Question

Difficulty: HardData Representation and Charts

A continuous grouped frequency distribution consists of four class intervals: 10−1410 - 14, 15−2415 - 24, 25−2925 - 29, and 30−4430 - 44. The total frequency of the distribution is 160160, and the frequency of the class interval 25−2925 - 29 is 2222.

In a histogram representing this data, the height of the rectangle for the interval 15−2415 - 24 corresponds to a frequency density of 66. In a pie chart representing the same distribution, the sector angle for the class interval 30−4430 - 44 is 108∘108^\circ.

What is the frequency density of the class interval 10−1410 - 14?

Answer: 6

Answer

The frequency density of the class interval 10−1410 - 14 is 66.
By converting the pie chart sector angle of 108∘108^\circ into a frequency of 4848 out of 160160, and using the frequency density of 66 with class width 1010 to find a frequency of 6060 for 15−2415 - 24, the remaining frequency for 10−1410 - 14 is 3030. Dividing this by the true class width of 55 (from boundaries 9.59.5 to 14.514.5) gives a frequency density of 66.

Step-by-Step Solution

1
Calculate the frequency of the class interval 30−4430 - 44 from the pie chart sector angle.
Frequency f30−44=108∘360∘×160=48f_{30-44} = \frac{108^\circ}{360^\circ} \times 160 = 48.
The sector angle in a pie chart is directly proportional to the frequency relative to the 360∘360^\circ total.
2
Determine the class width and frequency of the class interval 15−2415 - 24.
Class boundaries are 14.514.5 and 24.524.5, so width w=10w = 10. Frequency f15−24=6×10=60f_{15-24} = 6 \times 10 = 60.
Frequency density is defined as frequency divided by class width, so frequency equals frequency density multiplied by class width.
3
Determine the frequency of the class interval 10−1410 - 14.
Frequency f10−14=160−(60+22+48)=30f_{10-14} = 160 - (60 + 22 + 48) = 30.
The sum of all class frequencies must equal the total frequency of 160160.
4
Calculate the class width and frequency density of 10−1410 - 14.
Class width w10−14=14.5−9.5=5w_{10-14} = 14.5 - 9.5 = 5. Frequency density =305=6= \frac{30}{5} = 6.
Dividing the frequency of the class (3030) by its exact class boundary width (55) yields the frequency density.

Key Concept

Integration of Frequency Density and Pie Chart Sector Angles
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