NCERT Class 9 Solutions: Statistics (Chapter 14) Exercise 14.3-Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Q-4 The length of 40 leaves of a plant are measured correct to one millimeter and the obtained data is represented in the following table:

Table Shows the Length of Number of Leaves
Length (in mm)Number of leaves
118 - 1263
127 - 1355
136 - 1449
145 - 15312
154 - 1625
163 - 1714
172 - 1802

I. Draw a histogram to represent the given data.

II. Is there any other suitable graphical representation for the same data?

III. Is it correct to conclude that the maximum number of leaves is 153 mm long? Why?

Solution (I) :

Given frequency distribution is not continuous. So, first convert into frequency distribution.

So, Difference between lower limit of class and upper limit of preceding class:

Hence, added to each upper class limit and subtract from the lower class limits.

Obtain distribution is given as under:

Obtain Distribution as Per Given Data
Length (in mm)Number of leaves
117.5 - 126.53
126.5 - 135.55
135.5 - 144.59
144.5 - 153.512
153.5 - 162.55
162.5 - 171.54
171.5 - 180.52
Chart Representing the Data

Solution (II) : Other suitable graphical representation of this data is frequency polygon.

Solution (III) : No, maximum number of leaves has in between 144.5 mm and 153.5 mm. It is not necessary that all leaves length as 153 mm.

Q-5 The following table gives the life times of 400 neon lamps:

Life Times of Neon Lamps Are Shown in Table
Life time (in hours)Number of lamps
300 - 40014
400 - 50056
500 - 60060
600 - 70086
700 - 80074
800 - 90062
900 - 100048

I. Represent the given information with the help of a histogram.

II. How many lamps have a lifetime of more than 700 hours?

Solution (I) : Life time of x-axis and the number of lamps on y-axis.

Histogram:

Chart Representing the Data

Solution (II) :

Lifetime more than 700 is the sum of the number of neon lamps having their lifetime as , and

So, neon lamp having more than 700 hour՚s lifetime

Q-6 The following table gives the distribution of students of two sections according to the mark obtained by them:

Distribution of Students According to Marks
Section ASection B
MarksFrequencyMarksFrequency
0 - 1030 - 105
10 - 20910 - 2019
20 - 301720 - 3015
30 - 401230 - 4010
40 - 50940 - 501

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

Solution:

The class marks of the given class intervals by using the formula.

Class interval (Marks)

Frequency Polygons Data Table of Section a and B
Section ASection B
MarksClass marksFrequencyMarkClass marksFrequency
5355
1591519
25172515
35123510
459451

Class marks on x-axis and frequency on y-axis.

Choose appropriate scale.

Below drawn the frequency polygons:

Chart Representing the Data

Developed by: