IBDP Maths SL 4.12 Standardization of normal variables AA HL Paper 1- Exam Style Questions- New Syllabus

(a)
Since \( X \) and \( Y \) have the same variance \( a^2 \), their distributions are identical in shape, just shifted by a constant. \( Y \) has mean 19, \( X \) has mean 7, so \( Y = X + 12 \). Given \( P(Y > 22) \), we note \( Y > 22 \) corresponds to \( X + 12 > 22 \), i.e. \( X > 10 \). So \( b = 10 \).
\(\boxed{10}\)

(b)
For a normal distribution, about 68% of the data lies within one standard deviation of the mean[cite: 1383]. Here, \( 7 \pm a \) is exactly one standard deviation from the mean of \( X \). So \( P(7 – a < X < 7 + a) \approx 0.68 \).
\(\boxed{0.68}\)

(c)
Given \( a = 3 \), so \( Y \sim N(19, 3^2) \). The value \( 22 \) is exactly one standard deviation above the mean \( 19 \) (since \( 19 + 3 = 22 \)). Using the empirical rule: \( P(Y < 22) \) includes the area below the mean (0.50) plus the area between the mean and one standard deviation above (approximately 0.34). Thus, \( P(Y < 22) \approx 0.50 + 0.34 = 0.84 \).
\(\boxed{0.84}\)