Question 1202885
area between z-score of -1.82 and 1.82 = .9312.
that's the probabiity of getting a sample mean between 5 and 6.6 hours.


area between z-score of 0.91 and .91 = .6372.
that's the probability of getting a sample mean between 5.4 and 6.2 hours.


the following calcuilator results show these values to be true.


<img src = "http://theo.x10hosting.com/2023/062911.jpg">


<img src = "http://theo.x10hosting.com/2023/062912.jpg">


first you need to find the standard error.
standard error = standard devition / square root of sample size = 3.5 / sqrt(64) = .4375.
round to 2 decimal places = .44.


z-score for sample mean of 6.6 = (6.6 - 5.8) / .44 = 1.8181818.....
round to 2 decimal places = 1.82


z-score for sample mean of 5 = (5 - 5.8) / .44 = -1.8181818.....
round to 2 decimal places = -1.82


area to left of z-score of 1.82 = .9656 rounded to 4 decimal places.
area to left of z-score of -1.81 = .0344 rounded to 4 decimal places.
area in between = .9656 - .0344 = .9312


that's the probbiity of getting a sample mean between 5 and 6.6.


similar analysis was done for sample means between 5.4 and 6.2 hours.
z-scores were -.91 and .91
for example, z-score for sample mean of 5.4 was z = (5.4 - 5.8) / .44 = -.91 rounded to decimal places.

area in between was area to the left of z-score of .91 minus area to the left of z-score of -.91 which was equal to .6372.