(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, \(P(Z < -0.9157) = 1 - 0.8201 = 0.180\).