Question

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?

Answer 1

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