The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a .95 probability, the sample mean will provide a margin of error of ___.

We are given that the manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute.

That means; Sample size, n = 100

Population standard deviation = 1 minute

Sample mean = 3 minutes

Now, the Margin of error means how much deviation is there from the sample mean to calculate the confidence interval for true population mean.

Margin of error formula is given by = [tex]Z_\frac{\alpha}{2} \times \frac{\sigma}{\sqrt{n} }[/tex]

where, [tex]\sigma[/tex] = population standard deviation

n = sample size of customers

[tex]\alpha[/tex] = significance level = 5%

Now, the z probability area at 5% significance level is given to be 1.96 from z table.

So, Margin of error = [tex]1.96 \times \frac{1}{\sqrt{100} }[/tex] = [tex]\frac{1.96}{10}[/tex] = 0.196 .

Therefore, with a .95 probability, the sample mean will provide a margin of error of 0.196 .