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:

\(\Phi(1.2687) = 0.898\)

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

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

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

\(= 0.9915 - 0.6707 = 0.321\)

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 \(P(X > 60)\):

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

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

The number of times the gymnast performs the routine correctly follows a binomial distribution with parameters \(n = 50\) and \(p = 0.65\).

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

\(\mu = n \times p = 50 \times 0.65 = 32.5\)

\(\sigma^2 = n \times p \times (1-p) = 50 \times 0.65 \times 0.35 = 11.375\)

We use a continuity correction to find the probability of fewer than 29 occasions:

\(P(X < 29) \approx P\left(Z < \frac{28.5 - 32.5}{\sqrt{11.375}}\right)\)

\(= P\left(Z < \frac{-4}{\sqrt{11.375}}\right)\)

\(= P(Z < -1.186)\)

Using the standard normal distribution table, \(P(Z < -1.186) = 1 - P(Z < 1.186) = 1 - 0.8822 = 0.118\).

Let \(X\) be the number of years out of 56 where New Year's Day is on a Saturday. \(X\) follows a binomial distribution with parameters \(n = 56\) and \(p = \frac{1}{7}\).

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

\(\mu = 56 \times \frac{1}{7} = 8\)

\(\sigma^2 = 56 \times \frac{1}{7} \times \frac{6}{7} = 6.857\)

We need to find \(P(X > 7)\). Using a continuity correction, this becomes \(P(X > 7.5)\).

Standardizing, we have:

\(P(X > 7.5) = 1 - \Phi \left( \frac{7.5 - 8}{\sqrt{6.857}} \right)\)

\(= 1 - \Phi(-0.1909)\)

\(= \Phi(0.1909)\)

Using standard normal distribution tables, \(\Phi(0.1909) \approx 0.576\).

Let the number of small bands be a random variable, denoted by \(X\). Given that 60% of the bands are small, \(X\) follows a binomial distribution \(B(n=150, p=0.6)\).

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

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

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

Thus, the standard deviation \(\sigma\) is \(\sqrt{36} = 6\).

We use a normal approximation to find \(P(88 < X < 97)\). Applying continuity correction, this becomes \(P(87.5 < X < 97.5)\).

Standardizing, we calculate:

\(P(87.5 < X < 97.5) = \Phi\left(\frac{97.5 - 90}{6}\right) - \Phi\left(\frac{87.5 - 90}{6}\right)\)

\(= \Phi(1.25) - \Phi(-0.4166)\)

Using standard normal distribution tables, \(\Phi(1.25) = 0.8944\) and \(\Phi(-0.4166) = 1 - 0.6616 = 0.3384\).

Thus, the probability is:

\(0.8944 - 0.3384 = 0.556\)

Let the random variable \(X\) represent the number of adults wearing a watch on their left wrist. Given that 76% wear a watch on their left wrist, \(X\) follows a binomial distribution \(B(n=200, p=0.76)\).

We approximate this binomial distribution with a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are calculated as follows:

\(\mu = np = 200 \times 0.76 = 152\)

\(\sigma^2 = np(1-p) = 200 \times 0.76 \times 0.24 = 36.48\)

We use a continuity correction to find \(P(X > 155)\), which is approximated by \(P(X > 155.5)\) in the normal distribution.

Standardizing, we have:

\(Z = \frac{155.5 - 152}{\sqrt{36.48}} = \frac{3.5}{6.036} \approx 0.5795\)

Thus, \(P(X > 155) = 1 - \Phi(0.5795) \approx 1 - 0.7188 = 0.281\).

(i) We know that 10% of the bulbs last longer than 5130 hours. This corresponds to a z-score of 1.282 in a standard normal distribution. Using the formula for the z-score:

\(z = \frac{X - \mu}{\sigma}\)

where \(X = 5130\), \(\sigma = 40.6\), and \(z = 1.282\), we solve for \(\mu\):

\(1.282 = \frac{5130 - \mu}{40.6}\)

\(\mu = 5130 - 1.282 \times 40.6\)

\(\mu = 5080\) (rounded to the nearest hour)

(ii) To find the probability that a light bulb fails to last for 5000 hours, we calculate the z-score:

\(z = \frac{5000 - 5078}{40.6} = -1.921\)

The probability \(P(X < 5000) = \Phi(-1.921) = 0.0273\) or 2.73%.

