Almost all companies utilize some type of year-end performance review for their employees. Human Resources (HR) at the University of Texas H

Question

Almost all companies utilize some type of year-end performance review for their employees. Human Resources (HR) at the University of Texas Health Science Center provides guidelines for supervisors rating their subordinates. For example, raters are advised to examine their ratings for a tendency to be either too lenient or too harsh. According to HR, “if you have this tendency, consider using a normal distribution—top 10% of employees are rated as exemplary. Assume the ratings follow a normal distribution with a mean of 50 and a standard deviation of 15. a. What is the lowest rating you should give to an “exemplary” employee if you follow the University of Texas HR guidelines?b. B. What is the lowest rating you should give to a? “competent” employee if you follow the? university’s HR?guidelines?

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Cora 2 weeks 2022-01-06T04:01:24+00:00 1 Answer 0 views 0

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    2022-01-06T04:03:23+00:00

    Answer:

    a) a=50 +1.28*15=69.2

    So the value of height that separates the bottom 90% of data from the top 10% is 69.2.  

    b) a=50 -0.524*15=42.14

    So the value of height that separates the bottom 30% of data from the top 70% is 42.14.

    Step-by-step explanation:

    Assuming this complete question:

    “Almost all companies utilize some type of year-end performance review for their employees. Human Resources(HR) at the university of Texas health Science Center provides guidelines for supervisors rating their subordinates. For example raters are advised to examine their ratings for a tendency to be too lenient or too harsh. According to HR, if you have this tendency, consider using a normal distribution-10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable.

    Suppose you are rating an employees performance on a scale of 1(lowest) to 100(highest). Also assume the ratings follow normal distribution with a mean of 50 and a standard deviation of 15.

    a)What is the lowest rating you should give to an exemplary employee if you follow the Univ. of Texas HR guidelines?

    b) What is the lowest rating you should give to an competent employee if you follow the Univ. of Texas HR guidelines?”

    Solution to the problem

    Previous concepts

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    The Z-score is “a numerical measurement used in statistics of a value’s relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean”.  

    Part a

    Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

    X \sim N(50,15)  

    Where \mu=50 and \sigma=15

    For this part we want to find a value a, such that we satisfy this condition:

    P(X>a)=0.1   (a)

    P(X<a)=0.9   (b)

    Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

    As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it’s z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

    If we use condition (b) from previous we have this:

    P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90  

    P(z<\frac{a-\mu}{\sigma})=0.90

    But we know which value of z satisfy the previous equation so then we can do this:

    z=1.28<\frac{a-50}{15}

    And if we solve for a we got

    a=50 +1.28*15=69.2

    So the value of height that separates the bottom 90% of data from the top 10% is 69.2.  

    Part b

    For this part we want to find a value a, such that we satisfy this condition:

    P(X>a)=0.7   (a)

    P(X<a)=0.3   (b)

    Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

    As we can see on the figure attached the z value that satisfy the condition with 0.3 of the area on the left and 0.7 of the area on the right it’s z=-0.524. On this case P(Z<-0.524)=0.3 and P(z>-0.524)=0.7

    If we use condition (b) from previous we have this:

    P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.30  

    P(z<\frac{a-\mu}{\sigma})=0.30

    But we know which value of z satisfy the previous equation so then we can do this:

    z=-0.524<\frac{a-50}{15}

    And if we solve for a we got

    a=50 -0.524*15=42.14

    So the value of height that separates the bottom 30% of data from the top 70% is 42.14.  

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