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

Answer:

$z=\frac{9.3-9.2}{\frac{1.6}{\sqrt{49}}}= 0.4375$

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

$P(z>0.4375)= 1-P(z$

Step-by-step 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:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z=\frac{9.3-9.2}{\frac{1.6}{\sqrt{49}}}= 0.4375$

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

$P(z>0.4375)= 1-P(z$