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Insurance company records indicate that 10% of its policyholders file claims involving theft or robbery of personal property from their homes. Suppose a random sample of 400 policyholders is selected.
-What is probability that the sample proportion of policyholders filing claims involving theft or robbery of personal property from their homes exceeds 15%?
A) 99.96%
B) 0.04%
C) 42.07%
D) 57.93%
E) 38.21%

Question

Quizplus · Accepted Answer

The probability that the sample proportion exceeds 15% can be found using the normal approximation to the binomial distribution. The mean of the sample proportion ($\mu_{\hat{p}}$) is the population proportion, which is 0.10, and the standard deviation of the sample proportion ($\sigma_{\hat{p}}$) is $\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.10(0.90)}{400}}$. To find the probability that the sample proportion exceeds 15%, we calculate the z-score for 0.15 and then find the corresponding probability from the standard normal distribution.The z-score is calculated as follows:$$ z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.15 - 0.10}{\sqrt{\frac{0.10(0.90)}{400}}} $$After calculating the z-score, we look up this value in the standard normal distribution table or use a calculator to find the probability that z exceeds this value. This probability will be very small, indicating that it is highly unlikely for the sample proportion to exceed 15% if the true proportion is 10%. The correct answer is 0.04%, which corresponds to a very small likelihood of the sample proportion exceeding 15%.

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