Past Exam Questions

(i) Any sensible set of data, such as heights, weights, or times, can be modeled by a normal distribution.

(ii) We know \(P(X > 10.0) = 0.7389\). This corresponds to a standard normal variable \(z\) such that \(P(Z > z) = 0.7389\), which gives \(z = 0.64\). Using the standardization formula:

\(z = \frac{\mu - 10}{\sqrt{21}}\)

Substitute \(z = 0.64\):

\(0.64 = \frac{\mu - 10}{\sqrt{21}}\)

Solving for \(\mu\):

\(\mu - 10 = 0.64 \times \sqrt{21}\)

\(\mu = 10 + 0.64 \times \sqrt{21}\)

\(\mu = 12.9\)

(iii) To find how many observations are greater than 22.0, we first find the \(z\)-score for 22:

\(z = \frac{22 - 12.9}{\sqrt{21}} = 1.986\)

Using the standard normal distribution, \(P(X > 22) = 1 - \Phi(1.986) \approx 0.0235\).

For 300 observations, the expected number greater than 22 is:

\(300 \times 0.0235 = 7.05\)

Rounding to the nearest integer, we estimate 7 observations.

Let the random variable \(X\) represent the number of people who pass the test. \(X\) follows a binomial distribution with parameters \(n = 75\) and \(p = 0.3\).

We approximate the binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 75 \times 0.3 = 22.5\)

\(\sigma^2 = np(1-p) = 75 \times 0.3 \times 0.7 = 15.75\)

The standard deviation \(\sigma\) is:

\(\sigma = \sqrt{15.75} \approx 3.9686\)

We use a continuity correction for the normal approximation:

\(P(X > 20) \approx P\left(Z > \frac{20.5 - 22.5}{\sqrt{15.75}}\right)\)

\(= P\left(Z > \frac{-2}{3.9686}\right)\)

\(= P(Z > -0.504)\)

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

\(P(Z > -0.504) = \Phi(0.504) \approx 0.693\)

Thus, the probability that more than 20 people pass is approximately \(0.6925 < p < 0.693\).

Let the random variable \(X\) represent the number of students who play exactly one musical instrument in a sample of 90 students. Given that 52% of students play exactly one instrument, \(X\) follows a binomial distribution \(B(n=90, p=0.52)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are given by:

\(\mu = np = 90 \times 0.52 = 46.8\)

\(\sigma^2 = np(1-p) = 90 \times 0.52 \times 0.48 = 22.464\)

We use a continuity correction to find \(P(X < 40)\), which is approximated by \(P(X < 39.5)\) in the normal distribution.

Standardizing, we find the z-score:

\(z = \frac{39.5 - 46.8}{\sqrt{22.464}} = \frac{-7.3}{4.74} \approx -1.540\)

Using standard normal distribution tables, we find:

\(P(Z < -1.540) = 1 - 0.9382 = 0.0618\)

Thus, the probability that fewer than 40 students play exactly one musical instrument is approximately 0.0618.

Let the random variable \(X\) represent the number of households with no broadband service out of 120. \(X\) follows a binomial distribution with parameters \(n = 120\) and \(p = 0.35\).

To approximate, we use a normal distribution with mean \(\mu = np = 120 \times 0.35 = 42\) and variance \(\sigma^2 = np(1-p) = 120 \times 0.35 \times 0.65 = 27.3\).

We apply a continuity correction to approximate \(P(X > 32)\) as \(P(X > 32.5)\).

Standardize using \(Z = \frac{X - \mu}{\sigma}\):

\(P(X > 32.5) = P\left( Z > \frac{32.5 - 42}{\sqrt{27.3}} \right) = P(Z > -1.818)\)

Using the standard normal distribution table, \(\Phi(1.818) \approx 0.966\).

Thus, \(P(X > 32) = 1 - \Phi(-1.818) = \Phi(1.818)\).

Therefore, the probability is approximately \(0.965 < p < 0.966\).

The number of days the flight arrives early can be modeled by a binomial distribution with parameters \(n = 60\) and \(p = 0.15\).

First, calculate the mean and variance of the distribution:

Mean \(= np = 60 \times 0.15 = 9\)

Variance \(= np(1-p) = 60 \times 0.15 \times 0.85 = 7.65\)

Since \(n\) is large and \(p\) is small, we can use a normal approximation. The normal distribution has mean \(9\) and variance \(7.65\).

We need to find \(P(X \geq 12)\). Using continuity correction, this is approximated by \(P(X > 11.5)\).

Standardize using the normal distribution:

\(P\left(Z > \frac{11.5 - 9}{\sqrt{7.65}}\right)\)

Calculate the standardized value:

\(Z = \frac{11.5 - 9}{\sqrt{7.65}} \approx 0.9039\)

Find the probability from the standard normal distribution table:

\(1 - \Phi(0.9039) = 1 - 0.8169 = 0.183\)

Thus, the probability that Richard’s flight arrives early at least 12 times is approximately 0.183.

