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?
a 20.2 years
b 20.43 years
c 20.45 years
d 20.47 years
e 40.9 years

Explanation:

Step1: Assign weights to groups

Let the weight for men $w_M = 0.54$, weight for women $w_W = 0.46$

Step2: Calculate weighted mean

$\text{Combined Mean} = w_M \times \mu_M + w_W \times \mu_W$
$\text{Combined Mean} = 0.54 \times 20.7 + 0.46 \times 20.2$

Step3: Compute individual products

$0.54 \times 20.7 = 11.178$
$0.46 \times 20.2 = 9.292$

Step4: Sum the products

$11.178 + 9.292 = 20.47$

Answer:

D. 20.47 years