exercise:Edc8da013b

ABy Admin

Jan 15'24

Exercise

For a new light bulb, you are given:

i) [math]{ }_{t} q_{0}=\frac{t^{2}+t}{72}[/math] for [math]0 \leq t \leq 8[/math]

ii) [math]T_{0}[/math] is the random variable representing the future lifetime

Calculate [math]\operatorname{Var}\left[T_{0}\right][/math].

3.9
4.1
4.3
4.5
4.7

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ABy Admin

Jan 15'24

Answer: A

[math]E\left[T_{0}\right]=\int_{0}^{8}{ }_{t} p_{0} d t=\int_{0}^{8}\left(1-\frac{t^{2}+t}{72}\right) d t \rightarrow \frac{1}{72}\left[72 t-\frac{t^{3}}{3}-\frac{t^{2}}{2}\right]_{0}^{8}=5.1852[/math]

[math]E\left[T_{0}{ }^{2}\right]=2 \int_{0}^{8}\left({ }_{t} p_{0} \times t\right) d t=\frac{2}{72} \int_{0}^{8}\left(72 t-t^{3}-t^{2}\right) d t=\frac{2}{72}\left[36 t^{2}-\frac{t^{4}}{4}-\frac{t^{3}}{3}\right]_{0}^{8}=30.815[/math]

[math]\operatorname{Var}\left[T_{0}\right]=E\left[T_{0}^{2}\right]-\left(E\left[T_{0}\right]\right)^{2}=3.9287[/math]