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:

\(P(Z < -1.039) = \Phi(-1.039) = 1 - 0.8506 = 0.149\)