Question

section 2.2 - histograms, frequency polygons, and ogives
question 5 of 13 (1 point) | question attempt: 1 of unlimited
class limits frequency
70–109 2
110–149 6
150–189 4
190–229 3
230–269 2
270–309 1
total 18
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frequency
8
7
6
5
4
3
2
1
0
9.5 49.5 89.5 129.5 169.5 209.5 249.5 289.5 329.5 369.5
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Explanation:

Step1: Find class midpoints

For each class:

• $70$-$109$: $\frac{70+109}{2}=89.5$
• $110$-$149$: $\frac{110+149}{2}=129.5$
• $150$-$189$: $\frac{150+189}{2}=169.5$
• $190$-$229$: $\frac{190+229}{2}=209.5$
• $230$-$269$: $\frac{230+269}{2}=249.5$
• $270$-$309$: $\frac{270+309}{2}=289.5$

Step2: Add extra end classes

Add two additional classes with frequency 0:

• Lower class: $30$-$69$, midpoint $\frac{30+69}{2}=49.5$
• Upper class: $310$-$349$, midpoint $\frac{310+349}{2}=329.5$

Step3: Plot midpoints vs frequency

Plot the following points on the graph:

• $(49.5, 0)$
• $(89.5, 2)$
• $(129.5, 6)$
• $(169.5, 4)$
• $(209.5, 3)$
• $(249.5, 2)$
• $(289.5, 1)$
• $(329.5, 0)$

Step4: Connect plotted points

Draw straight lines between consecutive plotted points to form the frequency polygon.

Answer:

A completed frequency polygon for the given data