 0.05\).

Find the critical value: \(z = 1.645\)

Standardize: \(1.645 = \frac{x - 500}{91.5}\)

\(x = 1.645 \times 91.5 + 500 = 651\)

(iii) To find \(k\) such that \(P(k < X < 610) = 0.3\):

\(P(X > 610) = 0.1147\) (from symmetry)

\(0.3 + 0.1147 = 0.4147 \Rightarrow \Phi(x) = 0.5853\)

Find \(z\) such that \(\Phi(z) = 0.5853\): \(z = 0.215\)

Standardize: \(0.215 = \frac{k - 500}{91.5}\)

\(k = 0.215 \times 91.5 + 500 = 520\)

Given that 87.5% of the batteries last longer than 70 hours, we find the corresponding z-score for the 12.5th percentile (since 100% - 87.5% = 12.5%).

The z-score for the 12.5th percentile is approximately \(z = -1.151\).

Using the standardization formula:

\(z = \frac{X - \mu}{s}\)

Substitute the known values:

\(-1.151 = \frac{70 - 120}{s}\)

Solve for \(s\):

\(-1.151s = 70 - 120\)

\(-1.151s = -50\)

\(s = \frac{-50}{-1.151}\)

\(s \approx 43.4 \text{ or } 43.5\)

(i) To find the standard deviation \(\sigma\), use the standard normal distribution formula:

\(z = \frac{X - \mu}{\sigma}\)

Given \(X = 4.2\), \(\mu = 3.9\), and \(P(X > 4.2) = 0.18\), find \(z\) such that \(P(Z > z) = 0.18\). From standard normal tables, \(z \approx 0.916\).

\(0.916 = \frac{4.2 - 3.9}{\sigma}\)

\(\sigma = \frac{0.3}{0.916} \approx 0.328\)

(ii) To find the probability that the length differs from the mean by less than half a minute, calculate:

\(z = \frac{4.4 - 3.9}{0.328} \approx 1.5267\)

\(z = \frac{3.4 - 3.9}{0.328} \approx -1.5267\)

Using standard normal tables, \(\Phi(1.5267) \approx 0.9364\).

Probability = \(2 \times 0.9364 - 1 = 0.873\)

(iii) Since the standard deviation is doubled, the spread is larger, resulting in a smaller \(z\)-value for the same difference from the mean. Therefore, \(p\) is less than the probability in part (ii).

(i) To find μ, we use the standard normal distribution. Given P(X < 6.8) = 0.75, we find the corresponding z-value from the standard normal distribution table, which is approximately z = 0.674.

Using the standardization formula:

\(z = \frac{6.8 - μ}{0.25μ}\)

\(Substitute z = 0.674:\)

\(0.674 = \frac{6.8 - μ}{0.25μ}\)

Solving for μ:

\(0.674 imes 0.25μ = 6.8 - μ\)

\(0.1685μ = 6.8 - μ\)

\(1.1685μ = 6.8\)

\(μ = \frac{6.8}{1.1685} \approx 5.82\)

(ii) To find P(X < 4.7), we standardize using the mean and standard deviation:

\(P(X < 4.7) = P\left( z < \frac{4.7 - 5.82}{1.4548} \right)\)

\(= P(z < -0.769)\)

Using the standard normal distribution table, \(P(z < -0.769) = 1 - 0.7791 = 0.221\)

Given that 25% of women have fingers longer than 8.8 cm, the corresponding z-score is \(z = 0.674\) (since 75% is below 8.8 cm).

Using the z-score formula: \(z = \frac{X - \mu}{\sigma}\), we have:

\(0.674 = \frac{8.8 - \mu}{\sigma}\)

\(0.674\sigma = 8.8 - \mu\)

Similarly, for 17.5% having fingers shorter than 7.7 cm, the z-score is \(z = -0.935\).

\(-0.935 = \frac{7.7 - \mu}{\sigma}\)

\(-0.935\sigma = 7.7 - \mu\)

We now have two equations:

1. \(0.674\sigma = 8.8 - \mu\)
2. \(-0.935\sigma = 7.7 - \mu\)

Subtract the second equation from the first:

\(0.674\sigma + 0.935\sigma = 8.8 - 7.7\)

\(1.609\sigma = 1.1\)

\(\sigma = \frac{1.1}{1.609} \approx 0.684\)

Substitute \(\sigma = 0.684\) back into the first equation:

\(0.674 \times 0.684 = 8.8 - \mu\)

\(0.461 = 8.8 - \mu\)

\(\mu = 8.8 - 0.461\)

\(\mu = 8.34\)

(i) To find the proportion of bananas that are small, we standardize the weight 95 grams using the formula:

\(z = \frac{95 - 150}{50} = -1.1\)

We find \(P(z < -1.1)\) using the standard normal distribution table:

\(P(z < -1.1) = 0.136\)

Thus, the proportion of bananas that are small is 0.136.

(ii) To find the weight exceeded by 10% of bananas, we need the 90th percentile of the distribution. The z-value for 90% is approximately 1.282.

Using the standardization formula:

\(1.282 = \frac{x - 150}{50}\)

Solving for \(x\):

\(x = 1.282 \times 50 + 150 = 214 \text{ g}\)

(iii) (a) The probability that a banana is medium is given by:

\(P(\text{medium}) = 1 - 2 \times 0.1357 = 0.7286\)

(b) The expected cost per banana is calculated as follows:

\(0.1357 \times 10 + 0.1357 \times 25 + 0.7286 \times 20 = 19.3215 \text{ cents}\)

The total cost for 100 bananas is:

\(100 \times 19.3215 = 1930 \text{ cents} = \$19.30\)

Given that the time to cook an egg is normally distributed with mean \(\mu = 4.2\) minutes and standard deviation \(\sigma = 0.6\) minutes. We need to find the value of \(t\) such that 12% of people take more than \(t\) minutes.

This implies that 88% of people take less than \(t\) minutes. We find the corresponding \(z\)-score for 0.88 in the standard normal distribution table, which is approximately \(z = 1.175\).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute the known values:

\(1.175 = \frac{t - 4.2}{0.6}\)

Solving for \(t\):

\(t - 4.2 = 1.175 \times 0.6\)

\(t - 4.2 = 0.705\)

\(t = 4.2 + 0.705\)

\(t = 4.905\)

Rounding to two decimal places, \(t = 4.91\).

(i) To find the probability that a randomly chosen packet weighs less than 1 kg, we standardize the variable:

\(P(X < 1) = P\left( Z < \frac{1 - 1.04}{0.017} \right) = P(Z < -2.353)\)

Using the standard normal distribution table, \(P(Z < -2.353) = 0.0093\).

(ii) The expected number of packets from 1000 kg is calculated by dividing the total weight by the mean weight per packet:

\(\frac{1000}{1.04} \approx 961 \text{ or } 962\)

(iii) Given that the probability of a packet weighing less than 1 kg is 0.0388, we find \(\mu\) by setting up the equation:

\(P(Z < -1.765) = 0.0388\)

Standardizing gives:

\(-1.765 = \frac{1 - \mu}{0.017}\)

Solving for \(\mu\):

\(1 - \mu = -1.765 \times 0.017\)

\(\mu = 1.03\)

(iv) With the new mean, the expected number of packets is:

\(\frac{1000}{1.03} \approx 971 \text{ or } 970\)

Let the mean be \(\mu\) and the standard deviation be \(\sigma\).

