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6 December, 12:21

A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the likelihood that people were absent 0, 1, 2, 3, 4, or 5 days last month.

Number of Days Absent Probability

0 0.60

1 0.20

2 0.12

3 0.04

4 0.04

5 0.00

What is the variance of the number of days absent? Select one:

a. 1.1616

b. 1.41

c. 5.00

d. 55.52

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Answers (1)
  1. 6 December, 12:30
    0
    a. 1.1616

    Explanation:

    For computing the variance, first we have to determine the mean which is shown below:

    Mean = Number of Days * Absent Probability

    = 0 * 0.60 + 1 * 0.20 + 2 * 0.12 + 3 * 0.04 + 4 * 0.04

    = 0 + 0.20 + 0.24 + 0.12 + 0.16

    = 0.72

    Now the variance equal to

    = (Number of Days Absent - Mean) ^2 * Probability

    So,

    = (0 - 0.72) ^2 * 0.60 + (1 - 0.72) ^2 * 0.20 + (2 - 0.72) ^2 * 0.12 + (3 - 0.72) ^2 * 0.04 + (4 - 0.72) ^2 * 0.04

    = 1.1616
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