The probability that a sample of size n will have a mean, [tex]\bar{x}[/tex], less that a given value, X, where the population mean, [tex]\mu[/tex], and the population standard deviation, [tex]\sigma[/tex], is known is given by:
[tex]P(\bar{x}\ \textless \ X)=P\left(z\ \textless \ \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \right)[/tex]
Therefore, the required probability is given by:
[tex]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[/tex]
