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Company X is considering employing one of the two training programs. Group 1 was trained by program A, group 2 by program C. For the first group, the time required to train the employees had an average of of 32.11 hours and a variation of 68.09 hours, while for the second group, the mean is 19.75 hour while the variation is 71.14. Which training program has less relative variability in its performance? Program A (18.97 percent) Program B (19.87 percent) Program A (19.87 percent) Program B (18.97 percent)

Question

Company X is considering employing one of the two training programs. Group 1 was trained by program A, group 2 by program C. For the first group, the time required to train the employees had an average of of 32.11 hours and a variation of 68.09 hours, while for the second group, the mean is 19.75 hour while the variation is 71.14. Which training program has less relative variability in its performance? Program A (18.97 percent) Program B (19.87 percent) Program A (19.87 percent) Program B (18.97 percent)

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Solution

The relative variability of a data set is measured by the coefficient of variation (CV), which is the ratio of the standard deviation to the mean, expressed as a percentage. The lower the CV, the less variability there is relative to the mean.

First, we need to calculate the standard deviation for each group. The standard deviation is the square root of the variance.

For Group 1 (Program A): Variance = 68.09 hours Standard deviation = sqrt(68.09) = 8.25 hours

For Group 2 (Program C): Variance = 71.14 hours Standard deviation = sqrt(71.14) = 8.43 hours

Next, we calculate the CV for each group.

For Group 1 (Program A): CV = (Standard deviation / Mean) * 100 CV = (8.25 / 32.11) * 100 = 25.68%

For Group 2 (Program C): CV = (Standard deviation / Mean) * 100 CV = (8.43 / 19.75) * 100 = 42.68%

Based on these calculations, Program A has less relative variability in its performance. However, the percentages provided in the question (18.97% and 19.87%) do not match the calculated values. It's possible there may be a mistake in the question or the provided data.

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