A highway safety council reported that there were 3992 fatalities among drivers in auto accidents in a particular year. Following is a frequency distribution of their ages. Approximate the mean age. Round your answer to one decimal place.

Question

Ages 11-20 with a frequency of 327
ages 21-30 with a frequency of 1545
ages 31-40 with a frequency of 860
ages 41-50 with a frequency of 691
ages 51-60 with a frequency of 423
ages 61-70 with a frequency of 146

the mean is approximately [blank]

Answer 1

To calculate the mean age, we can use the midpoint of each age group, multiply it by the frequency of that age group, sum those products, and then divide by the total number of fatalities.

Here are the age groups and their midpoints:

• Ages 11-20: Midpoint = (11 + 20) / 2 = 15.5
• Ages 21-30: Midpoint = (21 + 30) / 2 = 25.5
• Ages 31-40: Midpoint = (31 + 40) / 2 = 35.5
• Ages 41-50: Midpoint = (41 + 50) / 2 = 45.5
• Ages 51-60: Midpoint = (51 + 60) / 2 = 55.5
• Ages 61-70: Midpoint = (61 + 70) / 2 = 65.5

Next, we multiply each midpoint by the corresponding frequency:

• Ages 11-20: \(15.5 \times 327 = 5088.5\)
• Ages 21-30: \(25.5 \times 1545 = 39472.5\)
• Ages 31-40: \(35.5 \times 860 = 30530.0\)
• Ages 41-50: \(45.5 \times 691 = 31480.5\)
• Ages 51-60: \(55.5 \times 423 = 23461.5\)
• Ages 61-70: \(65.5 \times 146 = 9583.0\)

Now, we sum these products:

\[ 5088.5 + 39472.5 + 30530.0 + 31480.5 + 23461.5 + 9583.0 = 109115.0 \]

Now, we find the mean age by dividing the total by the total frequency (which is 3992):

\[ \text{Mean age} = \frac{109115.0}{3992} \approx 27.4 \]

Thus, the mean age is approximately 27.4.