(iii) For 600 light bulbs, the mean number lasting longer than 5130 hours is \(\mu = 60\) and variance \(\sigma^2 = 54\). We approximate using a normal distribution:

\(P(\text{fewer than 65}) = \Phi\left(\frac{64.5 - 60}{\sqrt{54}}\right)\)

\(= \Phi(0.6123)\)

\(= 0.730\) or 0.73

Let the random variable \(X\) represent the number of overweight people in the sample of 400. Given that 2 out of 5 people are overweight, the probability \(p\) of a person being overweight is \(p = \frac{2}{5} = 0.4\).

The expected number of overweight people \(\mu\) is given by \(\mu = np = 400 \times 0.4 = 160\).

The variance \(\sigma^2\) is given by \(\sigma^2 = np(1-p) = 400 \times 0.4 \times 0.6 = 96\).

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is \(\sqrt{96} \approx 9.798\).

We apply continuity correction for the normal approximation: \(P(X < 165) \approx P(X \leq 164.5)\).

Standardizing, we find:

\(Z = \frac{164.5 - 160}{\sqrt{96}} = \frac{4.5}{9.798} \approx 0.4593\)

Using standard normal distribution tables, \(\Phi(0.4593) \approx 0.677\).

Thus, the probability that fewer than 165 people in the sample are overweight is approximately 0.677.

Let the random variable \(X\) represent the number of cars made by Ford in a sample of 50 cars. \(X\) follows a binomial distribution with parameters \(n = 50\) and \(p = 0.28\).

To use a normal approximation, we calculate the mean \(\mu\) and variance \(\sigma^2\) of the distribution:

\(\mu = 50 \times 0.28 = 14\)

\(\sigma^2 = 50 \times 0.28 \times 0.72 = 10.08\)

We apply a continuity correction to find \(P(X > 18)\), which is approximated by \(P(X > 18.5)\).

Standardize the variable:

\(Z = \frac{18.5 - 14}{\sqrt{10.08}} = 1.417\)

Using the standard normal distribution table, find \(\Phi(1.417) \approx 0.9218\) or \(0.9217\).

Thus, \(P(X > 18) = 1 - \Phi(1.417) = 1 - 0.9218 = 0.0782\) or \(1 - 0.9217 = 0.0783\).

Let the random variable \(X\) represent the number of students with blue eyes in a sample of 80 students. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.32\).

To approximate, we use a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are given by:

\(\mu = np = 80 \times 0.32 = 25.6\)

\(\sigma^2 = np(1-p) = 80 \times 0.32 \times 0.68 = 17.408\)

Using the normal approximation, we apply continuity correction for \(P(X < 20)\):

\(P(X < 20) = P\left(Z < \frac{19.5 - 25.6}{\sqrt{17.408}}\right) = P(Z < -1.462)\)

Using standard normal distribution tables or a calculator, find:

\(P(Z < -1.462) = 1 - \Phi(1.462) = 1 - 0.9282 = 0.0718\)

Thus, the probability that fewer than 20 students have blue eyes is approximately \(0.0718 \leq p < 0.0719\).

Let the random variable \(X\) represent the number of damaged tapes in a sample of 1600 tapes. Given that 1 in 5 tapes are damaged, the probability of a tape being damaged is \(p = 0.2\).

The expected number of damaged tapes is \(\mu = np = 1600 \times 0.2 = 320\).

The variance is \(\sigma^2 = np(1-p) = 1600 \times 0.2 \times 0.8 = 256\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(320, 256)\).

We need to find \(P(X \geq 290)\). Using continuity correction, this is \(P(X > 289.5)\).

Standardize: \(Z = \frac{289.5 - 320}{\sqrt{256}} = \frac{-30.5}{16} = -1.906\).

Thus, \(P(X \geq 290) = 1 - \Phi(-1.906) = \Phi(1.906)\).

Using standard normal distribution tables, \(\Phi(1.906) = 0.972\).

(i) The number of eggs laid by the hens follows a binomial distribution with parameters \(n = 30\) and \(p = 0.7\). We need to find \(P(X = 24)\).

Using the binomial probability formula:

\(P(X = 24) = \binom{30}{24} (0.7)^{24} (0.3)^{6}\)

\(= (0.7)^{24} \times (0.3)^{6} \times \binom{30}{24}\)

\(= 0.0829\)

(ii) To approximate the probability that fewer than 20 eggs are laid, we use a normal approximation to the binomial distribution.

