Question

Procter & Gamble reported that an American family of four washes an average of 1 ton (2000 pounds) of clothes each year. If the standard deviation of the distribution is 187.5 pounds, find the probability that the mean of a randomly selected sample of 50 families of four will be between 1,984 and 1,992 pounds.

Answer 1

Given that an American family of four washes an average of

$\mu=2000\text{ pounds}$

The standard deviation of the distribution is

$\sigma=187.5\text{ pounds}$

In order to find the probability that the mean off a randomnly selected sample of 50 families of 4 will be between 1,984 and 1992 pounds, you need to find the Z-scores.

$\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ \text{Where }X=\text{observed value} \\ Z_2=\frac{1984-2000}{187.5}=-\frac{16}{187.5}=-0.09 \\ Z_1=\frac{1992-2000}{187.5}=\frac{-8}{187.5}=-0.04 \end{gathered}$

Hence, the probability would be:

[tex]\begin{gathered} Pr(Z_1