Respuesta :
Answer:
a) 0.625
b) 0.2510
c) 0.2005
Step-by-step explanation:
Data provided in the question:
Sample size, n = 64
Mean = 40
Standard deviation [tex]\sigma[/tex] = 5
a) Mean of the sampling distribution = Sample mean = 40
Standard deviation of the sampling distribution = [tex]\frac \sigma \sqrt{n}[/tex]
= [tex]\frac 5 \sqrt{64}[/tex]
= 0.625
b) [tex]P=P(-0.2<\bar x -m u<0.2)[/tex]
[tex]=P(\frac{-0.2 }{ 0.625})<\frac{\bar x \text {-mean } }{\frac{s}{\sqrt n}}<\frac{0.2}{ 0.625}[/tex]
[tex]=P(-0.32<Z<0.32)[/tex]
[tex]=0.2510[/tex] (from standard normal table)
c) [tex]P=P(\bar x -m u>0.8)+P(\bar x -m u< -0.8)[/tex]
[tex]=P(Z>0.8 / 0.625)+P(Z<-0.8 / 0.625)[/tex]
[tex]=P(Z>1.28)+P(Z<-1.28)[/tex]
[tex]=0.2005[/tex] (from standard normal table)
