 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/14/b/iii/solution

Since

q

tq (t)

is an odd function and its integral over the symmetric interval vanishes. Thus

E [T] = 0

. Its variance is

E [T^{2}] = 2 \int_{0}^{1} t^{2} (1 - t) d t = 2 (\frac{1}{3} - \frac{1}{4}) = \frac{1}{6} .

Consequently

E [T] = 0, SD (T) = \frac{1}{6} minute .

The standard deviation is approximately

0.40825

24.49

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