Following table gives the distribution of two sections of a class according to the marks obtained by them: Section A Section B Marks Frequency Marks Frequency 0 - 10 9 0 - 10 5 10 - 20 3 10 - 20 15 20 - 30 17 20 - 30 19 30 - 40 9 30 - 40 10 40 - 50 12 40 - 50 4 Represent the marks of the students of both the sections on the same graph by two frequency polygons and thus, compare the performances of the two sections.

Given: The distribution of two sections of a class according to the marks obtained by them. Find- the marks of the students of both the sections on the same graph by two frequency polygons and thus, compare the performances of the two sections. To draw a frequency polygon we mark the class marks along x-axis. Therefore, the modified tables are: For section A : Marks Class Marks Frequency Ordered pairs 0 - 10 5 9 ( 5 , 9 ) 10 - 20 15 3 ( 15 , 3 ) 20 - 30 25 17 ( 25 , 17 ) 30 - 40 35 9 ( 35 , 9 ) 40 - 50 45 12 ( 45 , 12 ) For Section B: Marks Class Marks Frequency Ordered pairs 0 - 10 5 5 ( 5 , 5 ) 10 - 20 15 15 ( 15 , 15 ) 20 - 30 25 19 ( 25 , 19 ) 30 - 40 35 10 ( 35 , 10 ) 40 - 50 45 4 ( 45 , 4 ) We plot the ordered pair ( 5 , 9 ) , ( 15 , 3 ) , ( 25 , 17 ) , ( 35 , 9 ) and ( 45 , 12 ) and join the points by lines segments and obtain the frequency polygon of section A. Again to obtain the frequency polygon of section B, we plot the points ( 5 , 5 ) , ( 15 , 15 ) , ( 25 , 19 ) , ( 35 , 10 ) and ( 45 , 4 ) on the same coordinate axes and join these points by dotted line-segments. The two frequency polygons on the same graph are shown below:

Find the mean of the marks obtained by 30 students of class I X of a school. The marks are: 10 20 36 92 95 40 50 56 60 70 92 88 80 70 72 70 36 40 36 40 92 40 50 50 56 60 70 60 60 88

Sum of obtain marks = 10 + 20 + 36 + 92 + 95 + 40 + 50 + 56 + 60 + 70 + 92 + 88 + 80 + 70 + 72 + 70 + 36 + 40 + 36 + 40 + 92 + 40 + 50 + 50 + 56 + 60 + 70 + 60 + 60 + 88 = 1779 Total number of students = 30 Mean x - = Sum of all the observations Total number of observations Now, mean marks = 1779 30 = 59.3

Find the mean of the following data: x i 11 12 13 14 15 16 17 18 f i 10 1 3 4 8 10 3 1

Writing the given data as in the following table. X i f i f i × X i 11 10 11 × 10 = 110 12 1 12 × 1 = 12 13 3 13 × 3 = 39 14 4 14 × 4 = 56 15 8 15 × 8 = 120 16 10 16 × 10 = 160 17 3 17 × 3 = 51 18 1 18 × 1 = 18 Total ∑ i = 1 8 f i = 40 ∑ i = 1 8 ( f i x i ) = 566 We know that, X - = ∑ i = 1 n ( f i x i ) ∑ i = 1 n f i = 566 40 = 14.15

Find the value of P when the mean of the following data is 21.6 . x i 6 12 18 24 30 36 f i 5 4 P 6 4 6

Given: According to question, mean is 21.6 . Writing the given data as in the following table: X i f i f i × X i 6 5 6 × 5 = 30 12 4 12 × 4 = 48 18 P 18 × P = 18 P 24 6 24 × 6 = 144 30 4 30 × 4 = 120 36 6 36 × 6 = 216 Total ∑ i = 1 8 f i = 25 + P ∑ i = 1 8 f i x i = 558 + 18 P X - = ∑ i = 1 n ( f i x i ) ∑ i = 1 n f i 21.6 = 558 + 18 P 25 + P 25 + P × 21.6 = 558 + 18 P 540 + ( 21.6 ) P = 558 + 18 P 558 - 540 = ( 21 .6 ) P - 18 P 18 = ( 3.6 ) P P = 18 3.6 P = 18 × 10 36 = 180 36 = 10 2 = 5 P = 5

The following table gives a description of the marks obtained by 41 students of a class: Marks obtained 25 15 20 30 17 22 Number of students 8 2 10 4 5 12 Find the median of marks obtained.

