In a sample of 41 temperature readings taken from the freezer of a restaurant, the mean is 29.7 degrees and the population standard deviation is 2.7 degrees. What would be the 80% confidence interval for the temperatures in the freezer?

Step-by-step explanation:

Given: Sample size: n = 41

Sample mean degrees

Population standard deviation degrees

Confidence level (c) =  80% =0.80

Significance level (a) = 1- c = 1-0.80 = 0.20

z-score for 80% confidence level : z = 1.2816   [from z-table]

Confidence level for population mean :-

Hence, 80% confidence interval for the temperatures in the freezer

d

step-by-step explanation:

chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.