Question

the average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds (source: american dietetic association). if a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be greater than 200 pounds.
a. 0.8815
b. 0.7613
c. 0.9015
d. 0.0985

Explanation:

Step1: Calculate the standard error

The formula for the standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \). Given \( \sigma = 22 \) and \( n = 50 \), then \( \sigma_{\bar{x}}=\frac{22}{\sqrt{50}}\approx3.111 \).

Step2: Calculate the z - score

The formula for the z - score is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \). Here, \( \bar{x} = 200 \), \( \mu=196 \), and \( \sigma_{\bar{x}}\approx3.111 \). So \( z=\frac{200 - 196}{3.111}\approx1.286 \).

Step3: Find the probability

We want \( P(\bar{X}>200) \), which is equivalent to \( P(Z > 1.286) \). Using the standard normal distribution table \( P(Z\leq1.286)\approx0.9007 \), so \( P(Z>1.286)=1 - P(Z\leq1.286)=1 - 0.9007 = 0.0993\approx0.0985 \) (due to rounding differences in the table values).

Answer:

D. 0.0985