The distance travelled by a commuter driver in a day is normally distributed with mean $360$ km and standard deviation $60$ km. Find the probability that the distance travelled in a day exceeds $370$ km, to 3 decimal places.

Model answer

0.434

Also accepted: 0.4338, 0.43

Explanation

Standardise:

z = \dfrac{370 - 360}{60} = 0.1667

. Then

P(X > 370) = P(Z > 0.1667) = 1 - \Phi(0.1667) = 1 - 0.5662 = 0.4338

, that is

0.434

. Because

370

is only about a sixth of a standard deviation above the mean, the probability stays close to

0.5

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q8

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