A manager records the repair cost for 4 randomly selected stereos. A sample mean of $82.64 and standard deviation of $14.32 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the stereos. Assume the population is approximately normal. Step 2 of 2 : Construct the 90% confidence interval. Round your answer to two decimal places.

The 90% confidence interval for the mean repair cost for the stereos is between $70.86 and $94.42.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 82.64 - 11.78 = $70.86

The upper end of the interval is the sample mean added to M. So it is 82.64 + 11.78 = $94.42.

The 90% confidence interval for the mean repair cost for the stereos is between $70.86 and $94.42.

the median here is 25. your lowest value is 14 and highest value is 48. to find the third quartile, find the medium number in between 25 and 48 which would be 42 i believe.

answer: (-6,3)

step-by-step explanation: