In a certain country, the heights of the adult population are normally distributed with mean 1.64 m and standard deviation 0.25 m.
- (a)Find the probability that an adult chosen at random from this country will have height greater than 1.93 m.[3]
- (b)In another country, the heights of the adult population are also normally distributed. 33% of the adult population have height less than 1.56 m. 25% of the adult population have height greater than 1.86 m. Find the mean and the standard deviation of this distribution.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [M1] Using ± standardisation formula with 1.93, 1.64 and 0.25 substituted, not σ², not √σ, no continuity correction.
- [M1] Appropriate area Φ resulting from a standardisation, from final process, must be probability. Note: the appropriate probability answer implies this M1.
- [A1] 0.123 < p < 0.12303 If M0 M0, SC B1 if no standardisation shown.
- [B1] -0.441 < z1 < -0.439 or 0.439 < z1 < 0.441 seen.
- [B1] z2 = 0.674 or z2 = - 0.674 seen, CAO, critical value.
- [M1] Use of the standardisation formula once with μ, σ equating to a z-value (not 0.33, 0.67. 0.25, 0.75, 0.6293, 0.5987, 0.7486, 0.7734, (1-0.44), (1 – 0.674)). Condone continuity correct ±0.005, not σ², √σ.
- [M1] Solve two equations in μ and σ using the elimination method, substitution method or other appropriate approach to obtain values for both μ and σ.
- [A1] AWRT μ=1.68, σ = 0.269. If one or both of the M marks have not been awarded, SC B1 for both correct.
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