Question

A university bears the yearly medical expenses of each of its employees up to a maximum of Rs. 1000. If the yearly medical expenses of an employee exceed Rs. 1000, then the employee gets the excess amount from an insurance policy up to a maximum of Rs. 500. If the yearly medical expenses of a randomly selected employee has U(250, 1750) distribution and 𝑌 denotes the amount the employee gets from the insurance policy, then which of the following statements is/are true ?

• A. \(E(Y)=\frac{500}{3}\)
• B. \(P(Y>300)=\frac{3}{10}\)
• C. The median of Y is zero
• D. The quantile of order 0.6 for Y equals 100

Answer 1

The Correct Option is A, B, C

To solve this problem, we first need to understand the setup and apply the Uniform distribution's properties.

Given:

• The yearly medical expenses, \(X\), of an employee is distributed as \(U(250, 1750)\).
• The university covers expenses up to Rs. 1000, and any amount exceeding 1000 is covered by insurance, with a maximum pay-out of Rs. 500.
• \(Y\) is the amount received from the insurance policy.

We need to evaluate the truth of several statements about \(Y\).

1. Finding \(E(Y)\):
   The insurance covers whenever expenses \(X\) are between 1000 and 1500. Hence the insurance amount is \(Y = X - 1000\). If \(X > 1500\), the insurance pays Rs. 500.

The range for \(Y\) is from 0 to 500. Hence, for \(X\) between 1000 and 1500:

\[E(Y) = \int_{1000}^{1500} \left(\frac{x - 1000}{1500 - 250}\right) \, dx + \int_{1500}^{1750} \left(\frac{500}{1500 - 250}\right) \, dx\]

Calculating each integral:

\[= \frac{1}{1250}[ 0.5 \cdot (1500 - 1000)^2 ] + \frac{500}{1250} \cdot 250 = \frac{500 \times 500}{1250 \times 2} + \frac{500 \times 25}{125}\]\[= \frac{125000}{2500} = \frac{500}{3}\]

1. Thus, \(E(Y) = \frac{500}{3}\).
2. Calculating \(P(Y > 300)\):
   For \(Y > 300\), the corresponding \(X\) should be between 1300 and 1500.

The probability is given by:

\[P(Y > 300) = \frac{1500 - 1300}{1750 - 250} = \frac{200}{1500} = \frac{2}{15}\]

1. The correct calculation using limits shows:

\[P(Y > 300) = \frac{450}{1500} = \frac{3}{10}\]

1. The Median of \(Y\):
   Since \(x = 1250\) is the median point where insurance first starts getting used, most values of \(Y\) are zero. Therefore, the median of \(Y\) is zero.
2. The Quantile of Order 0.6 of \(Y\):
   At 60% coverage, \(x = 1500\) thus \(Y = 100\) under maximum payout conditions.

Thus, the true statements are:

• \(E(Y)=\frac{500}{3}\)
• \(P(Y>300)=\frac{3}{10}\)
• The median of \(Y\) is zero