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).
Solution :
Given data:
Mean, μ = $87,500
Standard deviation, σ = $26,000
Sample number, n = 63
a). The value of [tex]$\mu_{x}$[/tex] :
[tex]$\mu_x=\mu$[/tex]
= 87,500
b). The value of [tex]$\sigma_x$[/tex] :
[tex]$\sigma_x = \frac{\sigma}{\sqrt n}$[/tex]
[tex]$\sigma_x = \frac{26000}{\sqrt {63}}$[/tex]
= 3275.69
c). The shape of the sampling distribution is that of a normal distribution (bell curve).
d). The value z-score for the value =80,000.
[tex]$z-\text{score} =\frac{\overline x – \mu}{\sigma – \sqrt{n}}$[/tex]
[tex]$z-\text{score} =\frac{80000-87500}{26000 – \sqrt{63}}$[/tex]
= -2.2896
≈ -2.29
e). P(x > 80000) = P(z > -2.2896)
= 0.9890