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Answer:
"0.0749" is the correct solution.
Step-by-step explanation:
Given:
Mean,
[tex]\mu=5000 \ hours[/tex]
Standard deviation,
[tex]\sigma=500 \ hours[/tex]
Random sample,
[tex]n = 100[/tex]
∴ [tex]\mu_\bar x \ = \ \mu[/tex]
[tex]=5000 \ hours[/tex]
Now,
⇒ [tex]\sigma_\bar x[/tex] = [tex]\frac{\sigma}{\sqrt{n} }[/tex]
= [tex]\frac{500}{\sqrt{100} }[/tex]
= [tex]50 \ hours[/tex]
hence,
The probability will be:
= [tex]P(\bar X<4928)[/tex]
= [tex]P(\frac{\bar x-\mu_ \bar x}{\sigma_x} < \frac{4928-5000}{50} )[/tex]
= [tex]P(z<\frac{-72}{50} )[/tex]
= [tex]P(z<-1.44)[/tex]
= [tex]0.0749[/tex]