Let the random variable representing the time spent be denoted by \(X\), where \(X \sim N(96, 18^2)\).

We need to find \(P(85 < X < 100)\).

Standardize the variable: \(Z = \frac{X - 96}{18}\).

Thus, \(P(85 < X < 100) = P\left(\frac{85 - 96}{18} < Z < \frac{100 - 96}{18}\right)\).

Calculate the standardized values: \(\frac{85 - 96}{18} = -0.6111\) and \(\frac{100 - 96}{18} = 0.2222\).

Therefore, \(P(-0.6111 < Z < 0.2222) = \Phi(0.2222) + \Phi(0.6111) - 1\).

Using standard normal distribution tables, \(\Phi(0.2222) = 0.5879\) and \(\Phi(0.6111) = 0.7294\).

Thus, \(P(-0.6111 < Z < 0.2222) = 0.5879 + 0.7294 - 1 = 0.3173\).

Rounding to three decimal places, the probability is \(0.317\).

Let the random variable \(X\) represent the time taken to swim 100 metres. \(X\) follows a normal distribution with mean \(\mu = 62\) and standard deviation \(\sigma = 5\).

We need to find \(P(56 < X < 66)\).

Standardize the variable using the formula:

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

For \(X = 56\):

\(Z = \frac{56 - 62}{5} = -1.2\)

For \(X = 66\):

\(Z = \frac{66 - 62}{5} = 0.8\)

Thus, \(P(56 < X < 66) = P(-1.2 < Z < 0.8)\).

Using the standard normal distribution table, find:

\(\Phi(0.8) = 0.7881\)

\(\Phi(1.2) = 0.8849\)

Therefore, \(P(-1.2 < Z < 0.8) = \Phi(0.8) + \Phi(1.2) - 1 = 0.7881 + 0.8849 - 1 = 0.673\).

(a) To find the probability that Pia takes longer than 11.3 minutes, we use the standard normal distribution. First, we standardize the variable:

\(P(X > 11.3) = P\left( Z > \frac{11.3 - 10.1}{1.3} \right) = P(Z > 0.9231)\)

Using the standard normal distribution table, \(P(Z > 0.9231) = 1 - P(Z < 0.9231) = 1 - 0.822\)

Thus, \(P(X > 11.3) = 0.178\).

(c) To find the number of days Pia takes between 8.9 and 11.3 minutes, we calculate:

\(P(8.9 < X < 11.3) = 1 - 2 \times P(X > 11.3)\)

\(= 2 \times (0.5 - 0.178) = 0.644\)

The expected number of days is \(90 \times 0.644 = 57.96\), which rounds to 57 days.

(a) To find the probability that Davin plays for more than 4.2 hours, we use the standard normal distribution. First, we standardize the variable:

\(P(X > 4.2) = P\left( Z > \frac{4.2 - 3.5}{0.9} \right) = P(Z > 0.7778)\)

Using the standard normal distribution table, \(P(Z > 0.7778) = 1 - 0.7818 = 0.218\).

(c) To find the number of days Davin plays between 2.8 and 4.2 hours, we calculate:

\(P(2.8 < X < 4.2) = 1 - 2 \times P(X > 4.2)\)

\(= 2 \times (1 - 0.7818) - 1 = 2 \times 0.2182 - 1 = 0.5636\)

The number of days is \(365 \times 0.5636 = 205.7\), so approximately 205 days.

The time, X, is normally distributed with mean \(\mu = 15.8\) hours and standard deviation \(\sigma = 4.2\) hours.

We need to find \(P(X < 21)\).

First, calculate the z-score for 21 hours:

\(z = \frac{21 - 15.8}{4.2} = 1.238\)

Using the standard normal distribution table, find \(\Phi(1.238)\), which gives the probability that a standard normal variable is less than 1.238.

\(\Phi(1.238) = 0.892\)

Thus, \(P(X < 21) = 0.892\).