HumanPower, the temporary employment agency, has tested many people’s data entry skills. InfoTech needs a data entry person, and the person needs to be not only quick but also consistent. HumanPower pulls the speed records for 4 employees with the data given below in terms of number of correct entries per minute. Which employee is best for InfoTech based on relative dispersion?
John 63 66 68 62 69 72
Jeff 68 67 66 67 69
Mary 62 79 75 59 72 84
Tammy 64 68 58 57 59
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To determine which employee is best for InfoTech based on relative dispersion, we need to calculate the coefficient of variation (CV) for each employee.
The coefficient of variation (CV) formula is:
CV = (Standard Deviation / Mean) * 100
Let's calculate the CV for each employee:
John:
Mean = (63 + 66 + 68 + 62 + 69 + 72) / 6 = 66.5
Standard Deviation = 3.53
CV = (3.53 / 66.5) * 100 = 5.3%
Jeff:
Mean = (68 + 67 + 66 + 67 + 69) / 5 = 67.4
Standard Deviation = 1.14
CV = (1.14 / 67.4) * 100 = 1.69%
Mary:
Mean = (62 + 79 + 75 + 59 + 72 + 84) / 6 = 71.8
Standard Deviation = 8.9
CV = (8.9 / 71.8) * 100 = 12.36%
Tammy:
Mean = (64 + 68 + 58 + 57 + 59) / 5 = 61.2
Standard Deviation = 4.48
CV = (4.48 / 61.2) * 100 = 7.32%
Comparing the CV values:
- John has a CV of 5.3%
- Jeff has a CV of 1.69%
- Mary has a CV of 12.36%
- Tammy has a CV of 7.32%
The employee with the lowest CV is Jeff, with a relative dispersion of 1.69%. Therefore, Jeff would be the best choice for InfoTech based on relative dispersion.
To determine which employee is best for InfoTech based on relative dispersion, we need to calculate the coefficient of variation (CV) for each employee. The coefficient of variation is a measure of relative dispersion and shows the variability relative to the mean.
Here is how you can calculate the coefficient of variation for each employee:
1. Calculate the mean (average) of the data set for each employee. To do this, add up all the values and divide by the total number of values.
For John:
Mean = (63 + 66 + 68 + 62 + 69 + 72) / 6 = 400 / 6 = 66.67
For Jeff:
Mean = (68 + 67 + 66 + 67 + 69) / 5 = 337 / 5 = 67.4
For Mary:
Mean = (62 + 79 + 75 + 59 + 72 + 84) / 6 = 431 / 6 = 71.83
For Tammy:
Mean = (64 + 68 + 58 + 57 + 59) / 5 = 306 / 5 = 61.2
2. Calculate the standard deviation for each employee. The standard deviation measures the amount of variation or dispersion in the data set.
To calculate the standard deviation, you can use the following formula:
Standard Deviation = √(Σ(xi - mean)^2 / n)
Where:
- Σ(xi - mean)^2: Sum of the squared differences between each value and the mean
- n: Total number of values
For John:
Standard Deviation = √((63-66.67)^2 + (66-66.67)^2 + (68-66.67)^2 + (62-66.67)^2 + (69-66.67)^2 + (72-66.67)^2) / 6
= √(10.67 + 0.11 + 0.45 + 16.01 + 4.45 + 21.89) / 6
= √(53.58) / 6
= √(8.93) ≈ 2.99
For Jeff:
Standard Deviation = √((68-67.4)^2 + (67-67.4)^2 + (66-67.4)^2 + (67-67.4)^2 + (69-67.4)^2) / 5
= √(0.16 + 0.16 + 1.04 + 0.16 + 2.56) / 5
= √(4.08) / 5
= √(0.82) ≈ 0.91
For Mary:
Standard Deviation = √((62-71.83)^2 + (79-71.83)^2 + (75-71.83)^2 + (59-71.83)^2 + (72-71.83)^2 + (84-71.83)^2) / 6
= √(80.57 + 50.41 + 12.70 + 158.73 + 0.03 + 146.96) / 6
= √(449.40) / 6
= √(74.90) ≈ 8.66
For Tammy:
Standard Deviation = √((64-61.2)^2 + (68-61.2)^2 + (58-61.2)^2 + (57-61.2)^2 + (59-61.2)^2) / 5
= √(7.84 + 43.56 + 13.44 + 17.64 + 4.84) / 5
= √(87.32) / 5
= √(17.46) ≈ 4.18
3. Calculate the coefficient of variation (CV) for each employee using the formula:
Coefficient of Variation = (Standard Deviation / Mean) * 100
For John:
CV = (2.99 / 66.67) * 100 ≈ 4.48
For Jeff:
CV = (0.91 / 67.4) * 100 ≈ 1.35
For Mary:
CV = (8.66 / 71.83) * 100 ≈ 12.04
For Tammy:
CV = (4.18 / 61.2) * 100 ≈ 6.83
Based on the calculated coefficient of variation, the employee with the lowest relative dispersion (CV) is Jeff with a CV of 1.35. Therefore, Jeff would be the best choice for InfoTech.