Let the random variable \(X\) represent the number of adults who travel to work by car in a sample of 150. \(X\) follows a binomial distribution with parameters \(n = 150\) and \(p = 0.6\).

To approximate this binomial distribution, we use a normal distribution with the same mean and variance. The mean \(\mu\) is given by:

\(\mu = np = 150 \times 0.6 = 90\)

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

\(\sigma^2 = np(1-p) = 150 \times 0.6 \times 0.4 = 36\)

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

\(\sigma = \sqrt{36} = 6\)

We want to find \(P(X < 81)\). Using the normal approximation with continuity correction, we calculate:

\(P(X < 81) \approx P\left(Z < \frac{80.5 - 90}{6}\right)\)

\(= P\left(Z < -1.5833\right)\)

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

\(\Phi(-1.5833) = 1 - 0.9433 = 0.0567\)

Therefore, the probability that fewer than 81 adults travel to work by car is approximately 0.0567.

For part (c), the approximation is justified because both \(np = 90\) and \(nq = 60\) are greater than 5, satisfying the conditions for using the normal approximation to the binomial distribution.

Let the random variable \(X\) represent the number of students who do not like any sports. \(X\) follows a binomial distribution with parameters \(n = 90\) and \(p = 0.3\).

To approximate, we use a normal distribution with mean \(\mu = np = 0.3 \times 90 = 27\) and variance \(\sigma^2 = np(1-p) = 0.3 \times 90 \times 0.7 = 18.9\).

Thus, \(X \sim N(27, 18.9)\).

We want \(P(X < 32)\). Using continuity correction, this is approximated by \(P(X < 31.5)\).

Standardize: \(z = \frac{31.5 - 27}{\sqrt{18.9}} = \frac{4.5}{4.347} \approx 1.035\).

Using the standard normal distribution table, \(P(Z < 1.035) = \Phi(1.035) \approx 0.850\).

Let the random variable \(X\) represent the number of days the train arrives late out of 142 days. \(X\) follows a binomial distribution with parameters \(n = 142\) and \(p = 0.35\).

To approximate, we use a normal distribution with mean \(\mu\) and variance \(\sigma^2\).

Calculate the mean: \(\mu = np = 142 \times 0.35 = 49.7\).

Calculate the variance: \(\sigma^2 = npq = 142 \times 0.35 \times 0.65 = 32.305\).

Standard deviation: \(\sigma = \sqrt{32.305}\).

We approximate \(X\) by a normal distribution \(N(49.7, 32.305)\).

We need to find \(P(X > 40)\). Using continuity correction, this becomes \(P(X > 40.5)\).

Standardize: \(Z = \frac{40.5 - 49.7}{\sqrt{32.305}} \approx -1.619\).

Find \(P(Z > -1.619)\) using standard normal distribution tables or calculator.

\(P(Z > -1.619) = 0.947\).

The probability of obtaining a pair of tails in one throw of two fair coins is \(0.25\) (since there are four equally likely outcomes: HH, HT, TH, TT, and only TT is a pair of tails).

Let \(X\) be the number of times a pair of tails is obtained in 80 throws. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.25\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 80 \times 0.25 = 20\)

\(\sigma^2 = np(1-p) = 80 \times 0.25 \times 0.75 = 15\)

We need to find \(P(X > 25)\). Using the normal approximation,

\(P(X > 25) \approx P\left(Z > \frac{25.5 - 20}{\sqrt{15}}\right)\)

\(= P(Z > 1.42)\)

Using standard normal distribution tables, \(P(Z > 1.42) = 1 - P(Z \leq 1.42) = 1 - 0.9222 = 0.0778\).

The problem can be approximated using a normal distribution because the number of days is large (100 days), and the probability of messaging is not too close to 0 or 1.

First, calculate the mean and variance of the binomial distribution:

Mean: \(n \times p = 100 \times 0.72 = 72\)

Variance: \(n \times p \times (1-p) = 100 \times 0.72 \times 0.28 = 20.16\)

We approximate the binomial distribution with a normal distribution \(N(72, 20.16)\).

To find the probability of fewer than 64 days, use continuity correction: \(P(X < 64) \approx P(X < 63.5)\).

Standardize using the formula:

\(z = \frac{63.5 - 72}{\sqrt{20.16}}\)

\(z = \frac{63.5 - 72}{4.49} \approx -1.893\)

Using standard normal distribution tables or a calculator, find \(P(z < -1.893)\).

The probability is approximately 0.0292.

The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 120\)

- Probability of success, \(p = 0.7\)

We approximate the binomial distribution with a normal distribution using the Central Limit Theorem.

