The u. s. per capita chicken consumption for 2007 was 90.6 pounds. assume this consumption is normally distributed with a standard deviation of 17.2 pounds. assume n infinite. what is the probability that a sample taken of 100 individuals shows an average consumption of less than 90 pounds of chicken?

To solve this, we need to use the z statistic. The formula for z score is:

z = (x – u) / s

where x is sample value = less than 90, u is the sample mean = 90.6, s is the standard deviation = 17.2

z = (90 – 90.6) / 17.2

z = -0.035

From the standard distribution tables:

P (z = -0.035) = 0.4860

Therefore there is about 48.60 % chance that it will be less than 90 pounds

13.33%

say 50 oranges, then 100 apples

80 are red, 20 are yellow

total fruit = 150

20chances out of 150 for yellow

20/150 = 13.33%

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0