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Part 2.(6 polnts) Normal Distrlbutlons Some companies "grade on a bell curve" to compare the performance of their managers and professional workers. This forces the use of some low performance ratings so that not all workers are listed as "above average." Ford Motor Company's "performance management process" for this year assigned 10% A grades, 80% B grades, and 10% C grades to the company's managers. Suppose Ford's performance scores really are Normally distributed. This year, managers with scores less than 150 received C grades and those with scores of at least 356 received A grades. Performance Scores 10% 80% 10% (2 point) What are the z scores associated with the 10th and 90th percentiles from the standard normal distribution? a. 150 356 b. (2 point) What is the mean and standard deviation of the performance scores? Show work. Winter 2019 ST 314 Data Analysis 02 (2 point) Suppose the company adds grades D and Fso there are 5 categories to grade performance. If they want to give A's only to those in the top 396, what performance score must a manager exceed to get an A? c.

Answer 1

Solution:-

a) The z-scores for 10th and 90th percentile is -1.282 and 1.282.

b) The mean and standard deviation of the performance score is 253 and 80.343.

p-value for the 10th percentile = 0.10

z-score for the p-value = -1.282

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

$-1.282\ast \sigma + \mu =150$...................................(1)

p-value for the 90th percentile = 0.90

z-score for the p-value = 1.282

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

$1.282\ast \sigma + \mu =356$.....................................(2)

By solving (1) and (2) we get:-

$\mu =253,\sigma =80.343$

c) The performance score manager must exceed to get an A is 404.125.

p-value for the top 3% = 1 - 0.03 = 0.97

z-score for the p-value = 1.881

By applying normal distribution:-

$z = \frac{x-\mu }{\sigma }$

x = 404.125