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The National Sporting Goods Association (NSGA) conducted a survey of the ages of people that purchased athletic footwear in 2009. The ages are summarized in the following relative frequency distribution. Assume the survey was based on 100 individuals.

Age of Purchaser Percent
Under 14 years old 20
14 to 17 years old 5
18 to 24 years old 10
25 to 34 years old 13
35 to 44 years old 15
45 to 64 years old 25
65 years old and over 12

Required:
a. Calculate the average age of this distribution. Use 10 as the midpoint of the first class and 75 as the midpoint of the last class.
b. Calculate the sample standard deviation.

Sagot :

MrRoyal

The sample deviation of a distribution is the square root of the variance

  • The average age of the distribution is 37.26
  • The sample standard deviation is 21.27

How to calculate the average age

Using a midpoint of 10 for the first class, and 75 for the last, we have the following frequency distribution:

Age      Percentage

10           20

15.5        5

21           10

29.5       13

39.5       15

54.5      25

75        12

The average age is then calculated as:

[tex]E(x) = \sum x * P(x)[/tex]

So, we have:

[tex]E(x) =10 * 20\% + 15.5* 5\% + 21 * 10\% + 29.5*13\% + 39.5*15\% + 54.5 * 25\% + 75 *12\%[/tex]

[tex]E(x) =37.26[/tex]

Hence, the average age of the distribution is 37.26

How to calculate the sample standard deviation

This is calculated using:

[tex]\sigma = \sqrt{E(x^2) - E(x)^2}[/tex]

Where:

[tex]E(x^2) =10^2 * 20\% + 15.5^2* 5\% + 21^2 * 10\% + 29.5^2*13\% + 39.5^2*15\% + 54.5^2 * 25\% + 75^2 *12\%[/tex]

Evaluate the products

[tex]E(x^2) =1840.845[/tex]

The equation becomes

[tex]\sigma = \sqrt{1840.845- 37.26^2}[/tex]

[tex]\sigma = 21.27[/tex]

Hence, the sample standard deviation is 21.27

Read more about average and sample standard deviation at:

https://brainly.com/question/15858152

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