# Exam P Practice Problem 58 – Dental Care and Vision Care Expenses

Problem 58-A

A health plan offers dental care and vision care benefits. Let $X$ represents the total annual amount (in millions) paid in dental care benefits. Let $Y$ represents the total annual amount (in millions) paid in vision care benefits.

The health plan determined that

• $X=K^2$ where $K$ follows a normal distribution with mean 0 and variance 1,
• $Y=L^2$ where $L$ follows a normal distribution with mean 0 and variance 2, and
• $K$ and $L$ are independent.

Given that the total annual vision care benefits paid by the health plan exceeds 2.5 millions, what is the probability that the total annual dental care benefits paid by the health plan exceeds 2 millions? $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0228$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0793$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1586$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.8416$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.9207$ $\text{ }$ $\text{ }$ $\text{ }$ $\text{ }$

Problem 58-B

A health plan offers dental care and vision care benefits. Let $X$ represents the total annual amount (in millions) paid in dental care benefits. Let $Y$ represents the total annual amount (in millions) paid in vision care benefits.

The health plan determined that

• $X=2.5 K^2$ where $K$ follows a normal distribution with mean 0 and variance 1,
• $Y=5 L^2$ where $L$ follows a normal distribution with mean 0 and variance 1, and
• $K$ and $L$ are independent.

What is the probability that the total annual dental care benefits exceeds 3 millions and that the total annual vision care benefits exceeds 4 millions? $\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1013$ $\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.4565$ $\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.6266$ $\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.7286$ $\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ 0.7881$

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Answers

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