Most plants of a certain type have three leaves. However, it is known that, on average, 1 in 10000 of these plants have four leaves, and plants with four leaves are called ‘lucky'. The number of lucky plants in a random sample of 25 000 plants is denoted by X.
- (a)State, with a justification, an approximating distribution for X, giving the values of any parameters.[2]
- (b)Find P(X ≤ 3).[2]
- (c)Given that P(X = k) = 2P(X = k + 1), find k.[2]
- (d)The number of lucky plants in a random sample of n plants, where n is large, is denoted by Y. Given that P(Y ≥ 1) = 0.963, correct to 3 significant figures, use a suitable approximating distribution to find the value of n.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [B1] Po(2.5) Accept Poisson with mean = 2.5 not just np = 2.5
- [B1] n = 25 000 > 50 and np (or λ) = 2.5 which is < 5 or n = 25 000 > 50 and p = 0.0001 < 0.1
- [M1] e⁻²·⁵(1 + 2.5 + 2.5²/2 + 2.5³/3!) Any λ, accept one end error. FT binomial from part (a) scores M1 only for equivalent binomial expressions FT normal from part (a) must use correct continuity correction and tables scores M1 only for complete method
- [A1] 0.758 (3 sf) Unsupported answer of 0.758 scores B1 instead of M1A1
- [M1] e⁻²·⁵ × 2.5ᵏ/k! = 2e⁻²·⁵ × 2.5ᵏ⁺¹/(k+1)! Any λ FT binomial from (a) scores M1 only for equivalent binomial expression FT from (a) normal for equivalent expressions continuity correction must be included
- [A1] k = 4 No errors seen SC k = 4 unsupported scores B1 only, but see full Poisson expressions for P(4) and P(5) and 0.134 scores M1A1
- [M1] 1 - e⁻ˡ = 0.963 Accept their attempt at λ
- [M1] λ = -ln0.037 (= 3.2968 or 3.30 or 3.3) Correct use of lns
- [A1] n = 33 000 (3 sf) Allow n = 32 950 to 33 050 (must be an integer) SC use of binomial leading to 32 967 scores B1 for (0.9999)ⁿ = 0.037 B1 for 33 000 to 3sf (32 967)
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