Past Exam Questions

The probability that a person is neither a student, office worker, nor shop assistant is given by:

\(1 - (0.36 + 0.22 + 0.29) = 0.13\)

For a binomial distribution with parameters \(n = 300\) and \(p = 0.13\), the mean and variance are:

\(\text{mean} = 300 \times 0.13 = 39\)

\(\text{variance} = 300 \times 0.13 \times 0.87 = 33.93\)

We approximate the binomial distribution with a normal distribution \(N(39, 33.93)\).

Using continuity correction, we find:

\(P(31 \, \text{to} \, 49) = P(30.5 < x < 49.5)\)

Standardizing, we have:

\(P\left( \frac{30.5 - 39}{\sqrt{33.93}} < z < \frac{49.5 - 39}{\sqrt{33.93}} \right)\)

\(= P(-1.4592 < z < 1.8026)\)

Using the standard normal distribution table:

\(= \Phi(1.8026) + \Phi(1.4592) - 1\)

\(= 0.9643 + 0.9278 - 1\)

\(= 0.892\)

Let the random variable \(X\) represent the number of students with blue eyes. Given that one in five students has blue eyes, the probability \(p\) is \(0.2\).

The expected number of students with blue eyes is \(\mu = 120 \times 0.2 = 24\).

The variance is \(\sigma^2 = 120 \times 0.2 \times 0.8 = 19.2\).

We approximate \(X\) by a normal distribution \(N(24, 19.2)\).

To find \(P(X < 33)\), we use a continuity correction: \(P(X < 33) \approx P(Z < \frac{32.5 - 24}{\sqrt{19.2}})\).

Calculate the standard score: \(Z = \frac{32.5 - 24}{\sqrt{19.2}} = 1.9398\).

Using standard normal distribution tables, \(P(Z < 1.9398) = 0.974\).

Let the random variable \(X\) represent the number of people whose birthday falls on a Monday. Since each day of the week is equally likely, the probability \(p\) that a person's birthday is on a Monday is \(\frac{1}{7}\).

We have \(n = 350\) people, so the expected number of people with a birthday on Monday is \(np = 350 \times \frac{1}{7} = 50\).

The variance is \(npq = 350 \times \frac{1}{7} \times \frac{6}{7} = 42.857\).

We use a normal approximation to the binomial distribution. Applying continuity correction, we find:

\(P(X \geq 47) = P\left( z > \frac{46.5 - 50}{\sqrt{42.857}} \right)\)

Calculating the z-score:

\(z = \frac{46.5 - 50}{\sqrt{42.857}} = -0.5346\)

Thus, \(P(z > -0.5346) = 0.704\).

Let the probability that a value is less than 5 be denoted by \(p\). Since the probability that a value is greater than 5 is 0.15, we have \(p = 0.85\).

The mean number of values less than 5 in 200 trials is \(\mu = 200 \times 0.85 = 170\).

The variance is \(\text{var} = 200 \times 0.85 \times 0.15 = 25.5\).

We need to find \(P(\text{at least 160})\), which is equivalent to \(P(X \geq 160)\).

Using continuity correction, this becomes \(P(X > 159.5)\).

Standardizing, we have:

\(P\left( z > \frac{159.5 - 170}{\sqrt{25.5}} \right) = P(z > -2.079)\)

Using standard normal distribution tables, \(P(z > -2.079) = 0.981\).

The probability that Ana is on time on any given day is:

\(p = 1 - 0.05 - 0.75 = 0.2\)

We are looking for the probability that she is on time on fewer than 20 out of 96 days. This is a binomial distribution problem with parameters \(n = 96\) and \(p = 0.2\).

