Past Exam Questions

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