We will discuss here some of the problems on frequency polygon.

**1.** The monthly salaries of 55 workers of a
factory are displayed in the following ogive.

Answer the following.

(i) How many workers have a monthly salary under $ 4000?

(ii) How many workers have a monthly salary more than or equal to $ 8000 but less than $ 10000?

(iii) Construct the frequency distribution table of the salaries of 55 workers.

**Solution:**

(i) 35. {Since the point on the ogive has the coordinates (4000, 35) corresponding to the point representing 4000 on the monthly salary axis.}

(ii) 10 (i.e., 55 – 45). {Since, according to the ogive, 55 workers have a salary less than $ 10000 and 45 workers have a salary less than $ 8000.}

(iii)

Monthly

Salary (in $)

Under

2000

Under

4000

Under

6000

Under

8000

Under

10000

Cumulative Frequency

(Number of Workers)

10

35

40

45

55

Therefore, the frequency distribution table of the salaries of 55 workers will be as below.

Monthly

Salary (in $)

0 - 2000

2000 - 4000

4000 - 6000

6000 - 8000

8000 - 10000

Number of Workers

10

25

5

5

10

**2.** Construct an ogive for the following
distribution.

**Solution:**

Here, the distribution is not continuous. Changing the intervals to overlapping intervals, we get the following cumulative-frequency table.

Hours of Study in a day

Number of Students

(Frequency)

Cumulative Frequency

-0.5 - 3.5

10

10

3.5 - 7.5

25

35

(= 10 + 25)

7.5 - 11.5

15

50

(= 10 + 25 + 15)

11.5 - 15.5

5

55

(= 10 + 25 + 15 + 5)

15.5 - 19.5

1

56

(= 10 + 25 + 15 + 5 + 1)

Taking the upper limit of an interval and its corresponding cumulative frequency as x- and y-coordinates respectively, plot the points (3.5, 10)), (7.5, 35), (11.5, 50), (15.5, 55) and (19.5, 56).

**Scale:** On the x-axis, 1 cm = 4 hours.

On the y-axis, 1 mm = 1 student

**From Problems on Cumulative -Frequency Curve**

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