(a) The random variable X has a normal distribution with mean μ and standard deviation σ. You are given that σ = 0.25μ and P(X < 6.8) = 0.75. (b) The lengths of metal rods have a normal distribution with mean 16cm and standard deviation 0.2 cm. Rods which are shorter than 15.75 cm or longer than 16.25 cm are not usable. Find the expected number of usable rods in a batch of 1000 rods.
- (a)(i)Find the value of μ.[4]
- (a)(ii)Find P(X < 4.7).[3]
- (b)The lengths of metal rods have a normal distribution with mean 16cm and standard deviation 0.2 cm. Rods which are shorter than 15.75 cm or longer than 16.25 cm are not usable. Find the expected number of usable rods in a batch of 1000 rods.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] rounding to ±0.674 or 0.675
- [M1] standardising, no cc, no sq rt, no sq, σ may still be present on RHS
- [M1] subst and sensible solving for μ must collect terms, no z-value needed can be 0.75 or 0.7734 need a value for μ
- [A1]
- [M1] ± standardising no cc, no sq rt, no sq unless penalised in (a)(i)
- [M1] correct side for their mean i.e. 1-Φ (final solution)
- [A1]
- [*M1] Standardising for 15.75 or 16.25 no cc no sq no sq rt unless penalised in (a)(i) or (a)(ii)
- [B1] 2Φ-1 OE for required prob, (final solution)
- [DM1] Mult their prob by 1000 dep on recognisable attempt to standardise
- [A1]
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