A marketing solutions poll of mutual funds and fund owners asked fund owners what action they took after the September 11^th2001 market drop. Sixteen percent of respondents said they bought more funds. If 600 fund owners were polled, calculate the mean and standard deviation of the number of respondents who bought more funds. Considering as unusual a result that differs from the mean by more than two standard deviations, it (___ (is/is not) ___) unusual that in one of these polls of 600 fund owners, 100 respondents bought more mutual funds. A) 96, 9, is not B) 96, 8, is C) 96, 9, is D) 95, 8, is not E) 95, 8, is

Question

Quizplus · Accepted Answer

The mean number of respondents who bought more funds is calculated as 16% of 600, which is 0.16 * 600 = 96. The standard deviation for a binomial distribution is calculated using the formula $\sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials (600) and $p$ is the probability of success (0.16). Thus, the standard deviation is $\sqrt{600 \cdot 0.16 \cdot (1-0.16)} = \sqrt{600 \cdot 0.16 \cdot 0.84} = \sqrt{96 \cdot 0.84} = \sqrt{80.64} \approx 9$. Considering a result unusual if it differs from the mean by more than two standard deviations, the range for usual results is mean ± 2*standard deviation, which is 96 ± 2*9 = 96 ± 18. This range is 78 to 114. Since 100 respondents fall within this range, it is not unusual.

A Marketing Solutions Poll of Mutual Funds and Fund Owners