 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/18/c/solution

The three preceding probabilities satisfy

\frac{28}{45} + \frac{16}{45} + \frac{1}{45} = 1.

μ = \frac{16 + 2}{45} = \frac{2}{5}, E [X^{2}] = \frac{16 + 4}{45} = \frac{4}{9} .

Subtracting the squared mean gives

μ = \frac{2}{5}, σ^{2} = \frac{4}{9} - \frac{4}{25} = \frac{64}{225} .

The value

μ = 2 (2/10)

 New to topics? Read the docs here!