For the time less than 36 minutes, 23% corresponds to a \(z\)-score of approximately \(-0.739\). Thus,

\(z_1 = \frac{36 - \mu}{\sigma} = -0.739\)

For the time greater than 54 minutes, 10% corresponds to a \(z\)-score of approximately \(1.282\). Thus,

\(z_2 = \frac{54 - \mu}{\sigma} = 1.282\)

We have two equations:

1. \(\frac{36 - \mu}{\sigma} = -0.739\)
2. \(\frac{54 - \mu}{\sigma} = 1.282\)

Solving these equations simultaneously, we find:

\(\mu = 42.6\)

\(\sigma = 8.91\)

Given that X ~ N(20, 49), we have a normal distribution with mean bc = 20 and variance 3c2 = 49, so the standard deviation 3c = 7.

\(We need to find k such that P(X > k) = 0.25. This implies that P(X 3c k) = 0.75.\)

Using the standard normal distribution table, we find the z-score corresponding to 0.75 is approximately 0.674.

Standardizing, we have:

\(3cbr>z = 3cfrac{k - 20}{7}\)

Substituting the z-score, we get:

\(0.674 = 3cfrac{k - 20}{7}\)

Solving for k:

\(k - 20 = 0.674 3c 7\)

\(k = 0.674 3c 7 + 20\)

\(k = 4.718 + 20\)

\(k = 24.7\)

(i) We know that 15.5% of desks have a height greater than 70 cm. This corresponds to a z-score of approximately 1.015 (from standard normal distribution tables).

Using the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substitute the known values:

\(1.015 = \frac{70 - 69}{\sigma}\)

Solve for \(\sigma\):

\(\sigma = \frac{1}{1.015} = 0.985\)

(ii) Jodu's knees are at 58 cm, and they need to be at least 9 cm below the desk top, so the desk height must be greater than 67 cm (58 + 9).

Standardize 67 cm:

\(P(>67) = P\left( z > \frac{67 - 69}{0.985} \right)\)

\(= P(z > -2.03)\)

From standard normal distribution tables, \(P(z > -2.03) = 0.9788\).

Estimate the number of comfortable desks:

\(300 \times 0.9788 = 293.64\)

Therefore, approximately 293 desks are comfortable for Jodu.

We are given that the time is normally distributed with mean \(\mu = 9.5\) and standard deviation \(\sigma = 1.3\). We need to find the value of \(t\) such that 90% of the time, Peter takes longer than \(t\) minutes. This means that 10% of the time, he takes less than \(t\) minutes.

Using the standard normal distribution table, we find the z-score corresponding to the 10th percentile, which is \(z = -1.282\).

We use the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substituting the known values:

\(-1.282 = \frac{t - 9.5}{1.3}\)

Solving for \(t\):

\(t - 9.5 = -1.282 \times 1.3\)

\(t - 9.5 = -1.6666\)

\(t = 9.5 - 1.6666\)

\(t = 7.8334\)

Rounding to two decimal places, \(t = 7.83\).

Given that the mean \(\mu = 1.62\) m and the probability \(P(X > 1.8) = 0.15\), we need to find \(\sigma\).

First, find the z-score corresponding to a probability of 0.15 in the standard normal distribution table. This gives \(z = 1.037\).

Using the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substitute the known values:

\(1.037 = \frac{1.8 - 1.62}{\sigma}\)

Solve for \(\sigma\):

\(\sigma = \frac{0.18}{1.037} = 0.174\)

(i) To find the mean \(m\), we use the standard normal distribution. Given that 95% of times are longer than 0.9 hours, the corresponding z-value is \(z = -1.645\) (since 5% is in the left tail).

Using the standardization formula:

\(z = \frac{0.9 - m}{0.35}\)

Substitute \(z = -1.645\):

\(-1.645 = \frac{0.9 - m}{0.35}\)

Solving for \(m\):

\(0.9 - m = -1.645 \times 0.35\)

\(0.9 - m = -0.57575\)

\(m = 0.9 + 0.57575\)

\(m = 1.47575\)

Rounding to 3 significant figures, \(m = 1.48\).

(ii) We need to find the probability that none of the 4 cars takes more than 2 hours to fit. First, find the probability that a single car takes less than 2 hours.

Standardize 2 hours:

\(z = \frac{2 - 1.48}{0.35}\)

\(z = \frac{0.52}{0.35}\)

\(z = 1.4857\)

Using the standard normal distribution table, \(P(z < 1.4857) \approx 0.933\).

The probability that none of the 4 cars takes more than 2 hours is:

\((0.933)^4 \approx 0.758\)

(i) To find \(\sigma\), we use the standard normal distribution. Given that 13% of seeds take longer than 136 hours, we find the corresponding \(z\)-value for 0.87 (since 1 - 0.13 = 0.87) in the standard normal distribution table, which is approximately 1.127.

Using the formula for standardization:

\(z = \frac{136 - 125}{\sigma}\)

Substitute \(z = 1.127\):

\(1.127 = \frac{136 - 125}{\sigma}\)

\(\sigma = \frac{11}{1.127} \approx 9.76\)

(ii) To find the expected number of seeds taking between 131 and 141 hours, we standardize these values:

\(P(131 < x < 141) = P\left( \frac{131 - 125}{9.76} < z < \frac{141 - 125}{9.76} \right)\)

\(= P(0.6147 < z < 1.639)\)

Using the standard normal distribution table:

\(\Phi(1.639) - \Phi(0.6147) = 0.9493 - 0.7307 = 0.2186\)

The expected number of seeds is:

\(0.2186 \times 170 \approx 37.2\)

Thus, the expected number of seeds is 37 or 38.

(i) To find the number of days the sales exceed 3900 litres, we standardize the value:

\(P(x > 3900) = P\left( z > \frac{3900 - 4520}{560} \right) = P(z > -1.107) = \Phi(1.107)\)

Using the standard normal distribution table, \(\Phi(1.107) = 0.8657\).

The expected number of days is \(365 \times 0.8657 = 315.98\), so approximately 315 or 316 days.

(ii) Given \(P(X > 8000) = 0.122\), we find the z-score:

\(z = 1.165\)

Using the standardization formula:

\(1.165 = \frac{8000 - m}{560}\)

Solving for \(m\):

\(m = 8000 - 1.165 \times 560 = 7350\)

(iii) The probability that sales exceed 8000 litres on fewer than 2 of 6 days is calculated using the binomial distribution:

\(P(0, 1) = (0.878)^6 + \binom{6}{1}(0.122)^1(0.878)^5\)

Calculating gives \(0.840\), which can be accepted as 0.84.

Since \(P(X < 54.1) = 0.5\), the mean \(\mu = 54.1\) because the median of a normal distribution is the mean.

For \(P(X > 50.9) = 0.8665\), we find the corresponding \(z\)-score. The probability \(P(X > 50.9) = 0.8665\) implies \(P(X < 50.9) = 1 - 0.8665 = 0.1335\).

Using the standard normal distribution table, \(P(Z < z) = 0.1335\) corresponds to \(z = -1.11\).

Standardizing, we have:

\(-1.11 = \frac{50.9 - 54.1}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{50.9 - 54.1}{-1.11} = 2.88\)

Given that the probability \(P(X > 195) = 0.128\), we need to find the mean \(\mu\) of the normal distribution.

First, find the z-score corresponding to the probability 0.128. From standard normal distribution tables, \(P(Z > 1.136) = 0.128\), so \(z = 1.136\).

