Respuesta :
Answer:
[tex] z=\frac{44257-44111}{\frac{2438}{\sqrt{80}}}= 0.536[/tex]
And if we use the normal standard table we got this:
[tex] P(z<0.536) =0.7040[/tex]
Step-by-step explanation:
For this case we have the following info :
[tex]\mu = 44111[/tex] represent the true mean
[tex]\sigma= \sqrt{5943844}= 2438[/tex] represent the deviation
n= 80 represent the sample size
And we want to find the follwing probability:
[tex] P(\bar X< 44257)[/tex]
For this case since the sample size is larger than 30 we can apply the central limit theorem and we can use the z score formula given by:
[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The distribution for the sample mean would be:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
And if we find the z score for this case we got:
[tex] z=\frac{44257-44111}{\frac{2438}{\sqrt{80}}}= 0.536[/tex]
And if we use the normal standard table we got this:
[tex] P(z<0.536) =0.7040[/tex]
