Alethia models the length of time, in minutes, by which her train is late on any day by the random variable X with probability density function given by f(x) = { (3/8000)(x-20)² 0 ≤ x ≤ 20, { 0 otherwise. }
- (a)Find the probability that the train is more than 10 minutes late on each of two randomly chosen days.[4]
- (b)Find E(X).[4]
- (c)The median of X is denoted by m. Show that m satisfies the equation (m – 20)³ = −4000, and hence find m correct to 3 significant figures.[4]
- (d)State one way in which Alethia's model may be unrealistic.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [M1] Attempt integration of f(x), ignore limits.
- [M1] Substitute correct limits 10 to 20 or 1 - ... limits 0 to 10 in their integral
- [A1] 1/8 or 0.125 SC Unsupported answer of 1/8 scores B1 only
- [B1 FT] (1/8)² or 0.0156 (3 sf) FT their P(X>10) dependent on first M1 gained
- [M1] Attempt integration of xf(x). Ignore limits.
- [A1] (3/8000)(x⁴/4 - 40x³/3 + 400x²/2) [20,0] Correct integral (by expanding or by parts)
- [M1] Subst correct limits in their (4th degree) integral
- [A1] 5
- [M1] Attempt to integrate f(x) and equate to 0.5. Ignore limits.
- [M1] (3/8000) × (x-20)³/3 [m,0] = 0.5 or (3/8000) × (x³/3 - 40x²/2 + 400x) [m,0] = 0.5 Attempt integral and substitute limits 0 and m or m and 20 and = 0.5
- [A1] (m-20)³ = -4000 AG. Found convincingly.
- [B1] m = 4.13 (3 sf)
- [B1] Doesn't allow for trains > 20 mins late or Doesn't allow for trains being early or any relevant comment e.g. trains on Sun may be different to trains on Mon
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