(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, \(P(Z < -1.186) = 1 - P(Z < 1.186) = 1 - 0.8822 = 0.118\).