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 be at least 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 for this case we can find this probability using the normal standard table and with the complement rule we got:

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

Explanation:

We have the following information given:

`\mu = 9.2` represent the mean

`\sigma =1.6` the population deviation

`n =49` the sample size selected

We want to find the following probability:

`P(\bar X> 9.3)`

And for this case we can conclude that is a Right tail probability

And in order to solve it 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 for this case we can find this probability using the normal standard table and with the complement rule we got:

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