The lengths of the leaves of a particular type of tree are modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves from this type of tree and finds that 42 are less than 4 cm long and 100 are more than 10 cm long.
- (a)Find estimates for the mean and standard deviation of the lengths of leaves from this type of tree.[5]
- (b)The lengths, in cm, of the leaves of a different type of tree have the distribution Ν(μ, σ²). The scientist takes a random sample of 800 leaves from this type of tree. Find how many of these leaves the scientist would expect to have lengths, in cm, between µ – 2σ and μ + 2σ.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [5] B1 1.378 < z₁≤ 1.379 or –1.379 < z₁ < -1.378
- B1 0.841 < Z2 ≤ 0.842 or -0.842 < z2 ≤ −0.841
- M1 Use of standardisation formula once with μ, σ, a z-value and 4 or 10, allow continuity correction, not σ² or √σ
- M1 Use either the elimination method or the substitution method to solve two equations in µ and σ.
- A1 2.70 < σ≤2.71 7.72 <μ≤7.73
- [4] M1 Identifying 2 and -2 as the appropriate z-values
- B1 Calculating the appropriate area from stated phis of z-values which must be ± the same number
- A1 Accept AWRT 0.954
- B1 FT their 4SF (or better) probability, final answer must be positive integer
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