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The distribution of weekly salaries at a large company is reverse J-shaped with a mean of $1000 and a standard deviation of $370. Suppose you want to calculate the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 80 employees will be at most $75. Which distribution would you use to solve the problem?

Respuesta :

Answer:

7%

Step-by-step explanation:

Given Data:

mean =$1000

standard deviation s = $370

number of employee n 80

weekly salary at most of $75

the value of z can be determined

[tex]n = [z\times \frac{s}{E}]^2[/tex]

[tex]80 = [z \times \frac{370}{75}]^2[/tex]

[tex]\sqrt(80) = 4.93z[/tex]

z = 1.81

From standard table of z

P(z > 1.81) = 0.035

P(condition described) = 2*0.035 = 7%

Z DISTRIBUTION

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