Question

The American Angus Association reports that mature Angus cows have a mean weight of 1309 pounds with a standard deviation of 157 pounds. For a random sample of 100 cows, what is the probability that the sample mean of these cows is at least 1290?

Answer 1

You must standardize the values to use the standard normal table.

Say \(\bar{X}\) = sample mean of cows. You are interested in \(P(\bar{X} \ge 1290)\).
Now, recall that for the sample mean, the mean is \(E(\bar{X})=\mu\), and the variance is \(Var(\bar{X})=\sigma^2/n\implies StDev(\bar{X})=\sigma/\sqrt{n}\).

Thus:\[P(\bar{X} \ge 1290)=P\left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\ge\frac{1290-1309}{157/\sqrt{100}}\right)=P(Z\ge-1.21)\]