Respuesta :
Answer:
a) [tex] \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})[/tex]
With:
[tex]\mu_{\bar X}= 88[/tex]
[tex]\sigma_{\bar X}= 8[/tex]
b) [tex]z=\frac{89.7-88}{\frac{8}{\sqrt{64}}}= 1.7[/tex]
[tex]P(Z>1.7) = 1-P(Z<1.7) =1-0.955=0.0446[/tex]
c) [tex]z =\frac{85.7-88}{\frac{8}{\sqrt{64}}}= -2.3[/tex]
[tex]P(Z<-2.3) = 0.0107[/tex]
d) [tex]z =\frac{87.35-88}{\frac{8}{\sqrt{64}}}= -0.65[/tex]
[tex]z =\frac{90.5-88}{\frac{8}{\sqrt{64}}}= 2.5[/tex]
[tex]P(-0.65<z<2.5)=P(Z<2.5)-P(Z<-0.65) =0.994-0.258 = 0.736[/tex]
Step-by-step explanation:
For this case we know the following propoertis for the random variable X
[tex]\mu = 88, \sigma = 8[/tex]
We select a sample size of n = 64
Part a
Since the sample size is large enough we can use the central limit distribution and the distribution for the sample mean on this case would be:
[tex] \bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})[/tex]
With:
[tex]\mu_{\bar X}= 88[/tex]
[tex]\sigma_{\bar X}= 8[/tex]
Part b
We want this probability:
[tex]P(\bar X>89.7)[/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 89.7 we got:
[tex]z=\frac{89.7-88}{\frac{8}{\sqrt{64}}}= 1.7[/tex]
[tex]P(Z>1.7) = 1-P(Z<1.7) =1-0.955=0.0446[/tex]
Part c
[tex]P(\bar X<85.7)[/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 85.7 we got:
[tex]z =\frac{85.7-88}{\frac{8}{\sqrt{64}}}= -2.3[/tex]
[tex]P(Z<-2.3) = 0.0107[/tex]
Part d
We want this probability:
[tex]P(87.35 <\bar X< 90.5)[/tex]
We find the z scores:
[tex]z =\frac{87.35-88}{\frac{8}{\sqrt{64}}}= -0.65[/tex]
[tex]z =\frac{90.5-88}{\frac{8}{\sqrt{64}}}= 2.5[/tex]
[tex]P(-0.65<z<2.5)=P(Z<2.5)-P(Z<-0.65) =0.994-0.258 = 0.736[/tex]
