A supervisor has determined that the average salary of the employees in his department is $40,000...

A supervisor has determined that the average salary of the employees in his department is $40,000 with a standard deviation of $15,000. A sample of 25 of the employees’ salaries was selected at random. Assuming that the distribution of the salaries is normal, what is the probability that the average for this sample is between $36,000 and $42,000?

Solutions

Expert Solution

Solution:

Given, the Normal distribution with,

= 40,000

= 15,000

A sample of size n = 25 is taken from this population.

Let be the mean of this sample.

= = 40,000

= / n = 15000/25 = 3000

Find P( average for this sample is between $36,000 and $42,000)

i.e. P(36000 < < 42,000)

= P( < 42,000) - P( < 36000)

= P[( - )/ < (42000 - )/] - P[( - )/ < (36000 - )/]

= P[Z < (42000 - 40000)/3000] - P[Z < (36000 - 40000)/3000]

= P[Z < 0.667] - P[Z < -1.333]

= 0.7476 - 0.0913

= 0.6563

orchestra answered 1 year ago

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