NCERT Class 9 Solutions: Statistics (Chapter 14) Exercise 14.3-Part 2

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:

 Length (in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2

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:

 Length (in mm) Number of leaves 117.5-126.5 3 126.5-135.5 5 135.5-144.5 9 144.5-153.5 12 153.5-162.5 5 162.5-171.5 4 171.5-180.5 2

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 time (in hours) Number of lamps 300-400 14 400-500 56 500-600 60 600-700 86 700-800 74 800-900 62 900-1000 48

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:

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:

 Section A Section B Marks Frequency Marks Frequency 0-10 3 0-10 5 10-20 9 10-20 19 20-30 17 20-30 15 30-40 12 30-40 10 40-50 9 40-50 1

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)

 Section A Section B Marks Class marks Frequency Mark Class marks Frequency Equation 5 3 Equation 5 5 Equation 15 9 Equation 15 19 Equation 25 17 Equation 25 15 Equation 35 12 Equation 35 10 Equation 45 9 Equation 45 1

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

Choose appropriate scale.

Below drawn the frequency polygons:

Chart3.txt

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