Question

Americans consume on average 92 pounds of chicken per year, with a standard deviation of 10.3 pounds. What is the probability that a sample of 50 people will have consumed an average of 100 pounds or more?

Answer 1

Answer:

0.2187

Step-by-step explanation:

Americans consume on average 92 pounds of chicken per year, with a standard deviation of 10.3 pounds. What is the probability that a sample of 50 people will have consumed an average of 100 pounds or more?

We to find

x ≥ 100 pounds

To solve for the above question, we use the z-score formula

z = (x-μ)/σ, where

x is the raw score = 100 pounds

μ is the population mean = 92 pounds

σ is the population standard deviation = 10.3

Hence:

z = (100 - 92)/ 10.3

z = 8/1.46

z = 0.7767

Probability value from Z-Table:

P(x<100) = 0.78133

P(x>100) = 1 - P(x<100) = 0.21867

Approximately = 0.2187

Therefore, the probability that a sample of 50 people will have consumed an average of 100 pounds or more is 0.2187