(a) [5 marks]

(i) For \( X \sim N(200, 25^2) \), find \( P(X > 225) \).
Standardize: \( Z = \frac{225 – 200}{25} = 1 \), so \( P(X > 225) = P(Z > 1) \) (M1).
Using standard normal tables, \( P(Z < 1) \approx 0.8413 \), so \( P(Z > 1) = 1 – 0.8413 = 0.1587 \) (A1).
Expected number: \( 450 \times 0.1587 \approx 71.4 \) (A1).
(ii) Given \( P(X < m) = 0.7 \), find \( m \).
For \( P(Z < z) = 0.7 \), standard normal tables give \( z \approx 0.5244 \) (M1).
Then, \( m = \mu + z \cdot \sigma = 200 + 0.5244 \times 25 \approx 213 \) grams (A1).

(b) [6 marks]

For \( X \sim N(\mu, \sigma^2) \), given \( P(X > 270) = 0.08 \) and \( P(X < 250) = 0.15 \).
Standardize: \( \frac{270 – \mu}{\sigma} = z_1 \), where \( P(Z > z_1) = 0.08 \), so \( P(Z < z_1) = 0.92 \), \( z_1 \approx 1.405 \) (M1A1).
Also, \( \frac{250 – \mu}{\sigma} = z_2 \), where \( P(Z < z_2) = 0.15 \), so \( z_2 \approx -1.036 \) (A1).
Solve: \( 270 – \mu = 1.405\sigma \), \( 250 – \mu = -1.036\sigma \) (M1).
Subtract equations: \( 270 – 250 = 1.405\sigma – (-1.036\sigma) \), so \( 20 = 2.441\sigma \), \( \sigma \approx 8.19 \).
Substitute: \( 270 – \mu = 1.405 \times 8.19 \), so \( \mu \approx 258 \) (A1A1).

(c) [3 marks]

For \( X \sim N(80, 4^2) \), find \( P(X > 82) \).
Standardize: \( Z = \frac{82 – 80}{4} = 0.5 \), so \( P(X > 82) = P(Z > 0.5) \approx 1 – 0.6915 = 0.3085 \) (A1).
Use binomial: \( X \sim B(5, 0.3085) \), find \( P(X = 3) = \left( \begin{array}{c} 5 \\ 3 \end{array} \right) (0.3085)^3 (0.6915)^2 \approx 0.140 \) (M1A1).