Probability
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#1.) Dr.Radman has been collecting data on a simplified intake form. One measure that was taken was the amount of time (in minutes) it took people to complete this form. The time to complete the simplified form was Mean=22 minutes and Standard Deviation= 3 minutes (assume there were enough forms completed to justify using population parameters). Dr. Radman has randomly selected some scores and wants to transform them to the time it would have taken to complete the original form. That version had a Mean=50 minutes and a Standard Deviation=10 minutes. Transform the times for the simplified score completion times so they would correspond to the original form completion times for Dr.Radman. Simplified form times: 20,17,28.
#2.) You are the supervisor of your department and have just received the rankings for your department. however, the report failed to include the mean and standard deviation for your entire workplace. You really want to know more information about the distribution for the general workplace. The report indicates that a score of X=65 corresponds to a z-score of a=+2.00, and a score of X=44 corresponds to a z-score of z=-1.50.
a.) indicate the mean and standard deviation for this distribution.
b.) If your department received a z-score of 2.50, indicate the score your department received.
c.) You are going to determine the probability of your department achieving this score, what assumptions do you have to make about this distribution?
d.) Indicate the probability of your department achieving that score and explain what this score indicates about your department rating.
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Solution Summary
The solution contains certain problems using the normal probability law. The determination of mean and variance of a normal distribution from z-score and calculation of original data values from z-score are discussed in the solution.
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Solution 1
Let X denote the simplified form completion time and Y denote the original form completion time.
Here X is assumed to have a Normal distribution with mean 22 and s.d. 3. So the corresponding z-score is Z = (X-22)/3.
Also Y is assumed to have a Normal distribution with mean 50 and s.d. 10. So the corresponding z-score is Z = (Y-50)/10.
Equating the z-sores we have,
(X-22)/3 = ...
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