Respuesta :
Answer:
a) [tex] \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})[/tex]
With:
[tex]\mu_{\bar X}= 83[/tex]
[tex]\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3[/tex]
b) [tex] z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2[/tex]
[tex] P(Z>2) = 1-P(Z<2)= 1-0.97725= 0.02275[/tex]
c) [tex] z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45[/tex]
[tex] P(Z<-2.45) =0.0071[/tex]
d) [tex] z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1[/tex]
[tex] z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2[/tex]
[tex] P(-1.2<Z<2.1)= P(Z<2.1) -P(z<-1.2) = 0.982-0.115= 0.867[/tex]
Step-by-step explanation:
For this case we know the following propoertis for the random variable X
[tex]\mu = 83, \sigma = 27[/tex]
We select a sample size of n = 81
Part a
Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:
[tex] \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})[/tex]
With:
[tex]\mu_{\bar X}= 83[/tex]
[tex]\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3[/tex]
Part b
We want this probability:
[tex]P(\bar X>89)[/tex]
We can use the z score formula given by:
[tex]z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 89 we got:
[tex] z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2[/tex]
[tex] P(Z>2) = 1-P(Z<2)= 1-0.97725= 0.02275[/tex]
Part c
[tex]P(\bar X<75.65)[/tex]
We can use the z score formula given by:
[tex]z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 75.65 we got:
[tex] z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45[/tex]
[tex] P(Z<-2.45) =0.0071[/tex]
Part d
We want this probability:
[tex] P(79.4 < \bar X < 89.3)[/tex]
We find the z scores:
[tex] z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1[/tex]
[tex] z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2[/tex]
[tex] P(-1.2<Z<2.1)= P(Z<2.1) -P(z<-1.2) = 0.982-0.115= 0.867[/tex]
