In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 36 pieces of carry-on luggage was weighed. The average weight was 20 pounds. Assume that we know the standard deviation of the population to be 8 pounds.
assume for part be that we do not kow the standard deviation. a determine a 95% confidence interval estimate for the mean weight fo the carry-on luggage. b determine the 97% confidence interval

a) The 95% confidence interval estimate for the mean weight fo the carry-on luggage

(17.38 , 22.61)

b) The 97% confidence interval estimate for the mean weight of the carry-on luggage

(17.11 , 22.89)

Step-by-step explanation:

Step1:-

a sample of 36 pieces of carry-on luggage was weighed

n=36

The average weight was 20 pounds

x⁻ = 20

given standard deviation of the population to be 8 pounds.

σ = 8

Confidence intervals :-

The values are called 95% of Confidence intervals for the mean of the population corresponding to the given sample.

The values are called 97% of Confidence intervals for the mean of the population corresponding to the given sample.

Step:-(1)

The 95% confidence interval estimate for the mean weight of the carry-on luggage

on calculation , we get

(20 -2.613 ,20 +2.613)

(17.3866 , 22.613)

The 95% confidence interval estimate for the mean weight fo the carry-on luggage

(17.3866 , 22.613)

Step2:-

The 97% confidence interval estimate for the mean weight of the carry-on luggage

on calculation , we get

(20-2.89 , 20+2.89)

(17.11 , 22.89)

The 97% confidence interval estimate for the mean weight of the carry-on luggage

(17.11 , 22.89)

The answer is 28.07 or 61.93. The hypotenuse (17) is across the right angle. Let x is across one leg (8): sin(90)/17 = sin(x)/8. 1/17 = sin(x)/8. sinx = 8/17. x = sin-1(8/17) = 28.07. If x is across the other leg, then x = 90 - 28.1 = 61.93