We approximate the binomial distribution with a normal distribution:

\(\mu = np = 96 \times 0.2 = 19.2\)

\(\sigma^2 = np(1-p) = 96 \times 0.2 \times 0.8 = 15.36\)

Using continuity correction, we find:

\(P(X < 20) = P\left( Z < \frac{19.5 - 19.2}{\sqrt{15.36}} \right) = P(Z < 0.07654)\)

Using standard normal distribution tables or a calculator, we find:

\(P(Z < 0.07654) = 0.531\)

(b) We know that 72% of Eastern bluebirds have a length greater than 17.1 cm, which implies \(P(Z > \frac{17.1 - 18.4}{\sigma}) = 0.72\). This corresponds to a standard normal variable \(Z\) such that \(Z = -0.583\) (since \(P(Z > -0.583) = 0.72\)).

Thus, \(\frac{17.1 - 18.4}{\sigma} = -0.583\).

Solving for \(\sigma\), we get \(\sigma = \frac{17.1 - 18.4}{-0.583} = 2.23\).

(c) For a sample of 120 bluebirds, the mean number with length greater than 17.1 cm is \(120 \times 0.72 = 86.4\).

The variance is \(120 \times 0.72 \times 0.28 = 24.192\).

Using a normal approximation, \(P(X < 80) = P(Z < \frac{79.5 - 86.4}{\sqrt{24.192}})\) (using continuity correction).

\(Z = \frac{79.5 - 86.4}{\sqrt{24.192}} = -1.4029\).

Thus, \(P(Z < -1.4029) = 1 - \Phi(1.403) = 1 - 0.9196 = 0.0804\).

Let the random variable representing the number of people with blood group A+ be denoted by \(X\). Given that 35% of the population are group A+, we have \(p = 0.35\).

The sample size is \(n = 150\). The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are given by:

\(\mu = n \times p = 150 \times 0.35 = 52.5\)

\(\sigma^2 = n \times p \times (1-p) = 150 \times 0.35 \times 0.65 = 34.125\)

We use the normal approximation to the binomial distribution. Applying continuity correction, we find:

\(P(X > 60) = P\left( Z > \frac{60.5 - 52.5}{\sqrt{34.125}} \right)\)

Calculating the standard score:

\(Z = \frac{60.5 - 52.5}{\sqrt{34.125}} = 1.369\)

Using standard normal distribution tables, we find:

\(P(Z > 1.369) = 1 - \Phi(1.369) = 0.0854 \text{ or } 0.0855\)

Let the probability of landing on the blue side be \(p\). Then, the probability of landing on the red side is \(4p\), and the probability of landing on the green side is \(3p\).

Since the total probability must be 1, we have:

\(p + 4p + 3p = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Thus, the probability of landing on the blue side is \(\frac{1}{8}\).

For a binomial distribution with \(n = 136\) and \(p = \frac{1}{8}\), the mean \(\mu\) is:

\(\mu = np = 136 \times \frac{1}{8} = 17\)

The variance \(\sigma^2\) is:

\(\sigma^2 = np(1-p) = 136 \times \frac{1}{8} \times \frac{7}{8} = 14.875\)

Using a normal approximation, we standardize for \(P(X < 20)\) with continuity correction:

\(P(X < 20) = P\left( Z < \frac{19.5 - 17}{\sqrt{14.875}} \right)\)

\(= P(Z < 0.648)\)

Using standard normal distribution tables, \(\Phi(0.648) = 0.742\).

Therefore, the probability that the spinner lands on the blue side fewer than 20 times is 0.742.

Let the random variable \(X\) represent the number of seeds that germinate. \(X\) follows a binomial distribution with parameters \(n = 250\) and \(p = 0.86\).

We approximate the binomial distribution with a normal distribution since \(n\) is large and \(p\) is not too close to 0 or 1.

The mean \(\mu\) of the distribution is given by:

\(\mu = np = 250 \times 0.86 = 215\)

The variance \(\sigma^2\) is given by:

\(\sigma^2 = np(1-p) = 250 \times 0.86 \times 0.14 = 30.1\)

Thus, the standard deviation \(\sigma\) is:

\(\sigma = \sqrt{30.1}\)

We use a continuity correction for the normal approximation. We want \(P(X > 210)\), which is approximated by \(P(X > 210.5)\).

