Answer:
[tex]P =0.3998[/tex]
Step-by-step explanation:
Let [tex]{\displaystyle {\overline{x}}}[/tex] be the average of the sample, and the population mean will be [tex]\mu[/tex]
We know that:
[tex]\mu = 4095[/tex] gr
Let [tex]\sigma[/tex] be the standard deviation and n the sample size, then we know that the standard error of the sample is:
[tex]E=\frac{\sigma}{\sqrt{n}}[/tex]
Where
[tex]\sigma=569[/tex]
[tex]n=130[/tex]
In this case we are looking for:
[tex]P(|{\displaystyle{\overline{x}}}- \mu|>42)[/tex]
This is:
[tex]{\displaystyle{\overline{x}}}- \mu>42[/tex] or [tex]{\displaystyle{\overline{x}}}- \mu<-42[/tex]
[tex]P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu<-42)[/tex]
Now we get the z score
[tex]Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z<-\frac{42}{\frac{569}{\sqrt{130}}})[/tex]
[tex]P=P(z>0.8416) + P(z<-0.8416)[/tex]
Looking at the tables for the standard nominal distribution we get
[tex]P =0.1999+0.1999[/tex]
[tex]P =0.3998[/tex]
