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? Use a calculator to find the probability that the mean overtime hours for the sample will be at least 9.3 hours. Select the two correct answers that apply below.

Answer 1

Complete Question

Answer:

The probability is Right- tailed

The mean overtime hours for the sample will be at least 9.3 hours is

`P(X \ge 9.3 ) = 0.3333`

Explanation:

From the question we are told that

The mean is
`\mu = 9.2 \ hours`

The standard deviation is
`\sigma = 1.6 \ hours`

The sample size is n = 49

The sample mean is
`\= x = 9.3`

Generally the standard error of the mean is mathematically represented as

`\sigma_(\= x) = (\sigma )/(√(n) )`

=>
`\sigma_(\= x) = (1.6 )/(√(49) )`

=>
`\sigma_(\= x) = 0.22857`

This is a right -tailed probability because the sample mean is greater than the population mean

Generally the probability that the mean overtime hours for the sample will be at least 9.3 hours is mathematically represented as

`P(X \ge 9.3 ) =1 - P(X < 9.3 )`

So

`P(X < 9.3 ) = P( ( X - \mu )/( \sigma_(\= x )) < ( 9.3 - 9.2 )/( 0.22857) )`

Here
`(X - \mu )/(\sigma _(\= x)) = Z`

`P(X < 9.3 ) = P( Z < 0.43750)`

From the z-table

`P(X < 9.3 ) = P( Z < 0.43750) = 0.667`

So

`P(X \ge 9.3 ) =1 - 0.667`

`P(X \ge 9.3 ) = 0.3333`