00:01
Alright, so here our question is that the average number of product that the assembly line workers at stylex company assemble per day is 40 with a standard deviation of 15.
00:11
So assume that the distribution of units assembled is normal.
00:14
So if a sample of 25 were drawn from all the workers, the first part of the question is what range of units manufactured by this sample would contain the sample mean 90 % of the time, okay.
00:27
So for middle 90%, the z -score is plus minus 1 .645.
00:32
Now here the value of the mean is computed as mean plus minus the z -value into the standard error.
00:42
So this will be equal to 40 plus minus 1 .645 into 15 upon square root of 25.
00:50
So this is equal to 40 plus minus 4 .935 that is equal to 35 .065 and 44 .935, right.
01:02
So the middle 90 % of the values is between these two values, fine.
01:07
Now moving on to the next part, it says what is the probability that the sample mean will be between 35 and 49 units.
01:16
So this is equal to probability of 35 minus 40 upon 15 divided by square root of 25 less than sample mean minus population mean upon standard deviation divided by square root of sample size less than 49 minus 40 upon 15 divided by square root of 25.
01:35
So this is equal to, so this is minus 1 .67 less than z less than 3 that is equal to probability of z less than 3 minus probability of z less than minus 1 .67, right.
01:53
So this is equal to 0 .9987 minus 0 .0475.
01:59
So the desired probability is 0 .9512, right...