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Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.4

  • Last Updated : 03 Mar, 2021

Question 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:

According to the question, we convert the given distribution to a less than type cumulative frequency distribution, 

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

Now according to the table we plot the points that are corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50) on a graph paper. Here x-axis represents the upper limit and y-axis represent the frequency. The curve obtained from the graph is known as less than type ogive curve. 



Question 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:

According to the give table, we get the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). Now using these points we draw an ogive, where the x-axis represents the upper limit and y-axis represents the frequency. The curve obtained is known as less than type ogive.

Now, 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 this point, now we draw a perpendicular line to the x-axis and the intersection point which is perpendicular to x-axis is the median of the given data. After, locating point now we create a table to find the mode:

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, hence, this is the modal class

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

Now we find the mode:



Mode = l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

On substituting the values in the given formula, we get

= 46 + 0.95 = 46.95

Hence, the mode is verified.

Question 3. The following tables give the 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:

According to the question, we change the distribution to a more than type distribution.

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

Now, according to the table we draw the ogive by plotting the points. Here, the a-axis represents the upper limit and y-axis represents the cumulative frequency. And the points 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 ogive curve.

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