Answer: a. 0.1500 b. 0.0367
Explanation:
Let X is a random variable with a distribution that may be known or unknown:
[tex]\mu_x[/tex] = the mean of X
[tex]\sigma_x[/tex] = the standard deviation of X
If we draw random samples of size n, then the random samples contains sample means [tex]\overline{X}[/tex], tends to be normally distributed
[tex]\overline{X}\sim N(\mu_x,\dfrac{\sigma_x}{\sqrt{n}})[/tex]
Here, [tex]\mu_x[/tex] = 75
[tex]\sigma_x[/tex] =5
n=20
a.
[tex]P(76<\overline{X}<77)=P(\dfrac{76-75}{\dfrac{5}{\sqrt{20}}}<\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{77-75}{\dfrac{5}{\sqrt{20}}})\\\\=P(0.89<Z<1.79) \ \ [z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<1.79)-P(z<0.89)\\\\=0.963273-0.813267=0.150006\approx0.1500[/tex]
b.
[tex]P(\overline{X}>77)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{77-75}{\dfrac{5}{\sqrt{20}}})\\\\=P(Z>1.79) \ \ [z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(z<1.79)\\\\=1-0.963273=0.036727\approx0.0367[/tex]
