Given:
[tex]\mu=\text{ \$63000 ; n=30 ; }\sigma\text{=6300}[/tex][tex]P(\bar{x}<60000)=P(Z<\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}})[/tex][tex]P(\bar{x}<60000)=P(Z<\frac{60000-63000}{\frac{6300}{\sqrt[]{30}}})[/tex][tex]P(\bar{x}<60000)=P(Z<\frac{-3000}{\frac{6300}{\sqrt[]{30}}})[/tex][tex]P(\bar{x}<60000)=P(Z<-2.608)[/tex][tex]P(\bar{x}<60000)=P(Z<-2.61)[/tex]From the standard normal table
[tex]P(Z<-2.61)=0.00453[/tex]Therefore, 0.0045 is the probability of the given sample.