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

\(\mu = np = 120 \times 0.7 = 84\)

\(\sigma^2 = np(1-p) = 120 \times 0.7 \times 0.3 = 25.2\)

Using a normal approximation, we find the probability that more than 75 adults own a car. Apply continuity correction:

\(P(X > 75) \approx P\left(Z > \frac{75.5 - 84}{\sqrt{25.2}}\right)\)

Calculate the standard score (z-score):

\(Z = \frac{75.5 - 84}{\sqrt{25.2}} = \frac{-8.5}{5.02} \approx -1.693\)

Using the standard normal distribution table, find:

\(P(Z > -1.693) = 0.955\)

Thus, the probability that more than 75 of them own a car is approximately 0.955.

Let the probability that a person does not support either choir be denoted by \(p\). Given that 30% support Notes and 45% support Classics, the probability of not supporting either is:

\(p = 1 - 0.30 - 0.45 = 0.25\)

The mean number of people in the sample who do not support either choir is:

\(\text{Mean} = 240 \times 0.25 = 60\)

The variance is:

\(\text{Variance} = 240 \times 0.25 \times 0.75 = 45\)

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

\(\sigma = \sqrt{45}\)

We apply the continuity correction for \(X < 50\):

\(P(X < 50) = P\left( Z < \frac{49.5 - 60}{\sqrt{45}} \right) = P(Z < -1.565)\)

Using standard normal distribution tables, \(P(Z < -1.565)\) is approximately:

\(1 - 0.9412 = 0.0588\)

Thus, the probability that fewer than 50 do not support either choir is \(0.0588\).

Let the random variable \(X\) represent the number of faulty components in a sample of 200. \(X\) follows a binomial distribution with parameters \(n = 200\) and \(p = 0.15\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 200 \times 0.15 = 30\)

\(\sigma^2 = np(1-p) = 200 \times 0.15 \times 0.85 = 25.5\)

Thus, \(X\) is approximately \(N(30, 25.5)\).

We need to find \(P(X > 40)\). Using a continuity correction, this is equivalent to \(P(X > 40.5)\).

Standardizing, we have:

\(P(X > 40.5) = P\left(Z > \frac{40.5 - 30}{\sqrt{25.5}}\right)\)

\(= P\left(Z > \frac{10.5}{5.04975}\right)\)

\(= P(Z > 2.079)\)

Using standard normal distribution tables or a calculator, \(P(Z > 2.079) = 1 - \Phi(2.079) \approx 1 - 0.9812 = 0.0188\).

Let the random variable \(X\) represent the number of households satisfied with their wifi connection. \(X\) follows a binomial distribution with parameters \(n = 150\) and \(p = 0.66\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(np, npq)\), where \(q = 1 - p\).

Calculate the mean and variance:

\(np = 0.66 \times 150 = 99\)

\(npq = 0.66 \times (1 - 0.66) \times 150 = 33.66\)

Standard deviation \(\sigma = \sqrt{33.66} \approx 5.8017\)

We need to find \(P(X > 84)\). Using continuity correction, this is \(P(X > 84.5)\).

Standardize using the normal distribution:

\(P(X > 84.5) = P\left( Z > \frac{84.5 - 99}{\sqrt{33.66}} \right)\)

\(= P(Z > -2.499)\)

Using standard normal distribution tables or a calculator, \(P(Z > -2.499) = 0.994\).

Let the random variable representing the number of men be denoted by a binomial distribution, \(X \sim B(n, p)\), where \(n = 600\) and \(p = 0.34\).

The mean \(\mu\) is given by \(np = 600 \times 0.34 = 204\).

The variance \(\sigma^2\) is given by \(np(1-p) = 600 \times 0.34 \times 0.66 = 134.64\).

We approximate the binomial distribution with a normal distribution \(N(\mu, \sigma^2)\).

To find \(P(X < 190)\), we use a continuity correction: \(P(X < 190) \approx P(X < 189.5)\).

Standardize using \(Z = \frac{X - \mu}{\sigma}\):

\(P(X < 189.5) = P\left( Z < \frac{189.5 - 204}{\sqrt{134.64}} \right) = P(Z < -1.2496)\).

Using the standard normal distribution table, \(P(Z < -1.2496) = 1 - \Phi(1.2496)\).

\(\Phi(1.2496) \approx 0.8944\), so \(1 - 0.8944 = 0.106\).

The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 250\)

- Probability of success, \(p = 0.65\)

We approximate the binomial distribution with a normal distribution using the following parameters:

- Mean, \(\mu = np = 250 \times 0.65 = 162.5\)

- Variance, \(\sigma^2 = np(1-p) = 250 \times 0.65 \times 0.35 = 56.875\)

- Standard deviation, \(\sigma = \sqrt{56.875}\)

We need to find \(P(X < 179)\). Using the continuity correction, we approximate this as \(P(X < 178.5)\).

