The residents of Mahjing were asked to classify their local bus service: * 25% of residents classified their service as good. * 60% of residents classified their service as satisfactory. * 15% of residents classified their service as poor.
- (a)A random sample of 110 residents of Mahjing is chosen. Use a suitable approximation to find the probability that fewer than 22 residents classified their bus service as good.[5]
- (b)For a random sample of 10 residents of Mahjing, find the probability that fewer than 8 classified their bus service as good or satisfactory.[3]
- (c)Three residents of Mahjing are selected at random. Find the probability that one resident classified the bus service as good, one as satisfactory and one as poor.[2]
Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [B1] 27.5 and 20.625 (CAO) seen, allow unsimplified. May be in standardisation formula (4.541475... to at least 4sf or √165/8 or √330/4 implies correct variance). Penalise incorrect identification, condone no identification.
- [M1] Substituting their 27.5 and their 20.625 into the standardising formula (any number for 21.5), not σ², not √σ.
- [M1] Using continuity correction 21.5 or 22.5 in their standardisation formula.
- [M1] Appropriate probability area, from final process, must be a probability. May be implied by a sketch of the required probability area. Expect final answer < 0.5.
- [A1] 0.0932 ≤ p < 0.09325 If either M1 M1 not awarded for standardisation and/or M1 not awarded for finding probability area, SC B1 0.0932 ≤ p < 0.09325 WWW.
- [M1] One term ¹⁰Cx (p)ˣ (1-p)¹⁰⁻ˣ, 0 < p < 1, x ≠ 0 or 10.
- [A1] Correct unsimplified expression. Condone omission of last bracket only.
- [B1] 0.1795 < p ≤ 0.180
- [(M1)] One term ¹⁰Cx (p)ˣ (1-p)¹⁰⁻ˣ, 0 < p < 1, x ≠ 0 or 10.
- [(A1)] Correct unsimplified expression.
- [(B1)] 0.1795 < p < 0.180
- [M1] 0.25 × 0.6 × 0.15 × k, k an integer > 1.
- [A1]
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