Respuesta :
Answer:
[tex] z=\frac{9900-10000}{\frac{501}{\sqrt{40}}}= -1.262[/tex]
[tex] z=\frac{10200-10000}{\frac{501}{\sqrt{40}}}= 2.525[/tex]
And we can find the probability with this difference and using the normal standard distribution table and we got:
[tex] P(-1.262 < z< 2.525) =P(Z<2.525) -P(Z<-1.262) = 0.994-0.103= 0.891[/tex]
Step-by-step explanation:
For this problem we have the following info:
[tex] \mu = 10000[/tex] represent the true mean
[tex] \sigma = 501[/tex] represent the deviation
[tex] n = 40[/tex] representthe sample size selected
We want to find the following probability:
[tex] P(9900 <\bar X <10200)[/tex]
And for this case since the sample size is large enough we can use the central limit theorem and then we can use the z score formula given by:
[tex] z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Andreplacing we got:
[tex] z=\frac{9900-10000}{\frac{501}{\sqrt{40}}}= -1.262[/tex]
[tex] z=\frac{10200-10000}{\frac{501}{\sqrt{40}}}= 2.525[/tex]
And we can find the probability with this difference and using the normal standard distribution table and we got:
[tex] P(-1.262 < z< 2.525) =P(Z<2.525) -P(Z<-1.262) = 0.994-0.103= 0.891[/tex]
