Question 2.8: For the triangular life distribution given in Example 2.3, c......
For the triangular life distribution given in Example 2.3, calculate the E[T], V[T], and standard deviation
Step-by-Step
We have
f(t)= \begin{cases}\frac{t}{5,000}, & \text { for } 0 \leq t \leq 100 \\ 0, & \text { otherwise. }\end{cases}Now
E[T]=\int_0^{\infty} t f(t) d t \\=\int_0^{100} t \frac{t}{5,000} d t=\left.\frac{1}{5,000} \frac{t^3}{3}\right|_0 ^{100}=\frac{1}{5,000} \frac{100^3}{3}\\=\frac{2}{3} \cdot 100=66.67 hours
and
E\left[T^2\right]=\int_0^{\infty} t^2 f(t) d t \\=\int_0^{100} t^2 \frac{t}{5,000} d t=\left.\frac{1}{5,000} \frac{t^4}{4}\right|_0 ^{100}=\frac{100^4}{20,000}=5,000so,
V[T]=E\left[T^2\right]-(E[T])^2 \\=5,000-\left(\frac{200}{3}\right)^2=\frac{5,000}{9}=555.55 .The standard deviation, σ, is 23.57, and the coefficient of variation is 23.57/66.67 = 0.354.
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