Answer:
The appropriate solution is:
(a) 0.1056
(b) 0
(c) 0.9544
Step-by-step explanation:
The given values are:
Mean,
[tex]\mu = 75[/tex]
Standard deviation,
[tex]\sigma = 4[/tex]
(a)
⇒  [tex]P(x>80)=1-(x<80)[/tex]
           [tex]=1-P[\frac{x-\mu}{\sigma} <\frac{80-75}{4} ][/tex]
           [tex]=1-P[\frac{x-\mu}{\sigma} <\frac{5}{4} ][/tex]
           [tex]=1-P(z<1.25)[/tex]
By using the table, we get
           [tex]=1-0.8944[/tex]
           [tex]=0.1056[/tex]
(b)
According to the question, the values are:
[tex]n[/tex] = 64
[tex]\mu_\bar{x}[/tex] = 75
Now,
⇒  [tex]\sigma_\bar{x}[/tex] = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
     = [tex]\frac{4}{\sqrt{64} }[/tex]
     = [tex]\frac{4}{8}[/tex]
     = [tex]0.5[/tex]
⇒  [tex]P(\bar {x} >80 ) = 1 - P(\bar x <80 )[/tex]
           [tex]=1 - P[\frac{(\bar x-\mu_\bar x)}{\sigma \bar x} < \frac{80-75}{0.5} ][/tex]
           [tex]=1-P(z<10)[/tex]
By using the table, we get
           [tex]=1-1[/tex]
           [tex]=0[/tex]
(c)
As we know,
⇒ [tex]\sigma_\bar x[/tex] = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
    = [tex]\frac{4}{\sqrt{64} }[/tex]
    = [tex]\frac{4}{8}[/tex]
    = [tex]0.5[/tex]
then,
= [tex]P(74< \bar x <76)[/tex]
= [tex]P[\frac{74-75}{0.5} < \frac{\bar x-\mu \bar x}{\sigma \bar x} < \frac{76-75}{0.5} ][/tex]
= [tex]P(-2<z<2)[/tex]
= [tex]P(z<2)-P(z<-2)[/tex]
By using the table, we get
= [tex]0.9772-0.0228[/tex]
= [tex]0.9544[/tex]
