A supervisor records the repair cost for 17 randomly selected TVs. A sample mean of $76.76 and standard deviation of $24.02 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Step-by-step explanation:

We want to determine a 99% confidence interval for the mean repair cost for the TVs

Number of sample, n = 17

Mean, u = $76.76

Standard deviation, s = $24.02

For a confidence level of 99%, the corresponding z value is 2.58. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

76.76 ± 2.58 × 24.02/√17

= 76.76 ± 2.58 × 5.83

= 76.76 ± 15.03

The lower end of the confidence interval is 76.76 - 15.03 = $61.73

The upper end of the confidence interval is 76.76 + 15.03 = $91.79

Therefore, with 99% confidence interval, the mean repair cost for the TVs is between $61.73 and $91.79

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.