Exercise-4

Example: The annual profits earned by 30 shops in a locality give rise to the following distribution.Draw both ogives for the data above. Hence obtain the median profit.

Solution:

We first draw the coordinate axes, with the profit along the X- axis, and the cumulative frequency along the Y-axis.

Plot the points (5, 30), (10,28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3). We join these points with a smooth curve to get the more than ogive.

Plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30) on same graph and join the points with a smooth curve to get the less than ogive.

The abcissa of their point of intersection is nearly 17.5, which is the median

Hence, the median profit (in lakhs) is ₹17.5.

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1. The following distribution gives the daily income of 50 workers of a factory.Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:

We first draw the coordinate axes, with upper limits of the daily income along the X-axis, and the less than cumulative frequency along the Y-axis. Then, plot the points (300, 12), (350,26), (400, 34), (450, 40), (500, 50) and join these points with a smooth curve to get the less than ogive.

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2. During the medical check-up of 35 students of a class, their weights were recorded as follows.Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

We first draw the coordinate axes, with lower limits of the weight along the X-axis, and the less than cumulative frequency along the Y-axis. Then, plot the points (38, 0), (40,3), (42, 5), (44, 9), (46, 14),(48,28),(50,32),(52,35) and join these points with a smooth curve to get the less than ogive.

So, Median from the graph is 46.5

Number of observations, n =35

`n/2=35/2= 17.5

which lies in the class 46 - 48.

∴ 46 - 48 is median class

Then, `l` (the lower boundary) = 44,

`cf` (the cumulative frequency of the class preceding 46 - 48) =14,

`f` (the frequency of the median class 46 - 48) = 14,

h (the class size) = 2.

Median = `l+[(n/2-cf)/f]times h`

Median `= 46+[(17.5-14)/14]times 2`

`= 46+7/14=46+0.5=46.5`

So,median of the given data is 46.5.

It is verified that median of given data is same in two ways .

3. The following table gives production yield per hectare of wheat of 100 farms of a village.Change the distribution to a more than type distribution, and draw its ogive.

Solution:

We first draw the coordinate axes, with lower limits of the production yield along the X-axis, and the more than cumulative frequency along the Y-axis. Then, plot the points on the graph and join these points with a smooth curve to get the more than ogive.

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