Question

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

Answer 1

`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`