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

Answer:

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

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

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

Step-by-step 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:

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

And replacing we got:

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

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

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