Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 60 manufacturing companies located in the Southwest. Estimate the mean and the standard deviation of advertising expenses. (Round squared deviations to nearest whole number and final answers to 2 decimal places.) Advertising Expenditure ($ millions) Number of Companies 25 up to 35 5 35 up to 45 10 45 up to 55 21 55 up to 65 16 65 up to 75 8 Total 60
Mean
Standard deviation

Question

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MrRoyal MrRoyal · Answer 1 · 2021-03-05T03:41:53+01:00

Answer:

[tex]\bar x = 53.5[/tex]

[tex]SD = 4.21[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{cc}{'\$\ Million}&{Companies}&25-35&5&35-45&10&45-55&21&55-65&16&65-75&8&Total&60\end{array}[/tex]

Solving (a): The mean

First, we calculate the class midpoint by calculating the average of the intervals.

For instance:

The midpoint of 25 -35 is: [tex]x = \frac{1}{2}(25 +35) = \frac{1}{2} * 60 = 30[/tex]

So, the table becomes:

[tex]\begin{array}{ccc}{'\$\ Million}&{f}&x&25-35&5&{30}&35-45&10&40&45-55&21&50&55-65&16&60&65-75&8&70&Total&60\end{array}[/tex]

The mean is then calculated as:

[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]

[tex]\bar x = \frac{30 * 5 + 40 *10 + 50 * 21 + 60 * 16 + 70 * 8}{60}[/tex]

[tex]\bar x = \frac{3210}{60}[/tex]

[tex]\bar x = 53.5[/tex]

Solving (b): The standard deviation

This is calculated using:

[tex]SD = \sqrt{\frac{\sum (x - \bar x)^2}{\sum f}}[/tex]

So, we have:

[tex]SD = \sqrt{\frac{(30-53.5)^2+(40-53.5)^2+(50-53.5)^2+(60-53.5)^2+(70-53.5)^2}{60}[/tex]

[tex]SD = \sqrt{\frac{1061.25}{60}[/tex]

[tex]SD = \sqrt{17.6875[/tex]

[tex]SD = 4.20565096032[/tex]

[tex]SD = 4.21[/tex]