The times taken, in minutes, to complete a particular task by employees at a large company are normally distributed with mean 32.2 and standard deviation 9.6.
- (a)Find the probability that a randomly chosen employee takes more than 28.6 minutes to complete the task.[3]
- (b)20% of employees take longer than t minutes to complete the task. Find the value of t.[3]
- (c)Find the probability that the time taken to complete the task by a randomly chosen employee differs from the mean by less than 15.0 minutes.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [M1] 28.6, 32.2 and 9.6 substituted appropriately in ± Standardisation formula once, allow continuity correction of ±0.05, no σ², √σ.
- [M1] Appropriate numerical area, from final process, must be probability, expect > 0.5.
- [A1] 0.646 AWRT
- [B1] z = ±0.842
- [M1] Substituting 32.2 and 9.6 into ± standardisation formula, no continuity correction, allow σ², √σ, must be equated to a z-value.
- [A1] 40.28 < t < 40.3 WWW
- [M1] Identifying at least one of 15/9.6 and -15/9.6 as the appropriate z-values or substituting their (32.2 ±15) into ± Standardisation formula once, no continuity correction, σ² nor √σ. Condone ±1.563 for M1.
- [A1] p = 0.941 AWRT SOI
- [M1] Appropriate area 2Φ – 1 oe, (eg 1 – 2 × 0.0591, 2 × (0.9409 – 0.5) or 0.9409 – 0.0591), from final process, must be probability > 0.5.
- [A1] 0.882
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →