Question

Data were collected on the ages, in years, of the men and women enrolled in a large sociology course. Let the random variables M and W represent the ages of the men and women, respectively. The distribution of M has mean 20.7 years and standard deviation 1.73 years. The distribution of W has mean 20.2 years and standard deviation 1.60 years. Of all of those enrolled in the course, 54 percent are men and 46 percent are women. What is the mean age of the combined distribution of both men and women in the course?

Answer 1

Answer:

The mean age of the combined distribution of both men and women in the course is:

20.47 years.

Explanation:

a) Data and Calculations:

Random variable M = age of men

Random variable W = age of women

Distribution of                    M:                 W:

Mean =                            20.7 years     20.2 years

Standard Deviation =       1.73 years      1.60 years

Percentage of enrolled   54%                46%

Mean Age Proportion     11.178              9.292

Combined mean age =  20.47 (11.178 + 9.292)

b) Mean Age Proportion:

M = 20.7 * 54% =   11.178

W = 20.2 * 46% = 9.292

Total                     20.47

Answer 2

The mean age of the combined distribution of both men and women in the course is 20.47 years

• Mean age of men , M = 20.7 years

• Mean age of Women , W = 20.2 years

• Percentage of Men = 54%

• Percentage of Women = 46%

Recall, :

• Mean, μ = np

The mean age of men in the distribution = (20.7 × 54%) = 11.178 years

The mean age of women in the distribution = (20.2 × 46%) = 9.292 years

The mean age of the combined distribution is :

• 11.178 + 9.292 = 20.47 years

Therefore, the mean age of the combined distribution is 20.47 years

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