Using the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substitute the known values:

\(1.136 = \frac{195 - \mu}{22}\)

Solving for \(\mu\):

\(1.136 \times 22 = 195 - \mu\)

\(24.992 = 195 - \mu\)

\(\mu = 195 - 24.992\)

\(\mu = 170.008\)

Rounding to the nearest whole number, \(\mu = 170\).

(a)(i) To find the probability that Zak takes less than 30 minutes, we standardize the variable:

\(z = \frac{30 - 35.2}{4.7} = -1.106\)

Using the standard normal distribution table, \(P(z < -1.106) = 0.1345\).

The expected number of days in a year is \(0.1345 \times 52 = 6.99\), which rounds to 7 days.

(a)(ii) We know \(\Phi(t) = 0.648\) and \(z = 0.380\).

Standardizing, \(\frac{t - 35.2}{4.7} = 0.380\).

Solving for \(t\), we get \(t = 37.0\).

(b) We have two equations from the given probabilities:

\(\frac{7 - \mu}{\sigma} = -0.8\) and \(\frac{10 - \mu}{\sigma} = 0.44\).

Solving these equations simultaneously, we find \(\mu = 8.94\) and \(\sigma = 2.42\).

Let the random variable \(X\) represent the weights of the bags of sugar. We know \(X\) is normally distributed with mean \(\mu = 1.04\) kg and standard deviation \(\sigma = 0.06\) kg.

We are given that \(P(X < w) = 0.81\). To find \(w\), we standardize \(X\) to a standard normal variable \(Z\):

\(P\left(Z < \frac{w - 1.04}{0.06}\right) = 0.81\)

Using the standard normal distribution table, we find that the z-score corresponding to 0.81 is approximately 0.878.

Thus, we have:

\(\frac{w - 1.04}{0.06} = 0.878\)

Solving for \(w\):

\(w - 1.04 = 0.878 \times 0.06\)

\(w - 1.04 = 0.05268\)

\(w = 1.04 + 0.05268\)

\(w = 1.09268\)

Therefore, \(1.09 \leq w \leq 1.093\).

Given that the probability \(P(3.2 < X < \mu) = 0.475\), we can find the probability \(P(X < 3.2) = 0.5 - 0.475 = 0.025\).

Using the standard normal distribution table, \(P(Z < z) = 0.025\) corresponds to \(z = -1.96\).

Standardizing, we have:

\(z = \frac{3.2 - \mu}{0.714} = -1.96\)

Solving for \(\mu\):

\(3.2 - \mu = -1.96 \times 0.714\)

\(\mu = 3.2 + 1.96 \times 0.714\)

\(\mu = 4.6\)

(i) To find the proportion of gem stones in each category, we standardize the weights using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For Small (\(X < 1.2\)): \(z = \frac{1.2 - 1.9}{0.55} = -1.2727\)

\(P(X < 1.2) = P(z < -1.2727) = 0.1014\)

For Large (\(X > 2.5\)): \(z = \frac{2.5 - 1.9}{0.55} = 1.0909\)

\(P(X > 2.5) = P(z > 1.0909) = 0.138\)

For Medium (\(1.2 < X < 2.5\)): \(P(1.2 < X < 2.5) = 1 - P(X < 1.2) - P(X > 2.5) = 1 - 0.101 - 0.138 = 0.761\)

(ii) We need \(P(k < X < 2.5) = 0.8\). We know \(P(X < 2.5) = 0.8623\), so \(P(X > k) = 0.9377\).

Find \(z\) such that \(P(z) = 0.9377\), which gives \(z = -1.536\).

Using the standardization formula:

\(-1.536 = \frac{k - 1.9}{0.55}\)

Solving for \(k\):

\(k = 1.9 + (-1.536 \times 0.55) = 1.06\)

We are given that the weight of the tea packets follows a normal distribution with mean \(\mu = 260\) g and standard deviation \(\sigma\) g. A packet is considered underweight if it weighs less than 250 g, and 1% of packets are underweight.

We need to find the value of \(\sigma\) such that the probability of a packet being underweight is 0.01. This corresponds to the left tail of the normal distribution.

Using the standard normal distribution table, the z-score corresponding to the 1st percentile is approximately \(z = -2.326\).

We use the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substituting the known values:

\(-2.326 = \frac{250 - 260}{\sigma}\)

\(-2.326 = \frac{-10}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{-10}{-2.326}\)

\(\sigma \approx 4.30\)

(i) To find the probability that a randomly chosen person sleeps for less than 8 hours, we standardize the variable:

\(P(X < 8) = P\left( Z < \frac{8 - 7.15}{0.88} \right)\)

\(= P(Z < 0.9659)\)

Using the standard normal distribution table, \(\Phi(0.9659) = 0.833\).

Thus, the probability is 0.833.

(ii) To find the value of \(q\) such that \(P(X < q) = 0.75\), we find the z-value corresponding to 0.75 in the standard normal distribution table, which is approximately 0.674.

Using the standardization formula:

\(\frac{q - 7.15}{0.88} = 0.674\)

Solving for \(q\):

\(q = 0.674 \times 0.88 + 7.15\)

\(q = 7.74\)

Given that 25% of the sheep weigh more than 67.5 kg, this corresponds to the 75th percentile of the normal distribution. The z-score for the 75th percentile is approximately 0.674.

Using the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substitute the known values:

\(0.674 = \frac{67.5 - \mu}{4.92}\)

Solving for \(\mu\):

\(67.5 - \mu = 0.674 \times 4.92\)

\(67.5 - \mu = 3.31608\)

\(\mu = 67.5 - 3.31608\)

\(\mu = 64.18392\)

Rounding to one decimal place, \(\mu = 64.2\).

(i) To find the value of t, we use the standard normal distribution. We know that 90% of calls take longer than t, so 10% take less time. This corresponds to a z-score of -1.282 (since the left tail is 0.10).

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

\(-1.282 = \frac{t - 6.5}{1.76}\)

Solving for t gives:

\(t = 6.5 + (-1.282 \times 1.76) = 4.24\) minutes

(ii) We need to find the probability that more than 7 out of 9 calls are within 1 standard deviation of the mean. The probability of a call being within 1 standard deviation is given by:

\(P(\text{within 1 sd of mean}) = 2\Phi(1) - 1 = 0.6826\)

We use the binomial distribution with \(n = 9\) and \(p = 0.6826\). We need \(P(X > 7) = P(X = 8) + P(X = 9)\).

\(P(X = 8) = \binom{9}{8} (0.6826)^8 (0.3174)^1\)

\(P(X = 9) = \binom{9}{9} (0.6826)^9\)

Calculating these gives:

\(P(X = 8) = 9 \times 0.6826^8 \times 0.3174\)

\(P(X = 9) = 0.6826^9\)

Adding these probabilities gives \(P(X > 7) = 0.167\).

Given that the lengths are normally distributed, we use the standard normal distribution to find \(\mu\) and \(\sigma\).

For 4% longer than 12 cm, the z-score is:

\(z = \frac{12 - \mu}{\sigma}\)

From standard normal tables, \(z = 1.751\) for 4% in the upper tail.

Thus, \(1.751 = \frac{12 - \mu}{\sigma}\).

For 32% longer than 9 cm, the z-score is:

\(z = \frac{9 - \mu}{\sigma}\)

From standard normal tables, \(z = 0.468\) for 32% in the upper tail.

Thus, \(0.468 = \frac{9 - \mu}{\sigma}\).

We have two equations:

1. \(1.751 = \frac{12 - \mu}{\sigma}\)
2. \(0.468 = \frac{9 - \mu}{\sigma}\)

Subtract the second equation from the first:

\(1.751 - 0.468 = \frac{12 - \mu}{\sigma} - \frac{9 - \mu}{\sigma}\)

\(1.283 = \frac{3}{\sigma}\)

\(\sigma = \frac{3}{1.283} = 2.34\)

Substitute \(\sigma = 2.34\) into the first equation:

\(1.751 = \frac{12 - \mu}{2.34}\)

\(12 - \mu = 1.751 \times 2.34\)

\(12 - \mu = 4.1\)

\(\mu = 12 - 4.1 = 7.91\)

(a) We know that \(P(82-q < X < 82+q) = 0.44\). This implies \(P(X < 82+q) - P(X < 82-q) = 0.44\). Since the distribution is symmetric, \(P(X < 82+q) = 0.72\) and \(P(X < 82-q) = 0.28\). The standard normal variable \(Z\) corresponding to \(X\) is \(Z = \frac{X - 82}{7.4}\). For \(P(X < 82+q) = 0.72\), \(Z = 0.583\). Therefore, \(\frac{q}{7.4} = 0.583\) which gives \(q = 4.31\).

(b) We have \(5\mu = 2\sigma^2\) and \(P(Y < \frac{1}{2}\mu) = 0.281\). Standardizing, \(Z = \frac{\frac{1}{2}\mu - \mu}{\sigma} = \frac{-0.5\mu}{\sigma}\). From the standard normal table, \(Z = -0.580\). Thus, \(\frac{-0.5\mu}{\sigma} = -0.580\) gives \(\sigma = \frac{0.5\mu}{0.580}\). Substituting \(\sigma\) in \(5\mu = 2\sigma^2\), we solve to find \(\mu = 3.36\) and \(\sigma = 2.90\).

(i) Let \(X\) be the amount of fibre in a packet, \(X \sim N(160, \sigma^2)\). We know \(P(X > 190) = 0.19\). The z-score for 190 is given by:

\(z = \frac{190 - 160}{\sigma}\)

From the standard normal distribution table, \(P(Z > z) = 0.19\) corresponds to \(z = 0.878\).

Thus, \(\frac{30}{\sigma} = 0.878\).

Solving for \(\sigma\), we get:

\(\sigma = \frac{30}{0.878} = 34.2\)

(ii) Let \(Y\) be the number of packets with more than 190 grams of fibre. \(Y \sim \text{Binomial}(12, 0.19)\).

We need \(P(Y \geq 1) = 1 - P(Y = 0)\).

\(P(Y = 0) = (0.81)^{12}\)

Thus, \(P(Y \geq 1) = 1 - (0.81)^{12} = 0.920\).

(i) To find the value of c, we use the standard normal distribution. Given that 8% of carrots are shorter than c, we find the corresponding z-score for 0.08 in the standard normal distribution table, which is approximately -1.406.

Using the formula for standardization:

\(z = \frac{c - \mu}{\sigma}\)

\(-1.406 = \frac{c - 14.2}{3.6}\)

Solving for c gives:

\(c = 14.2 + (-1.406 \times 3.6)\)

\(c = 9.14 \text{ cm}\)

(ii) To find the probability that at least 2 of the 7 carrots have lengths between 15 and 16 cm, we first find the probability that a single carrot is in this range.

Standardizing the values 15 and 16:

\(z_1 = \frac{15 - 14.2}{3.6} = 0.222\)

\(z_2 = \frac{16 - 14.2}{3.6} = 0.5\)

Using the standard normal distribution table, find:

\(P(0.222 < z < 0.5) = \Phi(0.5) - \Phi(0.222)\)

\(= 0.6915 - 0.5879 = 0.1036\)

Now, use the binomial distribution for 7 trials with probability 0.1036:

\(P(\text{at least 2}) = 1 - P(0) - P(1)\)

\(= 1 - (0.8964)^7 - 7 \times (0.8964)^6 \times 0.1036\)

\(= 1 - 0.8413\)

\(= 0.159\)

(a) To find the probability that the SBP of a randomly chosen adult is less than 132, we use the standard normal distribution. The z-score is calculated as:

\(z = \frac{132 - 125.4}{18.6} = 0.3548\)

The probability \(P(Z < 0.3548)\) is approximately 0.639.

(b) For the SBP of 12-year-old children, we know that 88% have SBP more than 108. This corresponds to a z-score of -1.175 (since 12% have SBP less than 108). Using the formula:

\(\frac{108 - 117}{\sigma} = -1.175\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{108 - 117}{-1.175} = 7.66\)

(c) To find the probability that each of these three adults has SBP within 1.5 standard deviations of the mean, we calculate:

\(P(-1.5 < Z < 1.5) = \Phi(1.5) - \Phi(-1.5) = 2\Phi(1.5) - 1\)

Using \(\Phi(1.5) \approx 0.9332\), we have:

\(2 \times 0.9332 - 1 = 0.8664\)

The probability that each of the three adults has SBP within this range is:

\(0.8664^3 = 0.650\)

(i) To find the probability that a building is classified as tall, we need to calculate the probability that the height is greater than 70 metres. Using the standard normal distribution, we standardize the height:

\(z = \frac{70 - 50}{16} = 1.25\)

The probability \(P(z > 1.25)\) is found using the standard normal distribution table:

\(P(z > 1.25) = 1 - P(z \leq 1.25) = 1 - 0.8944 = 0.106\)

(ii) Let \(P(\text{short})\) be the probability that a building is classified as short. Since there are twice as many medium buildings as short ones, \(P(\text{medium}) = 2 \times P(\text{short})\). The total probability for medium and short buildings is:

\(P(\text{medium}) + P(\text{short}) = 1 - P(\text{tall}) = 1 - 0.106 = 0.894\)

Substituting \(P(\text{medium}) = 2 \times P(\text{short})\), we have:

\(2P(\text{short}) + P(\text{short}) = 0.894\)

\(3P(\text{short}) = 0.894\)

\(P(\text{short}) = \frac{0.894}{3} = 0.298\)

To find the height below which buildings are classified as short, we find the z-score corresponding to \(P(z < z_0) = 0.298\). From the standard normal distribution table, \(z_0 \approx -0.53\).

Standardizing, we have:

\(-0.53 = \frac{x - 50}{16}\)

Solving for \(x\):

\(x = 50 + (-0.53 \times 16) = 41.5\) metres

(i) To find the probability that a randomly chosen can contains less than 440 ml of juice, we standardize the variable using the formula:

\(z = \frac{x - \mu}{\sigma}\)

where \(x = 440\), \(\mu = 445\), and \(\sigma = 3.6\).

\(z = \frac{440 - 445}{3.6} = -1.389\)

The probability \(P(x < 440)\) is equivalent to \(P(z < -1.389)\).

Using the standard normal distribution table, \(P(z < -1.389) = 1 - \Phi(1.389) = 1 - 0.9176 = 0.0824\).

(ii) We know that 94% of the cans contain between \(445 - c\) ml and \(445 + c\) ml. This corresponds to the middle 94% of the normal distribution, leaving 3% in each tail.

The corresponding z-value for 3% in the lower tail is approximately \(z = 1.881\).

