Question

1) A survey was conducted to measure the number of hours per week adults spend on home computers. In the survey, the number of hours were normally distributed, with a mean of 7 hours and a standard deviation of 1.5 hours.

a) What is the probability that the sample mean of 9 participants exceeds 8 hours?

b) What is the probability that the sample mean of 9 participants is below 6.5 hours?


c) What is the probability that the sample mean of 25 participants is between 6.8 and 7.8 hours?

Answer 1

Answer on Question #59275 – Math – Statistics and Probability

Question

A survey was conducted to measure the number of hours per week adults spend on home computers. In the survey, the number of hours were normally distributed, with a mean of 7 hours and a standard deviation of 1.5 hours.

a) What is the probability that the sample mean of 9 participants exceeds 8 hours?

b) What is the probability that the sample mean of 9 participants is below 6.5 hours?

c) What is the probability that the sample mean of 25 participants is between 6.8 and 7.8 hours?

Solution

First of all, we note that if the random variables $\xi_1, \xi_2, \ldots, \xi_n$ are normally distributed, with a mean of $a$ and a standard deviation of $\sigma$ then their mean $\bar{\xi} := \frac{1}{n} \sum_{k=1}^{n} \xi_k$ is normally distributed, with a mean of $a$ and a standard deviation of $\frac{\sigma}{\sqrt{n}}$.

Note also that

$$\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{u^2}{2}} du \text{ is the function of Laplace.}$$

a) In this case $n = 9$, $\sigma_n = \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{9}} = \frac{1.5}{3} = 0.5$, $\bar{\xi} \sim N(7, 0.5)$. Then

$$P\{\bar{\xi} > 8\} = P\left\{\frac{\bar{\xi} - 7}{0.5} > \frac{8 - 7}{0.5}\right\} = P\left\{\frac{\bar{\xi} - 7}{0.5} > 2\right\} = 0.5 - \Phi(2) = \begin{cases} \text{from the table} \\ \text{of Laplace} \end{cases} = 0.5 - -0.47725 = 0.02275 \approx 0.0228.$$

b) Similarly to a) we have

$$P\{\bar{\xi} < 6.5\} = P\left\{\frac{\bar{\xi} - 7}{0.5} < \frac{6.5 - 7}{0.5}\right\} = P\left\{\frac{\bar{\xi} - 7}{0.5} < -1\right\} = 0.5 - \Phi(1) = 0.5 - 0.34134 = 0.15866 \approx 0.1587.$$

c) In this case $n = 25$, $\sigma_n = \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{25}} = \frac{1.5}{5} = 0.3$, $\bar{\xi} \sim N(7, 0.3)$. Then

$$P\{6.8 < \bar{\xi} < 7.8\} = P\left\{\frac{6.8 - 7}{0.3} < \frac{\bar{\xi} - 7}{0.3} < \frac{7.8 - 7}{0.3}\right\} = P\left\{-0.67 < \frac{\bar{\xi} - 7}{0.3} < 2.67\right\} = \Phi(2.67) + \Phi(0.67) = 0.49621 + 0.24857 = 0.74478 \approx 0.7448.$$

Answer: a) 0.0228; b) 0.1587; c) 0.7448.

www.AssignmentExpert.com