Past Exam Questions

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 \(s\):

\(-1.151s = 70 - 120\)

\(-1.151s = -50\)

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

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