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The annual salaries of all employees at a financial company are normally distributed with a mean Mu = $34,000 and a standard deviation Sigma = $4,000. What is the z-score of a company employee who makes an annual salary of $28,000? |
| Answer» CORE of a company employee who makes an annual salary of $28,000 is -1.5.Step-by-step explanation:We are given that the annual salaries of all employees at a financial company are normally DISTRIBUTED with a mean MU = 34,000 and a standard DEVIATION Sigma =34,000andastandarddeviationSigma=4,000.Let X = annual salaries of all employees at a financial companySO, X ~ Normal(\mu=\$34,000 , \sigma^{2} = \$4,000^{2}μ=$34,000,σ 2 =$4,000 2 )The z score probability distribution for the normal distribution is given by; Z = \frac{X-\mu}{\sigma} σX−μ ~ N(0,1)where, \muμ = mean annual salary of emloyees = $34,000 \sigmaσ = standard deviation = $4,000Now, the z-score of a company employee who makes an annual salary of $28,000 is given by ; Z = \frac{28,000-34,000}{4,000} 4,00028,000−34,000 = \frac{-6}{4} 4−6 = -1.5So, the required z-score is -1.5. | |