In a cycling event the times taken to complete a course are modelled by a normal distribution with mean 62.3 minutes and standard deviation 8.4 minutes.
- (a)Find the probability that a randomly chosen cyclist has a time less than 74 minutes.[2]
- (b)Find the probability that 4 randomly chosen cyclists all have times between 50 and 74 minutes.[4]
- (c)In a different cycling event, the times can also be modelled by a normal distribution. 23% of the cyclists have times less than 36 minutes and 10% of the cyclists have times greater than 54 minutes. Find estimates for the mean and standard deviation of this distribution.[5]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.
- [M1] Use of ± standardisation formula with 74, 62.3 and 8.4 substituted appropriately, not 8.4², not √8.4, no continuity correction.
- [A1] 0.918<p ≤ 0.9185 .
- [M1] Use of standardisation formula with both 74 (may be seen in 6(a) if their value seen) & 50, 62.3 and 8.4 substituted appropriately. Condone use of 8.4²,√8.4 and continuity correction ±0.5 (73.5 or 74.5 and 49.5 or 50.5).
- [M1] Calculating the appropriate probability area from stated Φ of z-values (leading to their final answer > 0.5) but not symmetrical values.
- [A1] 0.8465 < p < 0.8475 . SC B1 for 0.8465 < p <0.8475 if M0A0 awarded.
- [B1 FT] Accept 0.513 < p < 0.514 . FT (their 4-figure p)⁴, 0 <p< 1.
- [B1] -0.740 <z₁ < -0.738 or 0.738 < z1 < 0.740.
- [B1] Z2 = ±1.282 (critical value).
- [M1] Use of the standardisation formula once with μ, σ and a z-value (not 0.23, 0.77, 0.90, 0.10, ±0.261, ±0.282...). Condone continuity correction ±0.5, not σ²,√σ.
- [M1] Solve using the elimination method, substitution method or other appropriate approach to obtain values for both u and σ.
- [A1] 42.58 < μ ≤ 42.6, 8.90 < σ < 8.91 .
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