Examkin

Question

Difficulty: MediumMeasures of Dispersion

Five daily rainfall measurements (in mm) recorded in a city are 3,6,7,9,3, 6, 7, 9, and 1515. What is the standard deviation of these rainfall measurements?

  1. A
    3.23.2
  2. 44Answer
  3. C
    1616
  4. D
    2020

Answer

The standard deviation of the rainfall measurements is 44.
First compute the mean of the data set: xˉ=3+6+7+9+155=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8. Next, sum the squared deviations from the mean: (3−8)2+(6−8)2+(7−8)2+(9−8)2+(15−8)2=25+4+1+1+49=80(3-8)^2 + (6-8)^2 + (7-8)^2 + (9-8)^2 + (15-8)^2 = 25 + 4 + 1 + 1 + 49 = 80. Dividing by the number of observations (55) yields the variance σ2=805=16\sigma^2 = \frac{80}{5} = 16. Taking the square root gives the standard deviation σ=16=4\sigma = \sqrt{16} = 4.

Step-by-Step Solution

1
Calculate the arithmetic mean (xˉ\bar{x}) of the dataset
xˉ=3+6+7+9+155=405=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = \frac{40}{5} = 8
The mean is needed to find deviations for each data point.
2
Find the deviation of each number from the mean and square it
(3−8)2=25(3-8)^2 = 25, (6−8)2=4(6-8)^2 = 4, (7−8)2=1(7-8)^2 = 1, (9−8)2=1(9-8)^2 = 1, (15−8)2=49(15-8)^2 = 49
Squaring ensures all deviation values are non-negative.
3
Sum the squared deviations and divide by the total number of items (n=5n = 5) to find the variance
\text{Variance } (\sigma^2) = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16
Variance measures the average squared distance from the mean.
4
Take the square root of the variance to obtain standard deviation
\text{Standard Deviation } (\sigma) = \sqrt{16} = 4
Standard deviation returns the dispersion measure back to the original units.

Key Concept

Standard Deviation of Ungrouped Data
Rate this question

Topics

Measures of Dispersion

More questions from this topic