00:01
The weight of adult chickens have a normal distribution with a mean of 7 pounds and a standard deviation of 1 .4 pounds.
00:09
Brandon has 500 adult chickens on his farm.
00:14
Part a find the 90th percentile of the weight of the adult chickens.
00:18
So the 90th percentile corresponds to a z score of 1 .28.
00:23
So to calculate this value, we're going to take our mean of 7 plus our standard deviation of 1 .4 times that z scores.
00:30
Is that z score tells us how many standard deviations above or below the mean, in this case above the mean.
00:36
So 8 .792 would be the 90th percentile.
00:40
For part b, when an adult chicken is selected at random from brandon's farm, what's the probability that it weighs over seven pounds? well, seven pounds is our mean.
00:50
So being greater than seven pounds means that we would have a probability of zero, or not a probability, we'd have a z score of zero, which has a probability of 0 .4.
01:06
Part c, one of brandon's adult chickens is quite big and weighs 10 .57 pounds.
01:12
Is this extraordinarily big in terms of weight? well, we can calculate a z score, 10 .57 minus 7 pounds divided by 1 .4.
01:22
It gives us 2 .55 standard deviations...