Cambridge International AS & A Level Mathematics: Worked Solutions Manual
(
19.5 − 24 √16.8 = 1 − Φ(1.098) = 0.136
P(X < 19) ≈ P z <
) Provided np and nq are both greater than 5, the distribution B(n, p) can be approximated by N(np, npq) . Remember to make continuity corrections when approximating a discrete distribution by a continuous distribution.
0.136 > 0.10, so H0 is accepted. There is no evidence to suggest that the percentage of sheep deficient in a particular mineral has decreased.
(c + 0.5) − 24 < −1.282 √16.8 c < 18.245... The critical value is 18.
FT
b With a continuity correction at c + 0.5, we need to find the largest integer c for which P(X < c + 0.5) < 0.10.
4 a For X ∼ B(9, 0.5), we need to find the largest integer n for which P(X > n) > 0.05.
A
P(X > 8) = 9C 8 × 0.51 × 0.58 + 0.59 = 0.0195 < 0.05
P(X > 7) = 9C 7 × 0.52 × 0.57 + 9C 8 × 0.51 × 0.58 + 0.59 = 0.0898 > 0.05
R
Critical region is X > 7.
H0: population proportion of tails = 0.5 and H1: population proportion of tails > 0.5. 0.0898 > 0.05, so H0 is accepted.
There is insufficient evidence to support the claim that the coin has bias towards tails.
D
b Let the number of heads obtained be Y , then Y ∼ B(9, 0.5). H0: population proportion of heads = 0.5 and H1: population proportion of heads < 0.5.
Critical region is Y < 2.
P(Y < 2) = 0.59 + 9C 1 × 0.51 × 0.58 + 9C 2 × 0.52 × 0.57 = 0.0898 0.0898 > 0.05, so H0 is accepted. There is insufficient evidence to support the claim that the coin has bias towards tails.
2
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