Example 11

EXAMPLE 11Calculating the variance and standard deviation of a discrete random variable

The probability distribution for the number of credits is repeated here as Table 5. In Example 7, we calculated the mean number of credits as $μ = 15$ . Calculate the variance and standard deviation.

Table 6.10: Table 5Probability distribution of the number of credits

$X = number of credits taken$	$P (X)$
12	0.1
13	0.1
14	0.1
15	0.5
16	0.1
20	0.1

Solution

Refer to Table 6. The first two columns correspond to the probability distribution of $X = number of credits taken$ . The third column represents the calculations needed to find ${(X - μ)}^{2} \cdot P (X)$ . Summing the values in the rightmost column provides the variance $σ^{2} = 4$ . Taking the square root of the variance gives us the standard deviation $σ = \sqrt{σ^{2}} = \sqrt{4} = 2$ credits.

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credits

Table 6.11: Table 6Calculating

$σ^{2}$ and

$σ$

$X$	$P (X)$	${(X - μ)}^{2} \cdot P (X)$
12	0.1	${(12 - 15)}^{2} \cdot 0.1 = 0.9$
13	0.1	${(13 - 15)}^{2} \cdot 0.1 = 0.4$
14	0.1	${(14 - 15)}^{2} \cdot 0.1 = 0.1$
15	0.5	${(15 - 15)}^{2} \cdot 0.5 = 0.0$
16	0.1	${(16 - 15)}^{2} \cdot 0.1 = 0.1$
20	0.1	${(20 - 15)}^{2} \cdot 0.1 = 2.5$
		$σ^{2} = {\sum (X - μ)}^{2} \cdot P (X) = 4$

NOW YOU CAN DO

Exercises 53a,b–60a,b.