Past Exam Questions

Let the random variable \(X\) represent the number of people who choose a mobile phone made by Company B. \(X\) follows a binomial distribution with parameters \(n = 130\) and \(p = 0.35\).

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

\(\mu = np = 130 \times 0.35 = 45.5\)

\(\sigma^2 = np(1-p) = 130 \times 0.35 \times 0.65 = 29.575\)

Thus, the standard deviation \(\sigma\) is \(\sqrt{29.575} \approx 5.438\).

We need to find \(P(X \geq 50)\). Using the normal approximation, we standardize:

\(P(X \geq 50) = P\left( Z \geq \frac{49.5 - 45.5}{5.438} \right) = P(Z \geq 0.7355)\)

Using the standard normal distribution table, \(P(Z \geq 0.7355) = 1 - \Phi(0.7355) \approx 1 - 0.7691 = 0.231\).

Let the random variable \(X\) represent the number of families owning a dishwasher in a sample of 145 families. \(X\) follows a binomial distribution with parameters \(n = 145\) and \(p = 0.22\).

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

Calculate the mean \(\mu\) and variance \(\sigma^2\):

\(\mu = 145 \times 0.22 = 31.9\)

\(\sigma^2 = 145 \times 0.22 \times 0.78 = 24.882\)

We use a continuity correction for \(P(X > 26)\), which becomes \(P(X > 26.5)\).

Standardize the variable:

\(P(X > 26.5) = P\left( z > \frac{26.5 - 31.9}{\sqrt{24.882}} \right) = P(z > -1.08255)\)

Using the standard normal distribution table, \(\Phi(1.08255) = 0.861\).

Thus, the probability that more than 26 families own a dishwasher is \(0.861\).

Let \(p = 0.25\) and \(n = 160\). The mean of the distribution is \(n \times p = 160 \times 0.25 = 40\).

The variance is \(n \times p \times (1-p) = 160 \times 0.25 \times 0.75 = 30\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(40, 30)\).

To find \(P(X < 50)\), we use the continuity correction: \(P(X < 49.5)\).

Standardize: \(Z = \frac{49.5 - 40}{\sqrt{30}} \approx 1.734\).

Thus, \(P(Z < 1.734) = 0.959\).

(a) To find the percentage of eggs classified as small, we calculate the probability that an egg weighs less than 76 grams. Using the standardization formula:

\(P(X < 76) = P\left(Z < \frac{76 - 80.5}{6.6}\right)\)

\(= P(Z < -0.6818) = 1 - \Phi(0.6818)\)

\(= 1 - 0.7524 = 0.2476\)

Thus, the percentage is 24.8%.

(b) To find the least possible weight of a large egg, we first find the percentage of large eggs:

\(\text{Percentage of large eggs} = 100 - 40 - 24.76 = 35.24\)

We then find the weight corresponding to the cumulative probability of 0.6476:

\(P(Z > x) = 0.6476\)

\(\frac{x - 80.5}{6.6} = 0.378\)

\(x = 80.5 + 0.378 \times 6.6 = 83.0\)

(c) For 150 eggs, the mean number classified as medium is:

\(\text{Mean} = 150 \times 0.4 = 60\)

\(\text{Variance} = 150 \times 0.4 \times 0.6 = 36\)

We approximate the probability that more than 68 eggs are medium:

\(P(X > 68) = P\left(Z > \frac{68.5 - 60}{\sqrt{36}}\right)\)

\(= P(Z > 1.417) = 1 - \Phi(1.417)\)

\(= 1 - 0.9217 = 0.0783\)

(i) The number of faulty CDs in a box follows a binomial distribution with parameters \(n = 30\) and \(p = 0.04\). We need to find \(P(X > 2)\), where \(X\) is the number of faulty CDs.

\(P(X > 2) = 1 - P(X \leq 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))\)

\(P(X = 0) = \binom{30}{0} (0.04)^0 (0.96)^{30}\)

\(P(X = 1) = \binom{30}{1} (0.04)^1 (0.96)^{29}\)

\(P(X = 2) = \binom{30}{2} (0.04)^2 (0.96)^{28}\)

Calculating these probabilities:

\(P(X = 0) = 0.2938\)

\(P(X = 1) = 0.3673\)

\(P(X = 2) = 0.2219\)

\(P(X > 2) = 1 - (0.2938 + 0.3673 + 0.2219) = 0.117\)

(ii) Let \(Y\) be the number of rejected boxes out of 280. \(Y\) follows a binomial distribution with parameters \(n = 280\) and \(p = 0.117\) (from part (i)).

We approximate \(Y\) using a normal distribution with mean \(\mu = np = 280 \times 0.117 = 32.73\) and variance \(\sigma^2 = np(1-p) = 280 \times 0.117 \times 0.883 = 28.9\).

