Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
We are given the <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:
- Sample mean of

- Population standard deviation of

- Sample size of
.
The margin of error is:

In which z is the critical value.
We have to find the critical value, which is z with a p-value of
, in which
is the confidence level.
In this problem,
, thus, z with a p-value of
, which means that it is z = 1.96.
Then:

The confidence interval is the <u>sample mean plus/minutes the margin of error</u>, hence:


The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
A similar problem is given at brainly.com/question/22596713