We use the equation:

\(\frac{c}{3.6} = 1.881\)

Solving for \(c\), we get:

\(c = 1.881 \times 3.6 = 6.77\).

Let the mean be \(\mu\) and the standard deviation be \(\sigma\). We are given that \(\mu = 5\sigma\).

The probability \(P(Y > 20) = 0.0732\) corresponds to a standard normal variable \(Z\) such that \(P(Z > z) = 0.0732\). From standard normal distribution tables, \(z \approx 1.452\).

Using the transformation for a normal distribution, \(Z = \frac{Y - \mu}{\sigma}\), we have:

\(1.452 = \frac{20 - \mu}{\mu / 5}\)

Solving for \(\mu\):

\(1.452 = \frac{20 - \mu}{\mu / 5}\)

\(1.452 \times \frac{\mu}{5} = 20 - \mu\)

\(0.2904\mu = 20 - \mu\)

\(1.2904\mu = 20\)

\(\mu = \frac{20}{1.2904} \approx 15.5\)

First, calculate the probability of a bag weighing more than 2.1 kg:

\(P(x > 2.1) = \frac{253}{8000} = 0.031625\)

Thus, the probability of a bag weighing less than 2.1 kg is:

\(P(x < 2.1) = 1 - 0.031625 = 0.968375\)

This corresponds to the cumulative distribution function \(\Phi(z)\), so:

\(\Phi(z) = 0.968375\)

Using standard normal distribution tables, find \(z\) such that \(\Phi(z) = 0.968375\). This gives:

\(z = 1.857 \text{ or } 1.858 \text{ or } 1.859\)

Using the z-score formula:

\(z = \frac{2.1 - 2.04}{\sigma}\)

Substitute \(z = 1.857\) (or similar values) to solve for \(\sigma\):

\(1.857 = \frac{2.1 - 2.04}{\sigma}\)

\(\sigma = \frac{0.06}{1.857} \approx 0.0323\)

(ii) To find the probability that at most one of the five observations is greater than 87, we first find \(P(X > 87)\).

Standardize: \(P(X > 87) = 1 - \Phi \left( \frac{87 - 82}{\sqrt{126}} \right)\)

\(= 1 - \Phi(0.445) = 1 - 0.6718 = 0.3282\)

Using the binomial distribution for 5 trials, \(P(0, 1) = (0.6718)^5 + \binom{5}{1} (0.6718)^4 (0.3282)\)

\(= 0.471\)

(iii) We need to find \(k\) such that \(P(87 < X < k) = 0.3\).

First, find \(P(X < 87) = 0.6718\).

Then, \(P(X < k) = 0.6718 + 0.3 = 0.9718\).

Find the corresponding \(z\)-value: \(z = 1.908\) or \(1.909\).

Use the equation: \(1.909 = \frac{k - 82}{\sqrt{126}}\)

Solve for \(k\): \(k = 103\).

Let the mean be \(\mu = 9.3\) and the standard deviation be \(\sigma\). We are given that \(P(X > 5.6) = 0.85\).

Using the standard normal distribution, we find \(P(Z > z) = 0.85\), which implies \(P(Z < z) = 0.15\).

From the standard normal distribution table, \(z \approx -1.036\).

Using the formula for the standard normal variable, \(z = \frac{X - \mu}{\sigma}\), we have:

\(z = \frac{5.6 - 9.3}{\sigma} = -1.036\)

Solving for \(\sigma\):

\(-1.036 = \frac{5.6 - 9.3}{\sigma}\)

\(\sigma = \frac{5.6 - 9.3}{-1.036}\)

\(\sigma = 3.57\)

Let the mean be \(\mu\) and the standard deviation be \(s\). We know that \(\mu = 4s\).

The probability that a value is greater than 5 is 0.15, which corresponds to a z-score of approximately 1.036 or 1.037.

Using the z-score formula:

\(z = \frac{5 - \mu}{s}\)

Substitute \(\mu = 4s\) into the equation:

\(1.036 = \frac{5 - 4s}{s}\)

Solve for \(s\):

\(1.036s = 5 - 4s\)

\(5.036s = 5\)

\(s = \frac{5}{5.036} \approx 0.993\)

Substitute back to find \(\mu\):

\(\mu = 4s = 4 \times 0.993 = 3.97\)

(i) To find the standard deviation \(\sigma\), we use the z-score formula:

\(z = \frac{X - \mu}{\sigma}\)

Given \(X = 73\), \(\mu = 75\), and \(z = -1.036\) (from the z-table for 15%), we have:

\(-1.036 = \frac{73 - 75}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{2}{1.036} \approx 1.93\)

(ii) The probability that a roll is longer than 77 m is 0.15. We need to find the probability that fewer than 3 out of 8 rolls are longer than 77 m. This is a binomial probability problem with \(n = 8\) and \(p = 0.15\).

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(= (0.85)^8 + \binom{8}{1}(0.15)(0.85)^7 + \binom{8}{2}(0.15)^2(0.85)^6\)

\(= 0.895\)

(i) To find the mean \(\mu\) and standard deviation \(\sigma\), we use the standard normal distribution properties. Given that 96% of lengths are less than 34.1 cm, we find the z-score corresponding to 0.96, which is approximately 1.751. Thus,

\(\frac{34.1 - \mu}{\sigma} = 1.751\)

Similarly, 70% of lengths are more than 26.7 cm, meaning 30% are less than 26.7 cm. The z-score for 0.30 is approximately -0.524. Thus,

\(\frac{26.7 - \mu}{\sigma} = -0.524\)

Solving these two equations simultaneously, we find:

\(\mu = 28.4\), \(\sigma = 3.25\).

(iii) We need to find \(t\) such that \(P(31.8 < X < t) = 0.5\). The distribution of \(X\) is \(N(32.9, 2.4^2)\).

First, find the z-score for 31.8:

\(z = \frac{31.8 - 32.9}{2.4} = -0.4583\)

The cumulative probability for \(z = -0.4583\) is approximately 0.3235. Therefore, \(P(X > 31.8) = 1 - 0.3235 = 0.6765\).

We need \(P(31.8 < X < t) = 0.5\), so \(P(X < t) = 0.5 + 0.6765 = 0.8235\).

Find the z-score for 0.8235, which is approximately 0.929. Thus,

\(\frac{t - 32.9}{2.4} = 0.929\)

Solving for \(t\), we get:

\(t = 35.1\).

(i) To find the probability that the symphony takes longer than 42 minutes, we standardize the normal variable:

\(P(>42) = P\left( z > \frac{42 - 41.1}{3.4} \right) = P(z > 0.2647)\)

Using the standard normal distribution table, \(P(z > 0.2647) = 1 - 0.6045 = 0.3955\).

The probability that the symphony takes longer than 42 minutes on exactly 1 of 3 occasions is given by the binomial distribution:

\(\text{Prob} = (0.3955)(0.6045)^2 \times \binom{3}{1} = 0.433 \text{ or } 0.434\)

(ii) For Beethoven’s Fifth Symphony, we use the given probabilities to find the mean \(\mu\) and standard deviation \(\sigma\):

\(-1.282 = \frac{26.5 - \mu}{\sigma}\)

\(1.645 = \frac{34.6 - \mu}{\sigma}\)

Solving these equations simultaneously gives \(\mu = 30.0\) and \(\sigma = 2.77\).

(iii) To find the probability that both symphonies are less than 34.6 minutes:

For the Sixth Symphony:

\(P(B6 < 34.6) = P\left( z < \frac{34.6 - 41.1}{3.4} \right) = P(z < -1.912)\)

\(P(z < -1.912) = 1 - 0.9720 = 0.0280\)

For the Fifth Symphony, \(P(B5 < 34.6) = 0.95\).

The probability that both are less than 34.6 minutes is:

\(P(\text{both} < 34.6) = 0.028 \times 0.95 = 0.0266\)

The weights of the apples are normally distributed with mean \(\mu = 182\) and standard deviation \(\sigma = 20\). We need to find \(w\) such that \(P(X > w) = 0.72\).

First, convert the problem to a standard normal distribution problem:

\(P\left(Z > \frac{w - 182}{20}\right) = 0.72\)

This implies:

\(P\left(Z < \frac{w - 182}{20}\right) = 0.28\)

Using the standard normal distribution table, find the z-value for which \(P(Z < z) = 0.28\). This gives \(z = -0.583\).

Now, equate and solve for \(w\):

\(\frac{w - 182}{20} = -0.583\)

\(w - 182 = -0.583 \times 20\)

\(w = 182 - 11.66\)

\(w = 170.34\)

Therefore, \(170 \leq w < 170.35\).

(i) We use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).

For lengths less than 6 cm: \(Z = \frac{6 - \mu}{\sigma} = -1.253\)

For lengths more than 12 cm: \(Z = \frac{12 - \mu}{\sigma} = 0.648\)

Solving these equations simultaneously gives:

\(\mu = 9.9\)

\(\sigma = 3.15 \text{ or } 3.16\)

(ii) We need \(P(Z < -1 \text{ or } Z > 1)\).

This is equivalent to \(1 - \Phi(1) + \Phi(-1)\).

\(= 2 - 2 \times 0.8413\)

\(= 0.3174\)

In a sample of 1000, the expected number is \(0.3174 \times 1000 = 317\).

(i) Given that the mean μ is three times the standard deviation σ, we have μ = 3σ. The probability P(X < 25) = 0.648 corresponds to a standard normal variable z = 0.38. Using the standardization formula:

\(z = \frac{25 - \mu}{\sigma} = 0.38\)

Substitute \(\mu = 3\sigma\) into the equation:

\(\frac{25 - 3\sigma}{\sigma} = 0.38\)

Solving for \(\sigma\):

\(25 - 3\sigma = 0.38\sigma\)

\(25 = 3.38\sigma\)

\(\sigma = \frac{25}{3.38} \approx 7.40\)

Then, \(\mu = 3\sigma = 3 \times 7.40 = 22.2\).

(ii) The probability that a single value of X is greater than 25 is 1 - 0.648 = 0.352. We need the probability that exactly 4 out of 6 values are greater than 25. This follows a binomial distribution:

\(P(4) = \binom{6}{4} \times (0.352)^4 \times (0.648)^2\)

\(P(4) = 15 \times (0.0152) \times (0.419904)\)

\(P(4) \approx 0.0967\)

(ii) To find the proportion of winter days with a minimum temperature below zero, we standardize the variable:

\(z = \frac{0 - \mu}{2\mu} = -0.5\)

The probability \(P(z < -0.5) = 1 - 0.6915 = 0.309\) or 30.9%.

(iii) To find how many of the 70 days have a temperature more than three times the mean:

\(z = \frac{3\mu - \mu}{2\mu} = 1\)

The probability \(P(z > 1) = 1 - 0.8413 = 0.1587\).

Thus, the expected number of days is \(70 \times 0.1587 = 11.1\), so approximately 11 or 12 days.

(iv) Given the probability of the temperature being above 6 °C is 0.0735, we find \(\mu\):

\(z = 1.45\)

\(1.45 = \frac{6 - \mu}{2\mu}\)

Solving for \(\mu\), we get \(\mu = 1.54\).

(i) Given that 94% of the letters are within 12 g of the mean, we use the z-score for 47% (since 94% is the total for both tails, 47% for one tail) which is approximately 1.882. The equation is:

\(1.882 = \frac{32 - 20}{\sigma}\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{12}{1.882} \approx 6.38\)

(ii) To find the probability that a letter weighs more than 13 g, we standardize:

\(P(x > 13) = P\left( z > \frac{13 - 20}{6.38} \right)\)

\(= P(z > -1.0978)\)

Using the standard normal distribution table, \(P(z > -1.0978) = 0.864\).

(iii) The probability that a letter weighs more than 32 g is 0.03 (since 6% are more than 12 g above the mean). We use the binomial distribution for at least 2 out of 7:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - (0.97)^7 - (0.03)(0.97)^6 \times \binom{7}{1}\)

\(= 1 - (0.97)^7 - 7 \times (0.03)(0.97)^6\)

\(= 0.0171\)

(i) Given that 10% of the straws are shorter than 20 cm, we have \(P(x < 20) = 0.1\). Using the standard normal distribution, this corresponds to a \(z\)-value of approximately \(-1.282\).

Standardizing, \(z = \frac{20 - \mu}{0.8}\).

Thus, \(-1.282 = \frac{20 - \mu}{0.8}\).

Solving for \(\mu\), we get \(\mu = 20 + 1.282 \times 0.8 = 21.0256 \approx 21.0 \text{ cm}\).

(ii) We need to find \(P(21.5 < x < 22.5)\).

Standardizing, \(P\left( \frac{21.5 - 21.03}{0.8} < z < \frac{22.5 - 21.03}{0.8} \right)\).

This simplifies to \(P(0.5875 < z < 1.8375)\).

Using the standard normal distribution table, \(\Phi(1.8375) - \Phi(0.5875) = 0.9670 - 0.7217 = 0.2453\).

Now, we use the binomial distribution for 4 trials with probability \(p = 0.2453\).

We need \(P(X < 2) = P(0) + P(1)\).

\(P(0) = (0.7547)^4\).

\(P(1) = 4 \times (0.2453) \times (0.7547)^3\).

Calculating these gives \(P(0) + P(1) = 0.7547^4 + 4 \times 0.2453 \times 0.7547^3 = 0.746\).

(a) We start with the given equation 3μ = 7σ² and the probability P(X > 2μ) = 0.1016. Standardizing, we have:

\(z = \frac{2μ - μ}{σ} = \frac{μ}{σ}\)

\(From the standard normal distribution table, P(Z > 1.272) = 0.1016, so:\)

\(\frac{7σ}{3} = 1.272\)

Solving for σ, we get:

\(σ = 0.545\)

Substituting back to find μ:

\(μ = \frac{7σ^2}{3} = 0.693\)

(b) Given Y ~ N(33, 21), we need to find a such that P(33 − a < Y < 33 + a) = 0.5. This implies:

\(P(X < a + 33) = 0.75\)

\(From the standard normal distribution table, z = 0.674 for 0.75 probability. Standardizing:\)

\(\frac{a + 33 - 33}{\sqrt{21}} = 0.674\)

Solving for a gives:

\(a = 3.09\)

(i) Given that 79% of people have visits lasting less than 10 minutes, we find the z-score corresponding to 0.79, which is 0.807. Using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 10\), \(\mu = 8.2\), and \(z = 0.807\). Solving for \(\sigma\):

\(0.807 = \frac{10 - 8.2}{\sigma}\)

\(\sigma = \frac{1.8}{0.807} \approx 2.23\)

(ii) We need to find \(P(|X - 8.2| > 1)\), which is equivalent to \(P(X > 9.2) + P(X < 7.2)\). Standardizing these values:

\(P\left(\left| z \right| > \frac{1}{2.23}\right)\)

\(P(|z| > 0.4484) = 2 \times (1 - \Phi(0.4484))\)

\(= 2 \times (1 - 0.6729) = 0.654\)

(iii) Let \(p = 0.21\) be the probability of a visit lasting longer than 10 minutes. We want \(P(X > 2)\) for a binomial distribution with \(n = 6\) and \(p = 0.21\):

\(P(X > 2) = 1 - P(X \leq 2)\)

\(= 1 - \{ (0.79)^6 + 6 \times (0.21) \times (0.79)^5 + 15 \times (0.21)^2 \times (0.79)^4 \}\)

\(= 1 - \{ 0.2621 + 0.2371 + 0.3888 \}\)

\(= 0.112\)

(i) To find the probability that each car runs out of fuel before travelling 367 km, we standardize the normal distribution for each car.

