In a random sample of 50 refrigerators, the mean repair cost was $ 146.00 and the population standard deviation is $ 17.

In a random sample of 50 ​refrigerators, the mean repair cost was

​$146.00

and the population standard deviation is ​$17.80

.

Construct a 90​% confidence interval for the population mean repair cost. Interpret the results.

Construct a 90% confidence interval for the population mean repair cost.

The 90​% confidence interval is ( ? , ? )

​(Round to two decimal places as​ needed.)

I

nterpret you results. Choose the correct answer below.

A.With 90​% ​confidence, it can be said that the confidence interval contains the true mean repair cost.

B.With 90​% ​confidence, it can be said that the confidence interval contains the sample mean repair cost.

C.The confidence interval contains 90% of the mean repair costs.