ABy Admin
May 03'23
Exercise
The annual profit of a life insurance company is normally distributed.
The probability that the annual profit does not exceed 2000 is 0.7642. The probability that the annual profit does not exceed 3000 is 0.9066.
Calculate the probability that the annual profit does not exceed 1000.
- 0.1424
- 0.3022
- 0.5478
- 0.6218
- 0.7257
ABy Admin
May 03'23
Solution: C
[[math]]
\begin{align*}
\operatorname{P}(Z \leq 0.72) = 0.7642 = \operatorname{P}(X \leq 2000 ) = \operatorname{P}[Z \leq (2000-\mu)/\sigma] \\
0.72 = (2000-\mu)/\sigma \\
\operatorname{P}( Z ≤ 1.32) = 0.9066 = \operatorname{P}( X ≤ 3000) = \operatorname{P}[ Z ≤ (3000 − µ ) / σ ] \\
1.32 = (3000 − \mu ) / \sigma \\
1.32 / 0.72 = (3000 − \mu ) / (2000 − \mu ) \\
1.8333(2000 − \mu )= 3000 − \mu \\
\mu =[1.8333(2000) − 3000] / (1.8333 − 1) = 800 \\
\sigma =(3000 − \mu ) /1.32 = 1666.67 \\
\operatorname{P}( X ≤ 1000) = \operatorname{P}[ Z ≤ (1000 − 800) /1666.67] = \operatorname{P}( Z ≤ 0.12) = 0.5478.
\end{align*}
[[/math]]