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

\(\mu = np = 30 \times 0.7 = 21\)

\(\sigma^2 = np(1-p) = 30 \times 0.7 \times 0.3 = 6.3\)

Using the normal approximation, we standardize the variable:

\(P(X < 20) \approx P\left(Z < \frac{19.5 - 21}{\sqrt{6.3}}\right)\)

\(= P(Z < -0.5976)\)

Using standard normal distribution tables:

\(P(Z < -0.5976) = 1 - 0.7251 = 0.275\)

(i) Let the probability of choosing a chocolate biscuit be \(p = \frac{1}{4}\) and a cream biscuit be \(1 - p = \frac{3}{4}\). The probability of having 5 chocolate and 5 cream biscuits in a box of 10 is given by:

\(P(\text{equal}) = \left(0.25\right)^5 \times \left(0.75\right)^5 \times \binom{10}{5} = 0.0584\)

(ii) The probability that exactly 1 out of 8 boxes contains equal numbers of cream and chocolate biscuits is:

\((0.0584)^1 \times (0.9416)^7 \times \binom{8}{1} = 0.307\)

(iii) For a large box of 120 biscuits, use the normal approximation. Let \(X\) be the number of chocolate biscuits. Then \(\mu = 120 \times 0.25 = 30\) and \(\sigma^2 = 30 \times 0.75 = 22.5\).

Using continuity correction, find \(P(X < 35)\):

\(P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{22.5}}\right) = \Phi(0.949)\)

From standard normal tables, \(\Phi(0.949) = 0.829\).

Let the random variable \(X\) represent the number of plants producing pink flowers in a box of 100. \(X\) follows a binomial distribution with parameters \(n = 100\) and \(p = 0.3\).

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

Calculate the mean: \(\mu = np = 100 \times 0.3 = 30\).

Calculate the variance: \(\sigma^2 = np(1-p) = 100 \times 0.3 \times 0.7 = 21\).

Thus, \(X \sim N(30, 21)\).

We need to find \(P(X < 35)\). Using a continuity correction, this becomes \(P(X < 34.5)\).

Standardize the variable: \(Z = \frac{34.5 - 30}{\sqrt{21}} = \frac{4.5}{\sqrt{21}} \approx 0.9820\).

Using the standard normal distribution table, \(\Phi(0.9820) \approx 0.837\).

Therefore, the probability that a box contains fewer than 35 plants which produce pink flowers is \(0.837\).

Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.9\).

To approximate, use a normal distribution: \(X \sim N(\mu, \sigma^2)\).

Calculate the mean: \(\mu = np = 80 \times 0.9 = 72\).

Calculate the variance: \(\sigma^2 = npq = 80 \times 0.9 \times 0.1 = 7.2\).

Use continuity correction for \(P(X > 69)\):

\(P(X > 69) = P(Z > \frac{69.5 - 72}{\sqrt{7.2}})\).

Calculate the standard score:

\(Z = \frac{69.5 - 72}{\sqrt{7.2}} = -0.9317\).

Find the probability using the standard normal distribution:

\(P(Z > -0.9317) = \Phi(0.9317) \approx 0.824\).

For part (c), justify the approximation:

Both \(np = 72\) and \(nq = 8\) are greater than 5, so the normal approximation is valid.

Let the random variable \(X\) represent the number of residents who rate the service as 'good'. \(X\) follows a binomial distribution with parameters \(n = 125\) and \(p = 0.52\).

To approximate this binomial distribution, we use a normal distribution with mean \(\mu = np = 0.52 \times 125 = 65\) and variance \(\sigma^2 = np(1-p) = 0.52 \times 0.48 \times 125 = 31.2\).

Thus, \(X \sim N(65, 31.2)\).

We need to find \(P(X > 72)\). Using a continuity correction, this is approximated by \(P(X > 72.5)\).

Standardize the variable: \(Z = \frac{X - \mu}{\sigma} = \frac{72.5 - 65}{\sqrt{31.2}} \approx 1.343\).

Thus, \(P(X > 72.5) = P(Z > 1.343) = 1 - P(Z \leq 1.343)\).

Using standard normal distribution tables, \(P(Z \leq 1.343) \approx 0.9104\).

Therefore, \(P(Z > 1.343) = 1 - 0.9104 = 0.0896\).

Hence, the probability that more than 72 residents rate the service as 'good' is approximately between 0.0896 and 0.0897.