Standardize the variable:

\(Z = \frac{178.5 - 162.5}{\sqrt{56.875}} = \frac{16}{\sqrt{56.875}} \approx 2.122\)

Using the standard normal distribution table, find \(P(Z < 2.122)\).

The probability is approximately \(0.983\).

We are given that 35% of customers shop online, so the probability of a customer shopping online is 0.35. For a sample of 100 customers, we can model the number of customers shopping online as a binomial distribution with parameters \(n = 100\) and \(p = 0.35\).

To approximate this binomial distribution, we use a normal distribution with the same mean and variance. The mean \(\mu\) is given by:

\(\mu = np = 100 \times 0.35 = 35\)

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

\(\sigma^2 = np(1-p) = 100 \times 0.35 \times 0.65 = 22.75\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{22.75}\).

We need to find \(P(X > 39)\). Using the normal approximation, we apply a continuity correction and find \(P(X > 39.5)\).

We standardize this to find the corresponding \(z\)-score:

\(z = \frac{39.5 - 35}{\sqrt{22.75}} = 0.943\)

We then find \(P(Z > 0.943)\) using the standard normal distribution table:

\(P(Z > 0.943) = 1 - P(Z < 0.943) = 1 - 0.8272 = 0.173\)

Therefore, the probability that more than 39 customers shop online is 0.173.

(ii) The probability that a light bulb has a lifetime of more than 2500 hours is 0.2. For 300 bulbs, the mean is calculated as:

\(\text{Mean} = 300 \times 0.2 = 60\)

The variance is:

\(\text{Variance} = 300 \times 0.2 \times 0.8 = 48\)

We use the normal approximation to the binomial distribution. The probability that fewer than 70 bulbs have a lifetime of more than 2500 hours is:

\(P(X < 70) = P\left(Z > \frac{69.5 - 60}{\sqrt{48}}\right)\)

\(= \Phi(1.371)\)

\(= 0.915\)

(iii) The conditions for using the normal approximation are met because:

\(np = 60\) and \(nq = 240\), both greater than 5.

Let the random variable \(X\) represent the number of students who pass the examination. \(X\) follows a binomial distribution with parameters \(n = 200\) and \(p = 0.8\).

To approximate this binomial distribution, we use a normal distribution because both \(np\) and \(n(1-p)\) are greater than 5.

The mean of the binomial distribution is \(np = 200 \times 0.8 = 160\).

The variance is \(np(1-p) = 200 \times 0.8 \times 0.2 = 32\).

Thus, the standard deviation is \(\sqrt{32}\).

We approximate \(X\) by a normal distribution \(N(160, 32)\).

We need to find \(P(X > 166)\). Using continuity correction, this is approximately \(P(X > 166.5)\).

Standardizing, we have:

\(z = \frac{166.5 - 160}{\sqrt{32}} \approx 1.149\)

Thus, \(P(X > 166.5) = P(Z > 1.149) = 1 - P(Z \leq 1.149)\).

Using standard normal distribution tables, \(P(Z \leq 1.149) \approx 0.8747\).

Therefore, \(P(Z > 1.149) = 1 - 0.8747 = 0.125\).

For part (iii), the justification for using the normal approximation is that both \(np = 160\) and \(nq = 40\) are greater than 5, which satisfies the conditions for the normal approximation to the binomial distribution.

(i) To find the probability that an apple can be used as a toffee apple, we calculate the probability that the diameter is between 4.1 cm and 5 cm. We standardize these values using the formula for the z-score:

\(z_1 = \frac{4.1 - 5.7}{0.8} = -2\)

\(z_2 = \frac{5 - 5.7}{0.8} = -0.875\)

The probability is given by:

\(P(4.1 < d < 5) = P(z < -0.875) - P(z < -2)\)

Using standard normal distribution tables:

\(\Phi(-0.875) = 0.1908\)

\(\Phi(-2) = 0.0228\)

Thus, \(P(4.1 < d < 5) = 0.1908 - 0.0228 = 0.168\)

(ii) For 250 apples, the number that can be used as toffee apples follows a binomial distribution with \(n = 250\) and \(p = 0.168\). We approximate this with a normal distribution:

\(np = 250 \times 0.168 = 42\)

\(npq = 34.944\)

Standard deviation \(\sigma = \sqrt{34.944} = 5.911\)

We find \(P(X < 50)\) using continuity correction:

\(P(X < 50) = P\left( z < \frac{49.5 - 42}{5.911} \right) = P(z < 1.2687)\)

Using standard normal distribution tables:

\(\Phi(1.2687) = 0.898\)