Using the normal approximation, we find \(P(Y \geq 30)\) with continuity correction:

\(P(Y \geq 30) = P\left( Z \geq \frac{29.5 - 32.73}{\sqrt{28.9}} \right) = P(Z \geq -0.6008)\)

Using standard normal distribution tables, \(P(Z \geq -0.6008) = 0.726\).

Let \(X\) be the number of days George goes swimming in 270 days. \(X\) follows a binomial distribution with parameters \(n = 270\) and \(p = \frac{1}{3}\).

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

Calculate \(np\) and \(npq\):

\(np = 270 \times \frac{1}{3} = 90\)

\(npq = 270 \times \frac{1}{3} \times \frac{2}{3} = 60\)

We use a continuity correction to approximate \(P(X > 100)\) as \(P(X > 99.5)\).

Standardize using the normal distribution:

\(P(X > 99.5) = P\left( z > \frac{99.5 - 90}{\sqrt{60}} \right) = P(z > 1.2264)\)

Using the standard normal distribution table, find \(P(z > 1.2264) = 1 - P(z < 1.2264) = 1 - 0.8899 = 0.110\).

Let the random variable \(X\) represent the number of people who enjoy watching Historical Drama. \(X\) follows a binomial distribution with parameters \(n = 160\) and \(p = 0.1\).

We approximate this binomial distribution with a normal distribution. First, calculate the mean \(np\) and variance \(npq\):

\(np = 160 \times 0.1 = 16\)

\(npq = 160 \times 0.1 \times 0.9 = 14.4\)

Using the normal approximation, we have \(X \sim N(16, 14.4)\).

We need to find \(P(X > 17)\). Using continuity correction, this is \(P(X \geq 17.5)\).

Standardize the variable:

\(P(X \geq 17.5) = P\left( Z \geq \frac{17.5 - 16}{\sqrt{14.4}} \right) = P(Z \geq 0.3953)\)

Using standard normal distribution tables, \(P(Z \geq 0.3953) = 1 - P(Z < 0.3953) = 1 - 0.6536 = 0.346\).

The problem can be modeled using a binomial distribution with parameters:

- Number of trials, \(n = 365\) (days in a year) - Probability of success, \(p = 0.15\) (probability of eating potato)

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

\(\mu = np = 365 \times 0.15 = 54.75\)

\(\sigma^2 = np(1-p) = 365 \times 0.15 \times 0.85 = 46.5375\)

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

To find the probability that Annabel eats potato on more than 44 days, we standardize the variable:

\(P(X > 44) = P\left( Z > \frac{44.5 - 54.75}{\sqrt{46.5375}} \right)\)

\(= P(Z > -1.5025)\)

Using the standard normal distribution table, \(P(Z > -1.5025) = 0.933\).

The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 100\)

- Probability of success (asleep), \(p = 0.2\)

We approximate the binomial distribution with a normal distribution:

- Mean, \(\mu = np = 100 \times 0.2 = 20\)

- Variance, \(\sigma^2 = np(1-p) = 100 \times 0.2 \times 0.8 = 16\)

- Standard deviation, \(\sigma = \sqrt{16} = 4\)

We want to find \(P(X \leq 30)\). Using continuity correction, we find \(P(X \leq 30.5)\).

Standardizing, we calculate the z-score:

\(z = \frac{30.5 - 20}{4} = 2.625\)

Using the standard normal distribution table, we find:

\(P(Z < 2.625) = 0.996\)

The probability of exactly 7 passengers carrying backpacks in a minibus is given by the binomial probability:

\(P(7) = \binom{12}{7} (0.65)^7 (0.35)^5 = 0.2039\)

For 250 minibuses, the mean and variance of the number of minibuses with exactly 7 passengers carrying backpacks are:

Mean: \(250 \times 0.2039 = 50.9798\)

Variance: \(250 \times 0.2039 \times (1 - 0.2039) = 40.5851\)

We approximate the distribution of the number of minibuses with exactly 7 passengers carrying backpacks by a normal distribution with the above mean and variance.

We need to find \(P(X > 54)\), where \(X\) is the number of minibuses with exactly 7 passengers carrying backpacks.

Using continuity correction, we find:

\(P(X > 54) = P\left( \frac{54.5 - 50.9798}{\sqrt{40.5851}} \right) = P(Z > 0.5526)\)

Using the standard normal distribution table, \(P(Z > 0.5526) = 1 - \Phi(0.5526) = 1 - 0.7098 = 0.2902\)

Thus, the probability is approximately \(0.290\).

(i) The probability of waiting between 5 and 10 minutes is given by the difference between the probability of waiting less than 10 minutes and the probability of waiting less than 5 minutes:

\(P(5 < X < 10) = P(X < 10) - P(X < 5) = 0.88 - 0.16 = 0.72\)

(ii) Let \(X\) be the number of people who wait between 5 and 10 minutes. \(X\) follows a binomial distribution \(X \sim B(180, 0.72)\).

