Question

Difficulty: EasyMeasures of Dispersion

A library recorded the number of books borrowed over five consecutive days as 22, 44, 55, 66, and 88. What is the standard deviation of the number of books borrowed?

  1. 22Answer
  2. B
    44
  3. C
    55
  4. D
    66

Answer

The standard deviation of the number of books borrowed is 22.
The mean of the five data points is 55. The sum of the squared deviations from the mean is 9+1+0+1+9=209 + 1 + 0 + 1 + 9 = 20. Dividing by 55 gives a variance of 44. Taking the square root of 44 yields the standard deviation of 22.

Step-by-Step Solution

1
Calculate the arithmetic mean (xˉ\bar{x}) of the given dataset.
xˉ=2+4+5+6+85=255=5\bar{x} = \frac{2 + 4 + 5 + 6 + 8}{5} = \frac{25}{5} = 5
The mean is required to determine the deviations of each data point.
2
Find the deviations from the mean (xxˉ)(x - \bar{x}) and square each deviation (xxˉ)2(x - \bar{x})^2.
(25)2=9,(45)2=1,(55)2=0,(65)2=1,(85)2=9(2-5)^2 = 9, \quad (4-5)^2 = 1, \quad (5-5)^2 = 0, \quad (6-5)^2 = 1, \quad (8-5)^2 = 9
Squaring ensures all deviation values are positive before summation.
3
Calculate the variance (mean of squared deviations).
Variance (σ2)=9+1+0+1+95=205=4\text{Variance } (\sigma^2) = \frac{9 + 1 + 0 + 1 + 9}{5} = \frac{20}{5} = 4
Variance measures the average squared spread around the mean.
4
Take the square root of the variance to find the standard deviation.
Standard Deviation (σ)=4=2\text{Standard Deviation } (\sigma) = \sqrt{4} = 2
Standard deviation expresses dispersion in the original unit of measurement.

Key Concept

Standard Deviation of Ungrouped Data
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