The weight of 50 workers is given below:
The weight of 50 workers is given below:
Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120
No. of workers 4 7 11 14 6 5 3

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis , and 2 cm = 5 workers along the other axis.

Use a graph to estimate the following:

 (i) the upper and lower quartiles. 

(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight. (2015)

Solution:

We write the given data in cumulative frequency table.

Weight in kg No of workers f Cumulative frequency c.f
50-60 4 4
60-70 7 11
70-80 11 22
80-90 14 36
90-100 6 42
100-110 5 47
110-120 3 50

To represent the data in the table graphically, we mark the upper limits of the class intervals on

the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis),

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) on the graph.

Join the points with the free hand. We get an ogive as shown:

(i)Here n = 50, which is even.

Upper quartile = 3n/4

3×50/4

= 150/4

= 37.5

Now mark a point A (37.5) on the Y-axis and from A draw a line parallel to X-axis meeting the curve at B. From B, draw a perpendicular on x-axis meeting it at C.

C = 92.5

Hence the upper quartile is 92.5 kg.

Lower quartile, Q1 = (n/4) th term

= 50/4

= 12.5

Now mark a point D(12.5) on the Y-axis and from D draw a line parallel to X-axis meeting the curve at E. From E, draw a perpendicular on x-axis meeting it at F.

F = 72

Hence the lower quartile is 72 kg.

(ii) Mark on the graph point P which is 95 kg on X axis.

Through P draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet y-axis at R.

The ordinate of point R represents 40 workers on the y-axis .

The number of workers who are 95 kg and above = Total number of workers – number of workers of weight less than 95 kg = 50-40 = 10