Using a normal approximation, \(X\) can be approximated by a normal distribution \(X \sim N(np, npq)\) where:

\(np = 180 \times 0.72 = 129.6\)

\(npq = 180 \times 0.72 \times 0.28 = 36.288\)

Thus, \(X \sim N(129.6, 36.288)\).

We need to find \(P(X > 115)\). Using continuity correction, this becomes \(P(X > 115.5)\).

Standardizing, we have:

\(P\left( z > \frac{115.5 - 129.6}{\sqrt{36.288}} \right) = P(z > -2.341)\)

Using standard normal distribution tables, \(P(z > -2.341) = 0.990\).

(i) To find the probability that a straw fits into the hole, we need to calculate the probability that the diameter is less than 3 mm. This is given by:

\(P(X < 3.0) = P\left( Z < \frac{3.0 - 2.6}{0.25} \right) = P(Z < 1.6)\)

Using the standard normal distribution table, \(P(Z < 1.6) = 0.945\).

(ii) For 500 straws, we use a normal approximation to the binomial distribution. Let \(X \sim B(500, 0.9452)\). The normal approximation is \(X \sim N(472.6, 25.898)\).

We need \(P(X \geq 480)\), which is approximated by:

\(P\left( Z > \frac{479.5 - 472.6}{\sqrt{25.898}} \right) = P(Z > 1.3558)\)

Using the standard normal distribution table, \(P(Z > 1.3558) = 1 - 0.9125 = 0.0875\).

(iii) The approximation is justified because both \(500 \times 0.9452\) and \(500 \times (1 - 0.9452)\) are greater than 5, satisfying the conditions for using the normal approximation to the binomial distribution.

(iii) Let the number of defective water pistols be a binomial random variable with parameters \(n = 1800\) and \(p = 0.08\).

Calculate \(np = 1800 \times 0.08 = 144\).

Calculate \(npq = 1800 \times 0.08 \times 0.92 = 132.48\).

Use a normal approximation: \(X \sim N(144, 132.48)\).

Apply continuity correction: \(P(X \geq 152) = P\left( Z > \frac{151.5 - 144}{\sqrt{132.48}} \right)\).

Calculate \(Z = \frac{151.5 - 144}{\sqrt{132.48}} = 0.6516\).

Find \(P(Z > 0.6516) = 1 - 0.7429 = 0.257\).

(iv) The approximation is justified because \(np = 1800 \times 0.08 = 144 > 5\) and \(n(1-p) = 1800 \times 0.92 = 1656 > 5\).

(i) Let the probability of throwing an even number be \(p\). Then the probability of throwing an odd number is \(2p\). Since there are 3 even numbers (2, 4, 6) and 3 odd numbers (1, 3, 5), we have:

\(3p + 6p = 1\)

\(9p = 1\)

\(p = \frac{1}{9}\)

Thus, the probability of throwing a 3 is \(2p = \frac{2}{9}\).

(ii) The probability of throwing a 5 is \(\frac{2}{9}\) and the probability of throwing a 4 is \(\frac{1}{9}\). The probability of throwing two 5s and one 4 in three throws is:

\(\binom{3}{2} \left( \frac{2}{9} \right)^2 \left( \frac{1}{9} \right) = 3 \times \frac{4}{81} \times \frac{1}{9} = \frac{4}{243}\).

(iii) Let \(X\) be the number of times an even number is thrown. \(X\) follows a binomial distribution with \(n = 100\) and \(p = \frac{1}{3}\). The mean \(\mu = np = 33.3\) and the standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{100 \times \frac{1}{3} \times \frac{2}{3}} = 22.2\).

Using a normal approximation, \(X \sim N(33.3, 22.2^2)\). We find \(P(X \leq 37)\) by standardizing:

\(P(X \leq 37) = P\left( Z \leq \frac{37.5 - 33.3}{22.2} \right) = P(Z \leq 0.8839) \approx 0.812\).

The number of votes Anil received follows a binomial distribution with parameters \(n = 120\) and \(p = 0.4\).

We approximate this binomial distribution with a normal distribution with mean \(\mu = np = 120 \times 0.4 = 48\) and variance \(\sigma^2 = np(1-p) = 120 \times 0.4 \times 0.6 = 28.8\).

Using the normal approximation, we apply a continuity correction. We want to find \(P(36 \leq X \leq 54)\), which becomes \(P(35.5 < X < 54.5)\).

Standardizing, we calculate:

\(P\left( \frac{35.5 - 48}{\sqrt{28.8}} < Z < \frac{54.5 - 48}{\sqrt{28.8}} \right)\)

\(P\left( \frac{-12.5}{\sqrt{28.8}} < Z < \frac{6.5}{\sqrt{28.8}} \right)\)

\(P(-2.3292 < Z < 1.211)\)

Using standard normal distribution tables, we find:

\(P(Z < 1.211) = 0.8871\) and \(P(Z < -2.3292) = 0.0099\).

