Answer:
[tex]P({\displaystyle {\overline {x}}}>11,000)=0.0033[/tex]
Step-by-step explanation:
We know that the mean [tex]\mu[/tex] is:
[tex]\mu=10,500[/tex]
The standard deviation [tex]\sigma[/tex] is:
[tex]\sigma=1,300[/tex]
The sample size is:
[tex]n=50[/tex]
Then the sample average [tex]{\displaystyle {\overline {x}}}[/tex] is:
[tex]{\displaystyle {\overline {x}}}=\frac{550,000}{50}=11,000[/tex]
In this case we look for
[tex]P({\displaystyle {\overline {x}}}>11,000)[/tex]
We calculate the Z score.
In this case:
[tex]Z=\frac{{\displaystyle {\overline {x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]Z=\frac{11,000-10,500}{\frac{1,300}{\sqrt{50}}}[/tex]
[tex]Z=2.72[/tex]
Therefore:
[tex]P({\displaystyle {\overline {x}}}>11,000)=P(Z>2.72)[/tex]
Looking at the normal table we have to
[tex]P(Z>2.72)=0.0033[/tex]