A mathematical puzzle is given to a large number of students. The times taken to complete the puzzle are normally distributed with mean 14.6 minutes and standard deviation 5.2 minutes.
- (a)In a random sample of 250 of the students, how many would you expect to have taken more than 20 minutes to complete the puzzle?[4]
- (b)All the students are given a second puzzle to complete. Their times, in minutes, are normally distributed with mean u and standard deviation σ. It is found that 20% of the students have times less than 14.5 minutes and 67% of the students have times greater than 18.5 minutes. Find the value of u and the value of σ.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [M1] Use of ± standardisation formula with 20, 14.6 and 5.2 not σ², not √σ, no continuity correction.
- [M1] Calculating the appropriate probability area (leading to their final answer).
- [A1] 0.1496, 0.149 < p < 0.15[0] . Only dependent on the 2nd M mark so M0M1A1 possible. SC B1 for 0.149 < p < 0.15[0] if M0M0A0 awarded.
- [B1 FT] Strict FT their at least 4-figure probability seen anywhere (give BOD if they go on to use 0.150). Final answer must be positive integer, no approximation or rounding stated.
- [B1] -0.843 < z₁ < -0.841 or 0.841 < z₁ < 0.843.
- [B1] -0.441 < z₂ < -0.439 or 0.439 < z₂ < 0.441.
- [M1] Use of the standardisation formula once with μ, σ and a z-value (not 0.20, 0.80, 0.67, 0.23, 0.5793, 0.7881, 0.7486, 0.591 or 1-z i.e. 0.158 etc.). Condone continuity correction ±0.05, not σ²,√σ.
- [M1] Solve using the elimination method, substitution method or other appropriate approach to obtain values for both μ and σ.
- [A1] AWRT 22.9, 9.95 .
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