For the Zotoc car:

\(z = \frac{367 - 320}{21.6} = 2.176\)

The probability \(P(Z < 2.176)\) is approximately 0.985.

For the Ganmor car:

\(z = \frac{367 - 350}{7.5} = 2.267\)

The probability \(P(Z < 2.267)\) is approximately 0.988.

(ii) We are given that the probability that a Zotoc car can travel at least \(320 + d\) km is 0.409. This corresponds to a z-score of 0.23 (since \(P(Z > 0.23) = 0.409\)).

Standardizing, we have:

\(0.23 = \frac{x - 320}{21.6}\)

Solving for \(x\), we get:

\(x = 0.23 \times 21.6 + 320 = 324.968\)

Thus, \(d = 324.968 - 320 = 4.97\).

(i) We have two conditions:

1. 8 out of 10 children can jump more than 127 cm, so \(P(X > 127) = 0.8\). This corresponds to a z-score of \(-0.842\).

2. 1 out of 3 children can jump more than 135 cm, so \(P(X > 135) = 0.333\). This corresponds to a z-score of \(0.431\).

Using the z-score formula \(z = \frac{x - \mu}{\sigma}\), we set up the equations:

\(0.431 = \frac{135 - \mu}{\sigma}\)

\(-0.842 = \frac{127 - \mu}{\sigma}\)

Solving these equations simultaneously gives \(\mu = 132\) and \(\sigma = 6.29\).

(ii) To find the probability that a child will not be able to jump 145 cm, we calculate \(P(X < 145)\).

Standardize: \(z = \frac{145 - 132}{6.29} = 2.023\).

Using normal distribution tables, \(P(z < 2.023) = 0.978\).

(iii) Let \(p = \frac{1}{3}\) be the probability that a child can jump more than 135 cm.

We need \(P(\text{at least 2 out of 8}) = 1 - P(0, 1)\).

\(P(0, 1) = \left(\frac{2}{3}\right)^8 + \binom{8}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^7\).

Calculate: \(P(0, 1) = 0.195\).

Thus, \(P(\text{at least 2}) = 1 - 0.195 = 0.805\).

(i) To find P(X < 2μ), we standardize the variable:

\(P(X < 2μ) = P\left( z < \frac{2μ - μ}{σ} \right) = P\left( z < \frac{μ}{σ} \right).\)

Given 5σ = 3μ, we have \(\frac{μ}{σ} = \frac{5}{3}\).

\(Thus, P(X < 2μ) = P\left( z < \frac{5}{3} \right) = 0.952.\)

(ii) We are given P(X < \(\frac{1}{3}μ\)) = 0.8524. Standardizing gives:

\(P\left( z < \frac{\frac{1}{3}μ - μ}{σ} \right) = P\left( z < \frac{-2μ}{3σ} \right).\)

From the standard normal distribution table, \(\frac{-2μ}{3σ} = -1.047\).

Solving for μ, we get \(\frac{-2μ}{3σ} = -1.047\) which gives \(μ = -1.57σ\).

First, calculate the probability that a bag weighs more than 2.6 kg:

\(P(X > 2.6) = \frac{134}{5000} = 0.0268\)

The probability that a bag weighs less than 2.6 kg is:

\(P(X < 2.6) = 1 - 0.0268 = 0.9732\)

Using the standard normal distribution, find the z-score corresponding to this probability:

\(\frac{2.6 - 2.55}{\sigma} = 1.93\)

Solving for \(\sigma\):

\(\sigma = \frac{2.6 - 2.55}{1.93} = 0.0259\)

Given that \(X \sim N(\mu, \sigma^2)\), we have two probabilities:

1. \(P(X > 30.0) = 0.1480\)

2. \(P(X > 20.9) = 0.6228\)

Using the standard normal distribution, we convert these to \(z\)-scores:

For \(X = 30.0\):

\(z = \frac{30 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > 1.045) = 0.1480\), so \(z = 1.045\).

Thus, \(\frac{30 - \mu}{\sigma} = 1.045\).

For \(X = 20.9\):

\(z = \frac{20.9 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > -0.313) = 0.6228\), so \(z = -0.313\).

Thus, \(\frac{20.9 - \mu}{\sigma} = -0.313\).

We now have two equations:

1. \(30 - \mu = 1.045 \sigma\)

2. \(20.9 - \mu = -0.313 \sigma\)

Solving these simultaneously:

Subtract the second equation from the first:

\((30 - \mu) - (20.9 - \mu) = 1.045 \sigma + 0.313 \sigma\)

\(9.1 = 1.358 \sigma\)

\(\sigma = \frac{9.1}{1.358} = 6.70\)

Substitute \(\sigma = 6.70\) into the first equation:

\(30 - \mu = 1.045 \times 6.70\)

\(30 - \mu = 7.0015\)

\(\mu = 30 - 7.0015 = 23.0\)

Thus, \(\mu = 23.0\) and \(\sigma = 6.70\).

(i) To find the probability that a randomly chosen bar of soap weighs more than 128 grams, we standardize the variable:

\(P(X > 128) = P\left( z > \frac{128 - 125}{4.2} \right)\)

\(= P(z > 0.7143)\)

Using the standard normal distribution table, \(P(z > 0.7143) = 1 - 0.7623 = 0.238\).

(ii) We need to find the value of \(k\) such that \(P(k < X < 128) = 0.7465\).

\(P(X < 128) = 0.7465 + 0.2377 = 0.9842\)

Using the standard normal distribution table, \(z = -2.15\) corresponds to \(P(X < k)\).

\(-2.15 = \frac{k - 125}{4.2}\)

Solving for \(k\), we get \(k = 116\).

(iii) For five bars of soap, we find the probability that more than two weigh more than 128 grams:

\(P(X > 2) = P(3, 4, 5) = 1 - P(0, 1, 2)\)

Using the binomial distribution:

\(= \binom{5}{3} (0.2377)^3 (0.7623)^2 + \binom{5}{4} (0.2377)^4 (0.7623)^1 + \binom{5}{5} (0.2377)^5\)

\(= 0.07804 + 0.01216 + 0.0007588\)

\(= 0.0910\)

(i) The mean \(\mu\) is estimated as the median of the data, which is the middle value of the box-and-whisker plot. From the diagram, the median is 51 km h\(^{-1}\).

(ii) To estimate the standard deviation \(\sigma\), we use the interquartile range (IQR) and the properties of the normal distribution. The IQR is the difference between the upper quartile (63 km h\(^{-1}\)) and the median (51 km h\(^{-1}\)), which is 12 km h\(^{-1}\).

