 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/18/a/iv/solution

Independence makes the variances of the attendance indicators additive. The binomial distribution therefore gives

Var (N) = n p (1 - p) = 24 (3/4) (1/4) = 9/2

σ_{N} = \frac{3}{2} \approx 2.12132 lectures .

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