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For employees at a large company, the mean number of overtime hours worked each week is 9.2 hours with a population standard deviation of 1.6 hours. A random sample of 49 employees was taken and the probability that the mean number of overtime hours will exceed 9.3 hours was determined. Was the probability a Left-tail, Right -tail or Interval Probability

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Answer:


z=(9.3-9.2)/((1.6)/(√(49)))= 0.4375

And we can use the normal table and the complement rule we got:


P(z>0.4375)= 1-P(z<0.4375) = 1-0.669= 0.331

Explanation:

For this case we have the following parameters given:


\mu = 9.2 , \sigma =1.6

We select a ample size of n=49. And we want to find this probability:


P(\bar X> 9.3)

And for this case is a right tail probability and we can use the z score formula given by:


z=(\bar X -\mu)/((\sigma)/(√(n)))

And replacing we got:


z=(9.3-9.2)/((1.6)/(√(49)))= 0.4375

And we can use the normal table and the complement rule we got:


P(z>0.4375)= 1-P(z<0.4375) = 1-0.669= 0.331

User Fatikhan Gasimov
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