In a certain country the probability that a child owns a bicycle is 0.65.
- (i)A random sample of 15 children from this country is chosen. Find the probability that more than 12 own a bicycle.[3]
- (ii)A random sample of 250 children from this country is chosen. Use a suitable approximation to find the probability that fewer than 179 own a bicycle.[4]
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- [3] (P >12) = P(13, 14, 15). Binomial term of form 15Cxpx(1 −p)15-x 0 < p < 1 any p, x ≠ 15,0 M1 = 15C13(0.65)13(0.35)2 + 15C14(0.65)14(0.35)¹ + (0.65)15. Correct unsimplified answer A1 = 0.0617. SC if use np and npq with justification give (12.5 – 9.75)/√3.41 M1 1-F(1.489) A1 0.0681 A0 A1
- [4] mean = 250 × 0.65 = 162.5 variance = 250 × 0.65 × 0.35 = 56.875. Correct unsimplified np and npq B1 P(< 179) = P(z < (178.5-162.5) / √56.875) = P(z < 2.122). Substituting their µ and σ (condone σ²) into the Standardisation Formula with a numerical value for '178.5'. Continuity correct not required for this M1. Condone + standardisation formula M1 Using continuity correction 178.5 or 179.5 M1 = 0.983. Correct final answer A1
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