29 November, 07:06

# The mean annual salary for intermediate level executives is about \$74000 per year with a standard deviation of \$2500. A random sample of 50 intermediate level executives is selected. What is the probability that the mean annual salary of the sample is between \$71000 and \$73500? A 0.079B. 0.500C. 0.487D. 0.306

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1. 29 November, 10:40
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Step-by-step explanation:

Assuming a normal distribution for the annual salary for intermediate level executives, the formula for normal distribution is expressed as

z = (x - u) / s

Where

x = annual salary for intermediate level executives

u = mean annual salary

s = standard deviation

From the information given,

u = \$74000

s = \$2500

We want to find the probability that the mean annual salary of the sample is between \$71000 and \$73500. It is expressed as

P (71000 lesser than or equal to x lesser than or equal to 73500)

For x = 71000,

z = (71000 - 74000) / 2500 = - 1.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.1151

For x = 73500,

z = (73500 - 74000) / 2500 = - 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.4207

P (71000 lesser than or equal to x lesser than or equal to 73500) is

0.4207 - 0.1151 = 0.306