Question

the average number of pounds of red meat a person consumes each year is 196 with a standard deviation of 22 pounds. if a sample of 50 individuals is randomly selected, find the probability that the mean of the sample will be less than 200 pounds.

a. 0.0985
b. 0.9015
c. 0.8815
d. 0.7613

Explanation:

Step1: Calculate the standard error

The standard error ($\sigma_{\bar{x}}$) is calculated by the formula $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 22$ (population standard deviation) and $n = 50$ (sample size).
$$\sigma_{\bar{x}}=\frac{22}{\sqrt{50}}\approx\frac{22}{7.071}\approx3.111$$

Step2: Calculate the z - score

The z - score is calculated by the formula $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x} = 200$ (sample mean), $\mu = 196$ (population mean), and $\sigma_{\bar{x}}\approx3.111$.
$$z=\frac{200 - 196}{3.111}=\frac{4}{3.111}\approx1.286$$

Step3: Find the probability

Using the standard normal distribution table (or a calculator with a normal - distribution function), we find $P(Z<1.286)$.
Looking up the value in the standard normal table, $P(Z < 1.29)=0.9015$ (since the value of $z = 1.286\approx1.29$)

Answer:

B. 0.9015