When a child completes an online exercise called a Mathlit, they might be awarded a medal. The publishers claim that the probability that a randomly chosen child who completes a Mathlit will be awarded a medal is ⅓. Asha wishes to test this claim. She decides that if she is awarded no medals while completing 10 Mathlits, she will conclude that the true probability is less than ⅓.
- (a)Use a binomial distribution to find the probability of a Type I error.[2]
- (b)The true probability of being awarded a medal is denoted by p. Given that the probability of a Type II error is 0.8926, find the value of p.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [1] (1-1/3)¹⁰
- [1] = 0.0173 (3 sf) No working scores SC B1.
- [1] 1 - (1-p)¹⁰ = 0.8926 Accept 1- q¹⁰ = 0.8926. Equation must be in p or in q but not both.
- [1] 1 − p = 0.1074⁰.¹ [= 0.800] For valid attempt to solve their (binomial) equation in p¹⁰ or q¹⁰.
- [1] p = 0.200 (3 sf) or 0.2
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