The lengths, in cm, of the leaves of a particular type are modelled by the distribution N(5.2, 1.5²).
- (a)Find the probability that a randomly chosen leaf of this type has length less than 6cm.[2]
- (b)The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than 3 cm long and 95 are more than 8 cm long. Find estimates for the mean and standard deviation of the lengths of leaves of this type.[5]
- (c)In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean?[4]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [M1] 6, 5.2, 1.5 substituted into z standardisation formula, condone 1.5^2, continuity correction ±0.5
- [A1] 0.703
- [B1] 1.328 < z1 < 1.329 or -1.329 < z1 < -1.328
- [B1] 0.877 < z2 < 0.878 or -0.878 < z2 < -0.877
- [M1] Use of the z standardisation formula once with μ, σ, a z-value (not 0.8179, 0.7910, 0.5367, 0.5753, 0.19, 0.092 etc.) and 3 or 8, condone continuity correction but not σ² or √σ
- [M1] Use either the elimination method or the substitution method to solve their two equations in μ and σ
- [A1] 2.26 < σ <= 2.27, 6.01 < μ <= 6.02
- [M1] Identify 1 and -1 as the appropriate z-values.
- [M1] Calculating the appropriate area from stated phis of z-values which must be ± the same number
- [A1] Accept AWRT 0.317
- [B1 FT] FT their 4 s.f. (or better) probability, final answer must be positive integer no approximation or rounding stated
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