Answer to Question #254822 in Statistics and Probability for Adona

Question

1. Records of a health insurance company show that 30% of its policyholders over 50 submit a claim during the year.  15 policyholders over 50 are selected at random; what is the probability that at least 10 will submit a claim during the year?  What is the probability that 4 will submit a claim during the year?  How many do you expect to submit a claim? What is the standard deviation?

Answer 1

$n=15 \\ p=0.30 \\ q = 1 - 0.3 = 0.7$

X~Binomial(15,0.3)

$P(X) = C^{15}_x(0.3)^x(0.7)^{15-x}$

The probability that at least 10 will submit a claim during the year

$P(X≥10) = P(X=10) + P(X=11) +P(X=12) +P(X=13) + P(X=14) + P(X=15) \\ P(X=10) = C^{15}_{10}(0.3)^{10}(0.7)^{15-10} = 0.002980 = 298 \times 10^{-5} \\ P(X=11) = C^{15}_{11}(0.3)^{11}(0.7)^{15-11} = 0.000580 = 58 \times 10^{-5} \\ P(X=12) = C^{15}_{12}(0.3)^{12}(0.7)^{15-12} = 8.29 \times 10^{-5} \\ P(X=13) = C^{15}_{13}(0.3)^{13}(0.7)^{15-13} = 0.82 \times 10^{-5} \\ P(X=14) = C^{15}_{14}(0.3)^{14}(0.7)^{15-14} = 0.05 \times 10^{-5} \\ P(X=15) = C^{15}_{15}(0.3)^{15}(0.7)^{15-15} = 0.0014 \times 10^{-5} \\ P(X≥10) = 298 \times 10^{-5} + 58 \times 10^{-5} + 8.29 \times 10^{-5} + 0.82 \times 10^{-5} + 0.05 \times 10^{-5} + 0.0014 \times 10^{-5} = 365.1614 \times 10^{-5} \\ P(X≥10) = 0.00365$

The probability that 4 will submit a claim during the year

$P(X=4) = C^{15}_{4}(0.3)^{4}(0.7)^{15-4} = 0.2186$

Expected $= 15 \times 0.30 = 4.5 ≈ 5$

The standard deviation

$SD = \sqrt{\frac{p(1-p)}{n}} \\ = \sqrt{\frac{0.3(1-0.3)}{15}} \\ = 0.1183$