ABy Admin
May 09'23
Exercise
Claims filed under auto insurance policies follow a normal distribution with mean 19,400 and standard deviation 5,000.
Calculate the probability that the average of 25 randomly selected claims exceeds 20,000.
- 0.01
- 0.15
- 0.27
- 0.33
- 0.45
ABy Admin
May 09'23
Solution: C
Let [math]X_1,\ldots,X_n[/math] denote the 25 collision claims, and let
[[math]]
\overline{X} = \frac{1}{25}(X_1 \cdots + X_n).
[[/math]]
We are given that each [math]X_i \, (i=1,\ldots,25)[/math] follows a normal distribution with mean 19,400 and standard deviation 5,000. As a result [math]\overline{X}[/math] also follows a normal distribution with mean 19,400 and standard deviation 25-1/2 5000 = 1000. We conclude that
[[math]]
\begin{align*}
\operatorname{P}[\overline{X} \gt 20000] &= \operatorname{P}[\frac{\overline{X}-19400}{1000} \gt \frac{20000-19400}{1000}] \\
&= \operatorname{P}[\frac{\overline{X}-19400}{1000} \gt 0.6] \\
&= 1 - \Phi(0.6) = 1-0.7257 \\
& = 0.2743.
\end{align*}
[[/math]]