Question

The average salary for a certain profession is ​$75,500. Assume that the standard deviation of such...

The average salary for a certain profession is ​$75,500. Assume that the standard deviation of such salaries is $30,500. Consider a random sample of 71 people in this profession and let x over bar x represent the mean salary for the sample.

a.

What is

mu Subscript x over barμ x ​?

mu Subscript x overbarμxequals=?

b.

What is sigma Subscript x over bar σ x​?

sigma Subscript x over bar σ x=________​(Round to two decimal places as​ needed.)

c.

Describe the shape of the sampling distribution of

x over bar x.

A. The shape is that of a uniform distribution.

B. The shape is that of a poisson distribution.

C. The shape is that of a normal distribution.

D.The shape is that of a binomial distribution.

d.

Find the​ z-score for the value x over bar x =68,000.

z=____________​(Round to two decimal places as​ needed.)

e.

Find Px>68,000 = ________(Round to three decimal places as​ needed.

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Answer #1

Answer

(a) mu is the population mean, which is given in the question. So, we can write it as mu = 75,500

(b) sigma is the population standard deviation, which is given in the question. So, we can write it as sigma = 30,500

(c) Option C is correct because data is normally distributed with mean value 75500 and standard deviation = 30500

(d) Formula for z score is given as z = (ar{x}-mu)/(sigma/sqrt{n})

where mu = 75,500, sigma = 30,500, n = 71 and ar{x} = 68000

setting the given values, we get

(68000-75500)/(30500/V71) -7500/3619 .68ーー2.07 (rounded to two decimals)

(e) Now, we have to find the value of P(x > 68000) = P(z >-2.07)

Using the identity P(z>-a)= P(z<a)

we can write it as

P(z > 68000) = P(z >-2.07) = P(z < 2.07) = 0.981 (using z distribution table)

(we find 2.00 in the left most column and 0.07 in the top most row in z distribution, then selected the intersection value)

So, p value is 0.981 (rounded to 3 decimals)

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