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=(X-\mu)/(\sigma) \\ \text{Where }X=\text{observed value} \\ Z_2=(1984-2000)/(187.5)=-(16)/(187.5)=-0.09 \\ Z_1=(1992-2000)/(187.5)=(-8)/(187.5)=-0.04 \end{gathered}`

Hence, the probability would be:

[tex]\begin{gathered} Pr(Z_1