Standardizing, we find:

\(P(X > 210.5) = 1 - \Phi\left( \frac{210.5 - 215}{\sqrt{30.1}} \right)\)

\(= 1 - \Phi(-0.820)\)

\(= \Phi(0.820)\)

Using standard normal distribution tables, \(\Phi(0.820) = 0.794\).

(i) To find \(\mu\), we use the standard normal distribution. Given \(\text{P}(X > 20) = 0.04\), we find the corresponding \(z\)-value: \(z = \pm 1.751\).

Using the standardization formula: \(\frac{20 - \mu}{\mu/4} = 1.751\).

Solving for \(\mu\), we get \(\mu = 13.9\).

(ii) To find \(\text{P}(10 < X < 20)\), we first find \(\text{P}(X < 10)\).

Standardizing: \(\text{P}(z < \frac{10 - 13.91}{13.91/4}) = \text{P}(z < -1.124)\).

\(\text{P}(z < -1.124) = 1 - 0.8694 = 0.131\).

Thus, \(\text{P}(10 < X < 20) = 0.96 - 0.131 = 0.829 \text{ or } 0.830\).

(iii) For 250 observations, \(\mu = 250 \times 0.96 = 240\) and \(\sigma^2 = 250 \times 0.96 \times 0.04 = 9.6\).

We need \(\text{P}(\geq 235) = 1 - \Phi\left(\frac{234.5 - 240}{\sqrt{9.6}}\right)\).

\(\Phi(1.775) = 0.962\).

(i) Let \(X\) be the number of cloudy days in November. \(X\) follows a binomial distribution with parameters \(n = 30\) and \(p = 0.8\).

We approximate \(X\) using a normal distribution with mean \(\mu = np = 30 \times 0.8 = 24\) and variance \(\sigma^2 = npq = 30 \times 0.8 \times 0.2 = 4.8\).

Standard deviation \(\sigma = \sqrt{4.8}\).

To find the probability of fewer than 25 cloudy days, we use a continuity correction: \(P(X < 25) \approx P(Y < 24.5)\) where \(Y\) is normally distributed.

Calculate the z-score: \(z = \frac{24.5 - 24}{\sqrt{4.8}} = 0.228\).

Using standard normal distribution tables, \(P(Z < 0.228) = 0.590\).

(ii) The normal approximation is justified because both \(np = 24\) and \(nq = 6\) are greater than 5.

Let the number of people with visits lasting less than 8.2 minutes be a binomial random variable with parameters:

\(n = 35\) (number of trials) and \(p = 0.5\) (probability of success for each trial).

The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = n \times p = 35 \times 0.5 = 17.5\)

\(\sigma^2 = n \times p \times (1-p) = 35 \times 0.5 \times 0.5 = 8.75\)

We approximate the binomial distribution with a normal distribution \(N(17.5, 8.75)\).

Using continuity correction, we find:

\(P(X < 16) = \Phi\left( \frac{15.5 - 17.5}{\sqrt{8.75}} \right)\)

\(= 1 - \Phi(0.676)\)

\(= 1 - 0.7505\)

\(= 0.2495 \approx 0.250\)

Alternatively, using the binomial distribution directly:

\(P(X < 16) = \sum_{x=0}^{15} \binom{35}{x} (0.5)^x (0.5)^{35-x}\)

\(= 0.250\)

The number of days Julie's train is late follows a binomial distribution with parameters:

- Number of trials, \(n = 90\)

- Probability of success (train being late), \(p = 0.3\)

The mean \(\mu\) and variance \(\sigma^2\) of the distribution are given by:

\(\mu = np = 90 \times 0.3 = 27\)

\(\sigma^2 = np(1-p) = 90 \times 0.3 \times 0.7 = 18.9\)

We approximate the binomial distribution with a normal distribution \(N(27, 18.9)\).

