(a) [3 marks]

EITHER
For \( X \sim N(10, 2^2) \), compute 1.5 standard deviations above the mean: \( \mu + 1.5\sigma = 10 + 1.5 \times 2 = 13 \) (M1).
Standardize: \( Z = \frac{13 – 10}{2} = 1.5 \), so \( P(X > 13) = P(Z > 1.5) \) (A1).
Using standard normal tables, \( P(Z < 1.5) \approx 0.9332 \), so \( P(Z > 1.5) = 1 – 0.9332 = 0.0668 \) (A1).
OR
Recognize \( X > 13 \) corresponds to \( Z > 1.5 \), then use \( P(Z > 1.5) = 1 – P(Z < 1.5) \approx 0.0668 \) (M1A1A1).
Answer: \( 0.0668 \) (A1).

(b) [2 marks]

EITHER
Given \( P(X > 10 + 2k) = 0.1 \), standardize: \( Z = \frac{(10 + 2k) – 10}{2} = k \), so \( P(Z > k) = 0.1 \) (M1).
Then \( P(Z < k) = 0.9 \). From standard normal tables, \( P(Z < 1.28) \approx 0.8997 \), \( P(Z < 1.29) \approx 0.9015 \), so \( k \approx 1.28 \) (A1).
OR
Use inverse normal: for \( P(Z < k) = 0.9 \), find \( k \approx 1.2816 \), so \( k = 1.28 \) (to 2 decimal places) (M1A1).
Answer: \( k = 1.28 \) (A1).