Past Exam Questions

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, \(h = 152\).