00:01
So we have the mean consumption of salmon is 12 pounds with a standard deviation of 3 .2 pounds, and the distribution is normal.
00:08
And we have this person usually buys 20 pounds, but they may want to change over to meeting the demand just 90 % of the day.
00:16
So that may run out of the fish.
00:19
And we want to know if she orders 20 pounds, how often will she end up running out of fish? so that is to convert this to a z value, 20 minus the mean of 12.
00:30
Divided by that standard deviation of 3 .2.
00:33
And 20 minus 12, which is 8, divided by the 3 .2, gives us a z value of 2 .5.
00:41
And the area above, or the probability of being above 2 .5, is 0 .0062.
00:53
Now we want to know what's the likelihood that the amount of salmon is less than 15 pounds.
01:00
Meaning that there'd be five pounds that would be waste.
01:03
So now we convert that to a z value, the 15 minus 12 over 3 .2.
01:08
And that z value, quick get a little second entry and change that for you, that z value comes out to be, it would round off to 0 .94...