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
