CLASS 10 MATH NCERT SOLUTIONS FOR CHAPTER – 14 STATISTICS EX – 14.4

Statistics

 Page: 293

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

Daily income in Rupees100-120120-140140-160160-180180-200
Number of workers12148610

Solution

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

Daily incomeFrequencyCumulative Frequency
Less than 1201212
Less than 1401426
Less than 160834
Less than 180640
Less than 2001050

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve

Ncert solutions class 10 chapter 14-9

2.During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight in kgNumber of students
Less than 380
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

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:

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.

Ncert solutions class 10 chapter 14-10

Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.

Class intervalNumber of students(Frequency)Cumulative Frequency
Less than 3800
Less than 403-0=33
Less than 425-3=28
Less than 449-5=49
Less than 4614-9=514
Less than 4828-14=1428
Less than 5032-28=432
Less than 5235-22=335

The class 46 – 48 has the maximum frequency, therefore, this is modal class

Here, = 46, h = 2, f1= 14, f0= 5 and f2 = 4

The mode formula is given as:

Now, Mode =

Ncert solutions class 10 chapter 14-11

= 46 + 0.95 = 46.95

Thus, mode is verified.

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.

Production Yield50-5555-6060-6565-7070-7575-80
Number of farms2812243816

Change the distribution to a more than type distribution and draw its ogive.

Solution:

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha)Number of farms
More than or equal to 50100
More than or equal to 55100-2 = 98
More than or equal to 6098-8= 90
More than or equal to 6590-12=78
More than or equal to 7078-24=54
More than or equal to 7554-38 =16

From the table obtained draw the 0 give by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type 0 give curve.

Ncert solutions class 10 chapter 14-12

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