Question

The average salary for a certain profession is $87,500. assume that the standard deviation of such salaries is $26,000. Consider a random sample of 63 people in this profession and let xbar represent the mean salary for the sample.a. What is ?
b. What is ?c. Describe the shape of the sampling distributio of ?
d. Find the z-score for the value =80,000.
e. Find P( > 80,000).

Answer 1

Solution :

Given data:

Mean, μ = $87,500

Standard deviation, σ =  $26,000

Sample number, n = 63

a). The value of $\mu_{x}$ :

   $\mu_x=\mu$

       = 87,500

b). The value of $\sigma_x$ :

    $\sigma_x = \frac{\sigma}{\sqrt n}$

   $\sigma_x = \frac{26000}{\sqrt {63}}$

        = 3275.69

c). The shape of the sampling distribution is that of a normal distribution (bell curve).

d). The value z-score for the value =80,000.

   $z-\text{score} =\frac{\overline x - \mu}{\sigma - \sqrt{n}}$

  $z-\text{score} =\frac{80000-87500}{26000 - \sqrt{63}}$

                = -2.2896

                ≈ -2.29

e). P(x > 80000) = P(z > -2.2896)

                           = 0.9890