Don't drink and drive: A highway safety council reported that there were 3972 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. Age Number of Fatalities 11-20 340 21-30 1527 31-40 866 41-50 693 51-60 423 61-70 123 Send data to Excel The mean is approximately .

[tex]\begin{array}{cc}{Age} & {Fatalities} & {11 - 20} & {340} & {21-30} & {1527} & {31-40} & {866} & {41-50} & {693} & {51-60} & {423} & {61-70} & {123} \ \end{array}[/tex]

Required

Calculate the mean

The given data is a grouped data. So, we need to calculate the class mid -point first.

This is done by calculating the average of the class intervals.

So, we have:

[tex]\begin{array}{ccc}{Age} & {Fatalities} & {x} & {11 - 20} & {340} & {15.5} & {21-30} & {1527} &{25.5}& {31-40} & {866} & {35.5} & {41-50} & {693} & {45.5} & {51-60} & {423} & {55.5} & {61-70} & {123} & {65.5}\ \end{array}[/tex]

For 11 - 20, the midpoint is: [tex]x =\frac{1}{2}(11+20) = \frac{1}{2}*31=15.5[/tex]

For 21 - 30, the midpoint is: [tex]x =\frac{1}{2}(21+30) = \frac{1}{2}*51=25.5[/tex]

.....

For 61 - 70, the midpoint is: [tex]x =\frac{1}{2}(61+70) = \frac{1}{2}*131=65.5[/tex]

The mean is then calculated as:

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

[tex]\bar x = \frac{(15.5 * 340) + (25.5 * 1527) + (35.5 * 866) + (45.5 * 693) + (55.5 * 423) + (65.5 * 123)}{340 + 1527 + 866 + 693 + 423 + 123}[/tex]

[tex]\bar x = \frac{138016}{3972}[/tex]

[tex]\bar x = 34.7472306143[/tex]

[tex]\bar x = 34.7[/tex] --- approximated

The mean is approximately 34.7