To find \(P(X > 35)\), apply continuity correction:

\(P(X > 35) = 1 - \Phi \left( \frac{35.5 - 27}{\sqrt{18.9}} \right)\)

\(= 1 - \Phi(1.955) = 0.0253\)

To find \(P(X < 27)\), apply continuity correction:

\(P(X < 27) = \Phi \left( \frac{26.5 - 27}{\sqrt{18.9}} \right)\)

\(= \Phi(-0.115) = 0.4542\)

The total probability is:

\(P(X > 35) + P(X < 27) = 0.0253 + 0.4542 = 0.480\)

The probability of an apple being underweight is given by \(p = \frac{2}{15}\).

For a sample of 200 apples, the mean \(\mu\) and variance \(\sigma^2\) are calculated as follows:

\(\mu = 200 \times \frac{2}{15} = 26.67\)

\(\sigma^2 = 200 \times \frac{2}{15} \times \frac{13}{15} = 23.11\)

We need to find \(P(21 < X < 35)\).

Using continuity correction, this becomes \(P(21.5 < X < 34.5)\).

Standardizing, we have:

\(P\left( \frac{21.5 - 26.67}{\sqrt{23.11}} < z < \frac{34.5 - 26.67}{\sqrt{23.11}} \right)\)

\(P(-1.075 < z < 1.629)\)

Using the standard normal distribution table:

\(P(z < 1.629) = 0.9483\)

\(P(z < -1.075) = 0.1411\)

Thus, \(P(-1.075 < z < 1.629) = 0.9483 - 0.1411 = 0.807\).

(ii) For a normal approximation to be appropriate, both the expected number of successes and failures should be at least 5. Here, the expected number of successes is given by:

\(12 \times 0.25 = 3\)

Since 3 is less than 5, it is not appropriate to use a normal approximation.

(iii) Let the random variable \(X\) represent the number of days Martin is playing games when his friend phones in a 40-day period. \(X\) follows a binomial distribution with parameters \(n = 40\) and \(p = 0.25\).

The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are:

\(\mu = 40 \times 0.25 = 10\)

\(\sigma^2 = 40 \times 0.25 \times 0.75 = 7.5\)

Using a normal approximation, we find \(P(X \geq 13)\) with continuity correction:

\(P(X \geq 13) = P\left( Z > \frac{12.5 - 10}{\sqrt{7.5}} \right)\)

\(= P(Z > 0.913)\)

Using the standard normal distribution table:

\(= 1 - \Phi(0.913)\)

\(= 1 - 0.8194\)

\(= 0.181\)

Let the random variable \(X\) represent the number of pears. Each box has a probability of \(\frac{4}{11}\) of containing a pear.

The expected number of pears, \(\mu\), is given by:

\(\mu = 121 \times \frac{4}{11} = 44\)

The variance, \(\sigma^2\), is given by:

\(\sigma^2 = 121 \times \frac{4}{11} \times \frac{7}{11} = 28\)

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{28}\)

We apply continuity correction for \(X < 39\), using \(X < 38.5\).

Standardizing, we find:

\(Z = \frac{38.5 - 44}{\sqrt{28}} = -1.039\)

Using the standard normal distribution table, we find:

\(P(Z < -1.039) = \Phi(-1.039) = 1 - 0.8506 = 0.149\)

(b) To complete the probability distribution table, we need to find \(P(X = 1)\) and \(P(X = 2)\).

Using the binomial distribution formula, \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n = 4\) and \(p = \frac{1}{4}\).

\(P(X = 1) = \binom{4}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^3 = \frac{27}{64}\).

\(P(X = 2) = \binom{4}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^2 = \frac{27}{128}\).

The completed table is:

(d) Let \(Y\) be the number of times Eli obtains at least two 2s in 96 throws. \(Y\) follows a binomial distribution with \(n = 96\) and \(p = P(X \geq 2)\).

\(P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) = 1 - \left(\frac{81}{256} + \frac{27}{64}\right) = \frac{67}{256}\).

Using a normal approximation for \(Y\), \(Y \sim N(\mu, \sigma^2)\) where \(\mu = np = 96 \times \frac{67}{256} = 25.125\) and \(\sigma^2 = np(1-p) = 96 \times \frac{67}{256} \times \frac{189}{256} = 18.549\).

