Business, 17.07.2020 01:01 perezsamantha3oqr0za

Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 40 manufacturing companies. Estimate the mean, median, and standard deviation of advertising expense. Advertising Expenditure ($millions) Number of companies

$20 to under $30 9

30 to under 40 13

40 to under 50 21

50 to under 60 18

60 to under 70 14

Total 75

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Answer from: masonbugatti3477

Mean = 47

Median = 47.38

Standard Deviation = 12.73

Explanation:

Note: You wrote " 40 manufacturing companies, but the total number of companies you actually listed is 75, definitely you meant 75.

Let y represent the range of advertising expenditure, f represent the number of companies, x represent the midpoint of the range of advertising expenditure.

y f x fx fx²

$20 to under $30 9 25 225 5625

$30 to under $40 13 35 455 15925

$40 to under $50 21 45 945 42525

$50 to under $60 18 55 990 54450

$60 to under $70 14 65 910 59150

n = 75 $\sum fx = 3525$

$\sum fx^2 = 177675$

Mean, $\bar{X} = \frac{\sum fx}{n}$

$\bar{X} = \frac{3525}{75} \\\bar{X} = 47$

Standard Deviation:

$SD = \sqrt{\frac{n \sum fx^2 - (\sum fx)^2}{n(n-1)} } \\SD = \sqrt{\frac{(75*177675) - (3525)^2}{75(75-1)} }\\SD = 12.73$

Median:

Get the cumulative frequencies(cf)

y f cf

$20 to under $30 9 9

$30 to under $40 13 22

$40 to under $50 21 43

$50 to under $60 18 61

$60 to under $70 14 75

N = 75

Median = Size of (N/2)th item

Median = Size of (75/2)th item

Median = Size of (37.5)th item

The median class = 40 to under 50

Lower limit, L₁ = 40

Cumulative frequency, cf = 22

f = 21

Class Width, h = 10

Median = $L_1 + \frac{ (N/2) - cf}{f} * h\\$

Median = $40 + \frac{ (75/2) - 22}{21} * 10\\$

Median = 47.38

Answer from: Quest

a is the correct answer, tell me if im right

Answer from: Quest

im thinking a but im not 100% sure

explanation:

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