Question

emails times spent by americans a survey found that the average time per weekday americans spend checking their email is 5.4 hours. assume the variable is approximately normally distributed and the standard deviation is 1.8 hours. if a random sample of 58 people is selected, find the probability that the mean of the sample will be between 5.3 hours and 5.7 hours. use the standard normal distribution table. round intermediate z value calculations to two decimal places, and the final answer to at least four decimal places.

p(5.3 < \overline{x} < 5.7) = \square

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 1.8$ and $n = 58$, then $\sigma_{\bar{x}}=\frac{1.8}{\sqrt{58}}\approx\frac{1.8}{7.62}\approx0.24$.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
For $\bar{x}=5.3$, $z_1=\frac{5.3 - 5.4}{0.24}=\frac{- 0.1}{0.24}\approx - 0.42$.
For $\bar{x}=5.7$, $z_2=\frac{5.7 - 5.4}{0.24}=\frac{0.3}{0.24}=1.25$.

Step3: Find the probabilities

Using the standard normal distribution table, $P(Z\lt - 0.42)=0.3372$ and $P(Z\lt1.25)=0.8944$.
Then $P(-0.42\lt Z\lt1.25)=P(Z\lt1.25)-P(Z\lt - 0.42)$.
$P(-0.42\lt Z\lt1.25)=0.8944 - 0.3372=0.5572$.

Answer:

$0.5572$