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A worker in the automobile industry works an average of 43.7 hours per week. Assume the distribution is normal with a standard deviation of 1.6 hours.


(i) What is the probability that a randomly selected automobile worker works less than 40 hours per week?


(ii) If 15 automobile workers are randomly selected, what is the probability that the sample mean of working time is more than 45 hours per week?

Respuesta :

Cricetus

Answer:

The solution is:

(1) 0.0104

(2) 0.0008

Step-by-step explanation:

Given:

Mean,

[tex]\mu = 43.7[/tex]

Standard deviation,

[tex]\sigma = 1.6[/tex]

(1)

⇒ [tex]P(X<40) = P(\frac{x-\mu}{\sigma}<\frac{40-43.7}{1.6} )[/tex]

                      [tex]=P(z< - 2.3125)[/tex]

                      [tex]=P(z<-2.31)[/tex]

                      [tex]=0.0104[/tex]

(2)

As we know,

n = 15

⇒ [tex]P(\bar X > 45)= P(\frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n} } } >\frac{45-43.7}{\frac{1.6}{\sqrt{15} } } )[/tex]

                      [tex]=P(z> 3.15)[/tex]

                      [tex]=1-P(z<3.15)[/tex]

                      [tex]=1-0.9992[/tex]

                      [tex]=0.0008[/tex]

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