A-LevelMathematicsProbability and statistics9709/62 · May/June 2020 · Q412 Marks
The score on one spin of a 5-sided spinner is denoted by the random variable X with probability distribution as shown in the table. x | 0 | 1 | 2 | 3 | 4 P(X = x) | 0.1 | 0.2 | 0.4 | 0.2 | 0.1 The spinner is spun 200 times. The score on each spin is noted and the mean, X, of the 200 scores is found.
- (a)Show that Var(X) = 1.2.[2]
- (b)Given that P(X > a) = 0.1, find the value of a.[4]
- (c)Explain whether it was necessary to use the Central Limit theorem in your answer to part (b).[1]
- (d)Johann has another, similar, spinner. He suspects that it is biased so that the mean score is less than 2. He spins his spinner 200 times and finds that the mean of the 200 scores is 1.86. Given that the variance of the score on one spin of this spinner is also 1.2, test Johann's suspicion at the 5% significance level.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [B1] E(X) = 2
- [B1] 0.2 × 1 + 0.4 × 2^2 + 0.2 × 3^2 + 0.1 × 4^2 – 2^2 (= 1.2) AG
- [M1] (a-2) / √(1.2/200)
- [M1] φ⁻¹(0.9)
- [M1] a = 2 + √(1.2/200) × 1.282
- [A1] 2.10 (3 sf)
- [B1] Yes, because X is not normally distributed.
- [B1] H0: pop mean = 2; H1: pop mean < 2
- [M1] (1.86-2) / √(1.2/200)
- [A1] -1.807
- [M1] comp z = 1.645
- [A1] There is evidence that the spinner is biased so that mean is less than 2
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