Question

Scenario 7.5 Time spent using e-mail per session is normally distributed with H 8 minutes and o 2 minutes. If a random sample of 25 sessions is selected, answer the questions below. Question 16 1 pts Based on the information in Scenario 7.5, what is the probability that the sample mean is between 7.8 and 8.2 minutes? (4 decimals)

Answer 1

$\mu=8, \sigma=2, n=25$

16. Since we're given that the distribution is normal, we have to calculate the z value for the given range and compute its probability.

$7.8<\overline{x}<8.2 \rightarrow \frac{5(7.8-8)}{2}<\frac{5(x-8)}{2}<\frac{5(8.2-8)}{2} \rightarrow \newline -0.5<z<0.5$

To find this probability, we use the z table and subtract the probability associated with the z value (-0.5) from the probability associated with the z value (0.5).

Required probability = 0.6915 - 0.3085 = 0.383

17. Same steps from the previous question are followed:

$7.5<\overline{x}<8 \rightarrow \frac{5(7.5-8)}{2}<\frac{5(x-8)}{2}<\frac{5(8-8)}{2} \rightarrow -1.25<z<0$

Required probability = 0.5 - 0.1056 = 0.3944

18. Same as 16.

19. When the sample size increases, sample means will be closer to the population mean because sample mean is a consistent estimator of the population mean and approaches the population mean in probability as the sample size increases.

Standard error for n = 100 is

$\frac{2}{\sqrt{100}}=\frac{1}{25}$   

Comparing this to the standard error for n = 25,

$\frac{2}{\sqrt{25}}=\frac{10}{25}$

Hence, the standard error with n = 100 is lower than the standard error with n = 25.

With n = 100, question 16 is performed as follows:

$7.8<\overline{x}<8.2 \rightarrow \frac{10(7.8-8)}{2}<\frac{10(x-8)}{2}<\frac{10(8.2-8)}{2} \rightarrow \newline -1<z<1$

Required probability is 0.8413 - 0.1587 = 0.6826

Which is greater than the probability we found in question 16.

Hence, with a higher n, the probability that the sample mean lies within +/- 0.2 of the population mean is higher, which further proves the fact that the sample mean is a consistent estimator of the population mean.