Science:Math Exam Resources/Courses/MATH103/April 2016/Question 09 (b)/Solution 1

By the definition of variance for a continuous random variable,

$\sigma ^{2}=\int _{-\infty }^{\infty }(x-\mu )^{2}p(x)\,dx.$

Now since $x\in [0,100],$

$\sigma ^{2}=\color {blue}\int _{0}^{100}(x-\mu )^{2}p(x)\,dx.$