The time, X hours, for which students use a games machine in any given day has a normal distribution with mean 3.24 hours and standard deviation 0.96 hours. On how many days of the year (365 days) would you expect a randomly chosen student to use a games machine for less than 4 hours?
- (a)(i)The time, X hours, for which students use a games machine in any given day has a normal distribution with mean 3.24 hours and standard deviation 0.96 hours. On how many days of the year (365 days) would you expect a randomly chosen student to use a games machine for less than 4 hours?[3]
- (a)(ii)Find the value of k such that P(X > k) = 0.2.[3]
- (a)(iii)Find the probability that the number of hours for which a randomly chosen student uses a games machine in a day is within 1.5 standard deviations of the mean.[3]
- (b)The variable Y is normally distributed with mean μ and standard deviation σ, where 4σ = 3μ and μ ≠ 0. Find the probability that a randomly chosen value of Y is positive.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [M1] ±Standardisation attempt in terms of one variable, no cc, no sq rt, no square
- [A1] 0.7855 < p < 0.7858 or p = 0.786 Cao (implies M1A1 awarded), may be seen used in calculation
- [B1ft] Their probability × 365 provided 4sf probability seen. FT answer rounded or truncated to nearest integer. No approximation notation used.
- [B1] (z=) ± 0.842 seen
- [M1] z = ± (k - 3.24) / 0.96 allow cc, sq rt or square equated to a z-value (0.7881, 0.2119, 0.158, 0.8, 0.2 etc. are not acceptable)
- [A1] Correct final answer, www
- [M1] Φ(z = 1.5) or Φ(z = −1.5) seen used or p = 0.9332 seen
- [M1] Correct final area expression using their probabilities
- [A1] Correct final answer
- [M1] ±Standardisation attempt in terms of one variable no sq rt or square, condone ±0.5 as cc
- [A1] Correct unsimplified standardisation, no variables
- [A1] Correct final answer
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