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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).

User Moonbeam
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1 Answer

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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
$\sigma_x$ :


$\sigma_x = (\sigma)/(\sqrt n)$


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

= 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.


$z-\text{score} =(\overline x - \mu)/(\sigma - √(n))$


$z-\text{score} =(80000-87500)/(26000 - √(63))$

= -2.2896

≈ -2.29

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

= 0.9890

User Bobby Woolf
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