Exam P Practice Problem 50 – Payments for Hospitalization

Problem 50-A

An insurer sells a hospital expense plan that pays a fixed sum per day of hospitalization. Suppose that the number of days of hospitalization in a year for someone insured under this plan has the following probability function.

      \displaystyle P(X=x)=\frac{3}{4^{x+1}} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots

The plan pays 1,000 for each day of hospitalization up to 4 days a year.

What is the expected payment for hospitalization under this hospital expense plan?

      \displaystyle (A) \ \ \ \ \ \ \ 250.00

      \displaystyle (B) \ \ \ \ \ \ \ 328.13

      \displaystyle (C) \ \ \ \ \ \ \ 332.03

      \displaystyle (D) \ \ \ \ \ \ \ 333.33

      \displaystyle (E) \ \ \ \ \ \ \ 444.44

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Problem 50-B

An insurer sells a hospital expense plan that pays cash for each day of hospitalization. Suppose that the number of days of hospitalization in a year for someone insured under this plan has the following probability function.

      \displaystyle P(X=x)=\frac{3}{4^{x+1}} \ \ \ \ \ \ \ \ x=0,1,2,3,\cdots

The plan pays for each day of hospitalization up to 4 days a year subject to the condition that the plan pays 500 for each of the first 3 days and 2500 for the fourth day.

What is the expected amount paid to the insured under this hospital expense plan?

      \displaystyle (A) \ \ \ \ \ \ \ 150.00

      \displaystyle (B) \ \ \ \ \ \ \ 166.67

      \displaystyle (C) \ \ \ \ \ \ \ 169.92

      \displaystyle (D) \ \ \ \ \ \ \ 173.83

      \displaystyle (E) \ \ \ \ \ \ \ 175.00

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Answers

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\copyright \ 2013

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One response

  1. For 50 -A I used 1000 * P[x=1]+ 2000*P[x=2] +3000*P[x=3] + 4000*P[x=4]) and I ended up getting B, slightly short of the correct answer, C. For 50 – B I basically did the same thing 500 * P[x=1]+ 1000*P[x=2] +1500*P[x=3] + 4000*P[x=4]) and I get C, once again, slightly short of the correct answer, D. I’m not sure what I’m missing in my calculation.

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