To find the probability that a randomly chosen salmon is 34 cm long, we need to consider the range from 33.5 cm to 34.5 cm due to rounding to the nearest centimetre.

We standardize the values using the formula for the standard normal distribution:

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

For 34.5 cm: \(Z = \frac{34.5 - 32.9}{2.4} = 0.667\)

For 33.5 cm: \(Z = \frac{33.5 - 32.9}{2.4} = 0.25\)

We find the probabilities using the standard normal distribution table:

\(\Phi(0.667) = 0.7477\)

\(\Phi(0.25) = 0.5987\)

The probability that a salmon is 34 cm long is the difference between these probabilities:

\(0.7477 - 0.5987 = 0.149\)

To find the probability that \(X\) lies between 25 and 30, we standardize the values using the formula for the standard normal variable \(Z\):

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

where \(\mu = 28.3\) and \(\sigma = \sqrt{4.5}\).

First, calculate \(z_1\) for \(x = 30\):

\(z_1 = \frac{30 - 28.3}{\sqrt{4.5}} = 0.8014\)

Next, calculate \(z_2\) for \(x = 25\):

\(z_2 = \frac{25 - 28.3}{\sqrt{4.5}} = -1.5556\)

The probability that \(X\) lies between 25 and 30 is given by:

\(\Phi(z_1) - \Phi(z_2) = \Phi(0.8014) - (1 - \Phi(-1.5556))\)

Using standard normal distribution tables or a calculator, we find:

\(\Phi(0.8014) = 0.7884\)

\(\Phi(-1.5556) = 0.0599\)

Thus, the probability is:

\(0.7884 + 0.9401 - 1 = 0.729\)

We are given that the daily minimum temperature follows a normal distribution \(N(8, 24)\), where the mean \(\mu = 8\) and the variance \(\sigma^2 = 24\). The standard deviation \(\sigma = \sqrt{24}\).

To find the probability that the temperature is between 7°C and 12°C, we standardize these values to find the corresponding \(z\)-scores.

For 12°C:

\(z_1 = \frac{12 - 8}{\sqrt{24}} = \frac{4}{\sqrt{24}} = 0.816\)

Using the standard normal distribution table, \(\Phi(0.816) = 0.7926\).

For 7°C:

\(z_2 = \frac{7 - 8}{\sqrt{24}} = \frac{-1}{\sqrt{24}} = -0.204\)

Using the standard normal distribution table, \(\Phi(-0.204) = 1 - 0.5808 = 0.4192\).

The probability that the temperature is between 7°C and 12°C is:

\(P(7 < X < 12) = \Phi(0.816) - \Phi(-0.204) = 0.7926 - 0.4192 = 0.373\)

The weights of female 18-year-old students can be modeled using a normal distribution. This is a common assumption for biological measurements, which tend to be symmetrically distributed around a mean.

For the normal distribution, we need to specify the mean and the variance. According to the mark scheme, a suitable mean is 60 kg, and a suitable variance is 90 kg².

These values are within the sensible range provided: mean between 40 kg and 80 kg, and variance between 16 kg² and 225 kg².

(i) To find the number of students expected to take longer than 20 minutes, we first standardize the variable:

\(P(X > 20) = P\left(Z > \frac{20 - 26.4}{3.7}\right) = P(Z > -1.730)\)

Using the standard normal distribution table, \(P(Z > -1.730) = 0.9582\).

For a sample of 350 students, the expected number is \(350 \times 0.9582 \approx 335 \text{ or } 336\).

(ii) A ‘very slow’ student is more than 1.645 standard deviations above the mean:

\(P(\text{very slow}) = P(Z > 1.645) = 0.05\).

We use the binomial distribution for 8 students with \(p = 0.05\):

\(P(0, 1, 2) = (0.95)^8 + 8 \times (0.05)^1 \times (0.95)^7 + \binom{8}{2} \times (0.05)^2 \times (0.95)^6\)

\(= 0.6634 + 0.2793 + 0.0515 = 0.994\).