Question

The average yearly cost per household of owning a cat is $183.80. suppose that we randomly select 36 households that own a cat. what is the probability that the sample mean of these 36 households is more than $175.00? assume standard deviation of the population is $32.

Answer 1

The probability that that a sample of size, n, has a mean more that a given value, X, is given by:

$P(\bar{x}\ \textgreater \ X)=1-P(\bar{x}\leq X)=1-P\left(z\leq \frac{X-\mu}{\sigma/\sqrt{n}} \right)$

Thus, the required probability is given by:

$P(\bar{x}\ \textgreater \ 175)=1-P\left(z\leq \frac{175-183.80}{32/\sqrt{36}} \right) \\ \\ =1-P\left(z\leq \frac{-8.8}{32/6} \right)=1-P\left(z\leq \frac{-8.8}{5.333} \right) \\ \\ =1-P(z\leq-1.65)=1-0.04947=0.9505$