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In: Statistics and Probability

The incomes of a population of lawyers have a normal distribution with mean US$88,000 and standard...

The incomes of a population of lawyers have a normal distribution with mean US$88,000 and standard deviation US$60,000. Forty-nine lawyers are selected at random from the above population to serve as a sample in a research project. Use the z for a sample mean formula and determine the probability that the one sample of 49 lawyers drawn at random will have a mean salary of US$90,000 or greater.

a. .03

b. .23

c. -.03

d.-.23

e. .5120

f..4880

g. .5910

h. .4090

Solutions

Expert Solution

The incomes of a population of lawyers have a normal distribution with mean = $88,000 and a standard deviation = $60,000. n = 49 lawyers are selected at random from the above population to serve as a sample in a research project.

Using the Z for a sample mean formula we determine the probability that the one sample of 49 lawyers drawn at random will have a mean salary of M = $90,000 or greater.

so, Z score is calculated as:

So, P( >=90000) =P(Z>0.233), thus the probability can be computed using excel formula or the Z table shown below, the formula used is =1-NORM.S.DIST(0.23, TRUE), this results in probability =0.4090.

h. 0.4090

The Z table used is shown below:

orchestra answered 2 years ago
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