We need \(P(Y < 20)\). Using continuity correction, \(P(Y < 20) \approx P(Z < \frac{19.5 - 25.125}{\sqrt{18.549}})\).

\(Z \approx \frac{19.5 - 25.125}{\sqrt{18.549}} = -1.306\).

\(P(Z < -1.306) = 1 - P(Z < 1.306) = 1 - 0.9042 = 0.0958\).

Thus, \(0.0957 \leq p \leq 0.0958\).

The percentage of cars is calculated as:

\(100 ext{%} - 20 ext{%} - 16 ext{%} = 64 ext{%}\)

Let \(X\) be the number of cars in the sample of 125 vehicles. \(X\) follows a binomial distribution \(B(n, p)\) where \(n = 125\) and \(p = 0.64\).

We approximate this binomial distribution with a normal distribution \(N(\mu, \sigma^2)\) where:

\(\mu = np = 125 \times 0.64 = 80\)

\(\sigma^2 = np(1-p) = 125 \times 0.64 \times 0.36 = 28.8\)

We need to find \(P(X > 73)\). Using continuity correction, this becomes \(P(X > 73.5)\).

Standardizing, we have:

\(P(X > 73.5) = 1 - \Phi\left( \frac{73.5 - 80}{\sqrt{28.8}} \right)\)

Calculating the standard score:

\(\frac{73.5 - 80}{\sqrt{28.8}} = -1.211\)

Thus:

\(P(X > 73.5) = 1 - \Phi(-1.211) = \Phi(1.211)\)

Using standard normal distribution tables, \(\Phi(1.211) \approx 0.887\).

The number of faulty toys follows a binomial distribution with parameters \(n = 200\) and \(p = 0.08\).

First, calculate the mean and variance:

\(\text{mean} = np = 200 \times 0.08 = 16\)

\(\text{variance} = np(1-p) = 200 \times 0.08 \times 0.92 = 14.72\)

Since \(n\) is large and \(p\) is not too close to 0 or 1, we can use a normal approximation.

Apply continuity correction for \(P(X \geq 15)\):

\(P(X \geq 15) = P(X > 14.5)\)

Standardize using the normal distribution:

\(P(X > 14.5) = 1 - \Phi \left( \frac{14.5 - 16}{\sqrt{14.72}} \right)\)

Calculate the standardized value:

\(\frac{14.5 - 16}{\sqrt{14.72}} = -0.391\)

Using the standard normal distribution table, find:

\(\Phi(-0.391) = 0.391\)

Thus,

\(P(X \geq 15) = 1 - 0.391 = 0.652\)

(i) The probability of throwing a 1 is equal to the probability of throwing a 2, 3, 4, or 6. Since the probability of throwing a 5 is 0.75, the probability of throwing any other number is:

\(1 - 0.75 = 0.25\)

There are 5 numbers (1, 2, 3, 4, 6) with equal probability, so the probability of throwing a 1 is:

\(\frac{0.25}{5} = 0.05\)

The probability of throwing a 5 is 0.75, and the probability of throwing an even number (2, 4, 6) is:

\(\frac{3}{5} \times 0.25 = 0.15\)

The probability of the sequence 1, 5, even is:

\(0.05 \times 0.75 \times 0.15 = 0.00563\)

(iii) Let \(X\) be the number of times a 5 is thrown in 90 trials. \(X\) follows a binomial distribution \(B(90, 0.75)\).

Using a normal approximation, \(X \sim N(\mu, \sigma^2)\) where:

\(\mu = 90 \times 0.75 = 67.5\)

\(\sigma^2 = 90 \times 0.75 \times 0.25 = 16.875\)

We use a continuity correction for \(P(X > 60)\):

\(P(X > 60) = 1 - \Phi \left( \frac{60.5 - 67.5}{\sqrt{16.875}} \right) = \Phi(1.704)\)

\(\Phi(1.704) \approx 0.956\)