00:01
Hello students, here from the question, they given that, for the first subpart 1, the mean mu which is equal to 10 .6 and the standard deviation sigma which is 1 .2, sample size n which is equal to 74.
00:21
So, here the mu x bar which is equal to mu that is 10 .6.
00:26
Then sigma x bar which is sigma by square root of n 1 .2 by square root of 74, it is equal to 0 .1395.
00:37
So, here our x bar follows the normal distribution of 10 .6 and 0 .1395.
00:46
So, here we need to find the probability of x bar which is greater than 10 .7 is equal to the probability of x bar minus mu x bar divided by the sigma x bar greater than 10 .7 minus 10 .6 divided by 0 .1395.
01:09
By simplifying this probability of z greater than 0 .72, by using the z table, we get 0 .2367, 0 .128.
01:50
Then for the subpart 3, here they given that our sample mean x bar which is 6 .9, sample standard deviation is 1 .9, sample size n is equal to 8, the confidence level as 99 percentage.
02:11
So, alpha which is 0 .01 and the degrees of freedom here n minus 1, we get 7...