Thus, \(P(-2.3292 < Z < 1.211) = 0.8871 - 0.0099 = 0.8772\).

Therefore, the probability is approximately \(0.877\).

Let the random variable \(X\) represent the number of substandard cameras in a sample of 300. \(X\) follows a binomial distribution with parameters \(n = 300\) and \(p = 0.072\).

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

\(\mu = np = 300 \times 0.072 = 21.6\)

\(\sigma^2 = np(1-p) = 300 \times 0.072 \times 0.928 = 20.0448\)

Using a continuity correction, we find \(P(X < 18)\) by calculating \(P(X < 17.5)\).

Standardizing, we have:

\(P(X < 17.5) = P\left( Z < \frac{17.5 - 21.6}{\sqrt{20.0448}} \right)\)

\(= P(Z < -0.9157)\)

Using the standard normal distribution table, \(P(Z < -0.9157) = 1 - 0.8201 = 0.180\).

(ii) To approximate the probability, we use the normal approximation to the binomial distribution. Given that the sample size is 500, we calculate:

Mean: \(np = 340\)

Variance: \(npq = 108.8\)

We standardize the variable using continuity correction:

\(P(x > 337) = P\left( z > \frac{337.5 - 340}{\sqrt{108.8}} \right)\)

\(= P(z > -0.2396)\)

Using the standard normal distribution table, we find:

\(P(z > -0.2396) = 0.595\)

(iii) The normal approximation is justified because both \(np = 340\) and \(nq = 160\) are greater than 5.

Let the number of adults who own a cell phone be a binomial random variable with parameters \(n = 160\) and \(p = 0.75\).

Calculate the mean and variance:

\(np = 160 \times 0.75 = 120\)

\(npq = 160 \times 0.75 \times 0.25 = 30\)

Use the normal approximation to the binomial distribution:

\(P(X > 114) = P\left( z > \frac{114.5 - 120}{\sqrt{30}} \right)\)

\(= P(z > -1.004)\)

\(= \Phi(1.004) = 0.842\)

Thus, \(P(X > 114) = 1 - 0.842 = 0.158\).

For part (iii), the approximation is justified because both \(np\) and \(nq\) are greater than 5.

Let \(X\) be the number of days Wenjie goes out with her friends. \(X\) follows a binomial distribution with \(n = 252\) and \(p = \frac{1}{7}\).

For normal approximation, calculate the mean \(\mu\) and variance \(\sigma^2\):

\(\mu = np = 252 \times \frac{1}{7} = 36\)

\(\sigma^2 = npq = 252 \times \frac{1}{7} \times \frac{6}{7} = 30.857\)

Standard deviation \(\sigma = \sqrt{30.857}\).

Using continuity correction, find:

\(P(X < 30) = P\left( z < \frac{29.5 - 36}{\sqrt{30.857}} \right)\)

\(P(X > 44) = P\left( z > \frac{44.5 - 36}{\sqrt{30.857}} \right)\)

Calculate the z-scores:

\(z_1 = \frac{29.5 - 36}{\sqrt{30.857}} = -1.170\)

\(z_2 = \frac{44.5 - 36}{\sqrt{30.857}} = 1.530\)

Using standard normal distribution tables:

\(P(z < -1.170) = 0.1209\)

\(P(z > 1.530) = 0.0630\)

Total probability: \(P(X < 30) + P(X > 44) = 0.1209 + 0.0630 = 0.184\)

For part (ii), the use of a normal approximation is justified because both \(np\) and \(nq\) are greater than 5.

(i) Given that 80% of the days, Rafa spends more than 1.35 hours, we have:

\(P(X > 1.35) = 0.8\)

Using the standard normal distribution, \(z = -0.842\) for 20% in the left tail:

\(P\left( \frac{1.35 - 1.9}{\sigma} < z \right) = 0.2\)

\(-0.842 = \frac{1.35 - 1.9}{\sigma}\)

\(\sigma = \frac{-0.55}{-0.842} = 0.653\)

(ii) To find \(P(X < 2)\):

\(P\left( z < \frac{2 - 1.9}{0.6532} \right) = P(z < 0.1531)\)

Using the standard normal table, \(P(z < 0.1531) = 0.561\).

(iii) For a sample of 200 days, \(X \sim N(160, 32)\):

We need \(P(162.5 < X < 173.5)\):

\(P\left( \frac{162.5 - 160}{\sqrt{32}} < z < \frac{173.5 - 160}{\sqrt{32}} \right)\)

\(P(0.442 < z < 2.386) = \Phi(2.386) - \Phi(0.442)\)

\(= 0.9915 - 0.6707 = 0.321\)