For a normal distribution, the IQR corresponds to approximately 1.34896 standard deviations (since \(z = \pm 0.674\) for the quartiles). Therefore, \(\sigma\) can be estimated by:

\(\sigma = \frac{63 - 51}{0.674} = \frac{12}{0.674} \approx 17.8\)

(i) To find the probability that a journey takes between 85 and 100 minutes, we standardize the values using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For \(x = 85\), \(z = \frac{85 - 100}{7} = -2.143\).

The probability \(P(85 < x < 100) = 0.5 - P(z < -2.143)\).

Using the standard normal distribution table, \(P(z < -2.143) = 1 - \Phi(2.143) = 0.0161\).

Thus, \(P(85 < x < 100) = 0.5 - 0.0161 = 0.484\).

(ii) For 'standard' journeys, which are the middle 34%, we find the z-scores corresponding to the cumulative probabilities of 0.33 and 0.67.

Using \(z = \Phi^{-1}(0.67) = 0.44\).

For the upper limit, \(0.44 = \frac{a - 100}{7}\), solving gives \(a = 103.1\) minutes.

For the lower limit, \(-0.44 = \frac{a - 100}{7}\), solving gives \(a = 96.9\) minutes.

(i) We know that 225 out of 900 cartons contain more than 1002 millilitres. Therefore, the probability \(P(X > 1002) = \frac{225}{900} = 0.25\).

Using the standard normal distribution, \(P(Z > z) = 0.25\) corresponds to \(z = 0.674\) (using standard normal tables or calculator).

Standardizing, we have:

\(\frac{1002 - \mu}{8} = 0.674\)

Solving for \(\mu\):

\(1002 - \mu = 0.674 \times 8\)

\(1002 - \mu = 5.392\)

\(\mu = 1002 - 5.392 = 996.608\)

Rounding to the nearest whole number, \(\mu = 997\).

(ii) The probability that a carton contains more than 1002 millilitres is \(p = 0.25\). We want the probability that exactly 2 out of 3 chosen cartons contain more than 1002 millilitres.

This is a binomial probability problem with \(n = 3\), \(k = 2\), and \(p = 0.25\).

\(P(2) = \binom{3}{2} \times (0.25)^2 \times (0.75)^1\)

\(P(2) = 3 \times 0.0625 \times 0.75\)

\(P(2) = 0.140625\)

Rounding to three significant figures, the probability is 0.140.

The problem states that the daily minimum temperature follows a normal distribution \(N(\mu, 40.0)\). We are given that the probability of the temperature being above 0°C is 0.8888.

Using the standard normal distribution table, a probability of 0.8888 corresponds to a \(z\)-score of approximately 1.22. However, since we are dealing with the probability above 0°C, we use \(z = -1.22\).

The \(z\)-score formula is:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 0\) (the temperature), \(\mu\) is the mean, and \(\sigma = \sqrt{40}\).

Substitute the values into the formula:

\(-1.22 = \frac{0 - \mu}{\sqrt{40}}\)

Solving for \(\mu\):

\(-1.22 \times \sqrt{40} = -\mu\)

\(\mu = 1.22 \times \sqrt{40}\)

\(\mu \approx 7.72\)

(i) We know that 25% of infections clear up in less than 7 days, which corresponds to a z-score of approximately -0.674. Using the standardization formula:

\(z = \frac{X - \mu}{\sigma}\)

Substitute the known values:

\(-0.674 = \frac{7 - \mu}{2.6}\)

Solving for \(\mu\):

\(-0.674 \times 2.6 = 7 - \mu\)

\(-1.7524 = 7 - \mu\)

\(\mu = 7 + 1.7524 = 8.75\)

(ii) We need to find \(P(X > 6.2)\) where \(X\) is normally distributed with mean 6.5 and standard deviation 2.6. Standardizing:

\(P(X > 6.2) = P\left( Z > \frac{6.2 - 6.5}{2.6} \right)\)

\(= P(Z > -0.1154)\)

Using the symmetry of the normal distribution, \(P(Z > -0.1154) = 1 - P(Z < -0.1154)\). Since \(P(Z < -0.1154) \approx 0.454\),

\(P(Z > -0.1154) = 1 - 0.454 = 0.546\)

(i) To find the standard deviation \(\sigma\), we use the standard normal distribution formula:

\(z = \frac{X - \mu}{\sigma}\)

Given \(P(X > 5.5) = 0.0465\), we find the corresponding \(z\)-value from the standard normal distribution table, which is approximately 1.68. Thus,

\(z = \frac{5.5 - 4.5}{\sigma} = 1.68\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{1}{1.68} = 0.595\)

(ii) To find the probability that \(X\) lies between 3.8 and 4.8, we calculate the \(z\)-values for 3.8 and 4.8:

\(z_1 = \frac{3.8 - 4.5}{0.595} = -1.176\)

\(z_2 = \frac{4.8 - 4.5}{0.595} = 0.504\)

The probability is given by:

\(P(3.8 < X < 4.8) = \Phi(0.504) - (1 - \Phi(1.176))\)

Using the standard normal distribution table, \(\Phi(0.504) \approx 0.6929\) and \(\Phi(1.176) \approx 0.8802\), so:

\(P(3.8 < X < 4.8) = 0.6929 - (1 - 0.8802) = 0.573\)

(a) Let the mean be \(\mu = 2s\). The standard normal variable \(Z\) is given by:

\(Z = \frac{X - \mu}{s} = \frac{5.2 - 2s}{s}\)

Given \(P(X > 5.2) = 0.9\), we have \(P(Z > z) = 0.9\), which implies \(P(Z < z) = 0.1\). From standard normal tables, \(z = -1.282\).

Thus, \(\frac{5.2 - 2s}{s} = -1.282\).

Solving for \(s\):

\(5.2 - 2s = -1.282s\)

\(5.2 = 0.718s\)

\(s = \frac{5.2}{0.718} \approx 7.24 \text{ or } 7.23\)

(b) The probability that a value lies between \(\mu - \sigma\) and \(\mu + \sigma\) is \(P(|Z| < 1)\).

From standard normal tables, \(P(|Z| < 1) = 0.6826\).

Thus, the expected number of observations is:

\(0.6826 \times 800 = 546.08\)

Therefore, approximately 546 observations (accept 547).

(i) Any sensible set of data, such as heights, weights, or times, can be modeled by a normal distribution.

(ii) We know \(P(X > 10.0) = 0.7389\). This corresponds to a standard normal variable \(z\) such that \(P(Z > z) = 0.7389\), which gives \(z = 0.64\). Using the standardization formula:

\(z = \frac{\mu - 10}{\sqrt{21}}\)

Substitute \(z = 0.64\):

\(0.64 = \frac{\mu - 10}{\sqrt{21}}\)

Solving for \(\mu\):

\(\mu - 10 = 0.64 \times \sqrt{21}\)

\(\mu = 10 + 0.64 \times \sqrt{21}\)

\(\mu = 12.9\)

(iii) To find how many observations are greater than 22.0, we first find the \(z\)-score for 22:

\(z = \frac{22 - 12.9}{\sqrt{21}} = 1.986\)

Using the standard normal distribution, \(P(X > 22) = 1 - \Phi(1.986) \approx 0.0235\).

For 300 observations, the expected number greater than 22 is:

\(300 \times 0.0235 = 7.05\)

Rounding to the nearest integer, we estimate 7 observations.