IB Mathematics SL 4.9 normal distribution AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Part (a)
Shade the region representing \( P(X > 105) \).
Draw a vertical line at \( X = 105 \) (to the right of the mean 100). Shade the area to the right of this line under the normal curve.
Answer: Vertical line at 105 with shading to the right, as shown below.
Part (b)
Find \( d \) such that \( P(X < d) = P(X > 105) \).
Since \( X \sim N(100, 5^2) \), the distribution is symmetric about the mean 100. \( X = 105 \) is 5 cm (one standard deviation) above the mean, so \( P(X > 105) \) is the right tail. By symmetry, \( P(X < d) = P(X > 105) \) implies \( d \) is one standard deviation below the mean: \( d = 100 – 5 = 95 \).
Answer: \( d = 95 \).
Part (c)
Given \( P(X > 105) = 0.16 \), find \( P(d < X < 105) \).
From part (b), \( d = 95 \). By symmetry, \( P(X < 95) = P(X > 105) = 0.16 \).
Use the complement: \( P(95 < X < 105) = 1 – P(X < 95) – P(X > 105) = 1 – 0.16 – 0.16 = 0.68 \).
Answer: \( 0.68 \).