Answer:
[tex] z = \frac{500-520}{\frac{90}{\sqrt{100}}}= -2.22[/tex]
And we can find this probability using the normal standard distribution and we got:
[tex] P(z<-2.22) =0.0132[/tex]
Step-by-step explanation:
For this case we have the foolowing parameters given:
[tex] \mu = 520[/tex] represent the mean
[tex] \sigma =90[/tex] represent the standard deviation
[tex] n = 100[/tex] the sample size selected
And for this case since the sample size is large enough (n>30) we can apply the central limit theorem and the distribution for the sample mean would be given by:
[tex] \bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}}) [/tex]
And we want to find this probability:
[tex] P(\bar X <500)[/tex]
We can use the z score formula given by:
[tex] z = \frac{500-520}{\frac{90}{\sqrt{100}}}= -2.22[/tex]
And we can find this probability using the normal standard distribution and we got:
[tex] P(z<-2.22) =0.0132[/tex]
