Question

A hotel downtown is trying to implement an employee recognition program based on a standardized problem solving test. The company administering the test indicated that the scores are normally distributed with a mean of 82 points, and a variance of 16. The hotel has decided that employees who score in the bottom 7.5% of the test scores will not receive any additional benefits. The manager would like to know:

a. The probability that an employee would score between 70 and 78 points

b. What cutoff score on the test should the hotel use to not give any additional benefits?

Answer 1

GIVEN:

Let X be the scores of employee problem solving test which follow normal distribution with a mean points, and a variance . That is the standard deviation

(a) PROBABILITY THAT AN EMPLOYEE WOULD SCORE BETWEEN 70 AND 78 POINTS:

To calculate the probability we convert the raw score (X) into standard score (Z) using the formula given by,

The probability that an employee would score between 70 and 78 points is,

  

  

  

{Since }

Uisng the z table, the first probability value is the value with corresponding row -1.0 and column 0.00 and the second probability value is the value with corresponding row -3.0 and column 0.00.

  

  

The probability that an employee would score between 70 and 78 points is .

(b) CUT OFF SCORE HOTEL USE TO NOT GIVE ADDITIONAL BENEFITS:

Given that employees who score in the bottom 7.5% of the test scores will not receive any additional benefits.

Let us find the z value corresponding to the probability 0.075 using the z table.

The z score is where the row value is -1.4 and column value is 0.04.

Now the cutoff score on the test should the hotel use to not give any additional benefits is calculated using,

Thus the cutoff score on the test should the hotel use to not give any additional benefits is .