Past Exam Questions

(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 \(k\):

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