Question

The mean per capita consumption of milk per year is 131 liters with a variance of 841. If a sample of 132 people is randomly selected, what is the probability that the sample mean would be less than 133.5 liters? Round your answer to four decimal places.

Answer 1

The probability that a sample of size n will have a mean, $\bar{x}$, less that a given value, X, where the population mean, $\mu$, and the population standard deviation, $\sigma$, is known is given by:

$P(\bar{x}\ \textless \ X)=P\left(z\ \textless \ \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \right)$

Therefore, the required probability is given by:

$P(\bar{x}\ \textless \ 133.5)=P\left(z\ \textless \ \frac{133.5-131}{\sqrt{841/132}} \right) \\ \\ =P\left(z\ \textless \ \frac{2.5}{\sqrt{6.3712}} \right)=P\left(z\ \textless \ \frac{2.5}{2.524} \right) \\ \\ =P(z\ \textless \ 0.99)=0.8390$