At a large manufacturing company managers are given ratings based upon a multitu

Question

At a large manufacturing company managers are given ratings based upon a multitude of metrics higher overall scores mean better performance. This year these performance scores are normally distributed with an average of 560 and a standard deviation of 80. a. Draw the normal distribution of performance scores. Label the mean, plusminus 1 sigma, plusminus 2 sigma, and plusminus 3 sigma. You can do this by hand, using R (see code in Data Analysis 1 in Canvas) or using the Statistics Interactives App 2. b. What proportion of performance scores are above 600? Use the z table and include your calculation of the z score. You may verify your answer with R or the Statistics Interactive App. c. Those with a performance score above the 90^th percentile will receive a promotion. What is the performance score cutoff for promotion?

Explanation / Answer

Solution

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then, pdf of X, f(x) = {1/(2)}e^-[(1/2){(x - µ)/}2] …………………………….(A)

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

The numerical values of pdf are available in standard tables of Normal Ordinates.

Now, to work out solution,

Let X = Performance score. Then, we are given X ~ (µ, 2), where µ = 560 and = 80.

Part (a)

Data for drawing the curve

Score, x along

x-axis

240

320

400

480

560

640

720

800

880

f(x) along

y-axis

0.0001

0.0044

0.0540

0.2420

0.3989

0.2420

0.0540

0.0044

0.0001

[y-values are obtained from Normal Ordinates Tables]

Part (b)

Proportion of performance score above 600 = P(X > 600)

= P[Z > {(600 - 560)/80}] [vide (2) under Back-up Theory]

= P(Z > 0.5) = 0.3085 [using Excel Function] ANSWER

Part (c)

If t be the 90th percentile, then, by definition, P(X > T) = 0.1

=> [vide (2) under Back-up Theory], P[Z > {(t - 560)/80}] = 0.1

=> (t - 560)/80 = 1.2815 => t = 662.52 [using Excel Function]

So, the cut-off score for promotion = 662.52 ANSWER

Score, x along

x-axis

240

320

400

480

560

640

720

800

880

f(x) along

y-axis

0.0001

0.0044

0.0540

0.2420

0.3989

0.2420

0.0540

0.0044

0.0001

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