The distance that car tyres of a certain make can travel before they need to be replaced has a normal distribution. A survey of a large number of these tyres found that the probability of this distance being more than 36 800 km is 0.0082 and the probability of this distance being more than 31 000 km is 0.6915. Find the mean and standard deviation of the distribution.
- (a)The distance that car tyres of a certain make can travel before they need to be replaced has a normal distribution. A survey of a large number of these tyres found that the probability of this distance being more than 36 800 km is 0.0082 and the probability of this distance being more than 31 000 km is 0.6915. Find the mean and standard deviation of the distribution.[5]
- (b)The random variable X has the distribution N(μ, σ²), where 3σ = 4μ and μ ≠ 0. Find P(X < 3μ).[3]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [B1] ± 2.4 seen accept 2.396
- [B1] ±0.5 seen
- [M1] Either standardisation eqn with z value, not 0.5082, 0.7565, 0.0082, 0.6915, 0.3085, 0.6209, 0.0032 or any other probability
- [M1] Sensible attempt to eliminate μ or σ by substitution or subtraction from their 2 equations (z-value not required), need at least 1 value stated
- [A1] Both correct answers
- [M1] Standardise, in terms of one variable, accept σ² or √σ
- [M1] 6/4 or 6/4σ seen
- [A1] Correct final answer
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