Given: A description of the marks obtained by 41 students of a class. First arranging the observations in ascending order, we have: Marks obtained 15 17 20 22 25 30 Frequency 2 5 10 12 8 4 Now, we prepare the cumulative frequency table as given below: Marks obtained Frequency Cumulative frequency 15 2 2 17 5 7 20 10 17 22 12 29 25 8 37 30 4 41 ∵ Total number of observations = 41 Here, 41 is an odd number ∴ Median of marks = marks obtained by n + 1 2 t h student ∴ Median of marks = 41 + 1 2 t h = 42 2 = 21 marks obtained by the 21 s t student ∵ 18 t h to 29 t h students get 22 marks. i.e. 21 s t student gets 22 marks ∴ The median of the marks = 22

MEDIUM

9th CBSE

IMPORTANT

Earn 100

Following frequency distribution table is representing the weights of $36$ students of a class:

Weight (in $kg$ ) Number of Students

$30.5 - 35.5$ $9$

$35.5 - 40.5$ $6$

$40.5 - 45.5$ $15$

$45.5 - 50.5$ $3$

$50.5 - 55.5$ $1$

$55.5 - 60.5$ $2$

Total $36$

Represent this data by a histogram.

Important Questions on Statistics

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 3

A teacher wanted to analyse the performance of two sections of students in a mathematics test of $100$ marks. Looking at their performances, she found that a few students got under $20$ marks and a few got $70$ marks or above. So, she decided to group them into intervals of varying sizes as follows: $0 - 20, 20 - 30, \dots . ., 60 - 70, 70 - 100$ . Then she formed the following table.

Marks	Number of Students
$0 - 20$	$7$
$20 - 30$	$10$
$30 - 40$	$10$
$40 - 50$	$20$
$50 - 60$	$20$
$60 - 70$	$15$
$70$ and above	$8$
Total	$90$

Represent this data by a histogram.

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 4

Following table gives the distribution of two sections of a class according to the marks obtained by them:

Section A		Section B
Marks	Frequency	Marks	Frequency
$0 - 10$	$9$	$0 - 10$	$5$
$10 - 20$	$3$	$10 - 20$	$15$
$20 - 30$	$17$	$20 - 30$	$19$
$30 - 40$	$9$	$30 - 40$	$10$
$40 - 50$	$12$	$40 - 50$	$4$

Represent the marks of the students of both the sections on the same graph by two frequency polygons and thus, compare the performances of the two sections.

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 5

Five people were asked about the time in a week they spend in doing social work in their community. They said

10, 7, 13, 20

and

15

hours, respectively. Find the mean (or average) time in a week devoted by them for social work.

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 6

Find the mean of the marks obtained by

30

students of class

I X

of a school. The marks are:

\begin{matrix} 10 & 20 & 36 & 92 & 95 & 40 & 50 & 56 & 60 & 70 \\ 92 & 88 & 80 & 70 & 72 & 70 & 36 & 40 & 36 & 40 \\ 92 & 40 & 50 & 50 & 56 & 60 & 70 & 60 & 60 & 88 \end{matrix}

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 7

Find the mean of the following data:

$x_{i}$	$11$	$12$	$13$	$14$	$15$	$16$	$17$	$18$
$f_{i}$	$10$	$1$	$3$	$4$	$8$	$10$	$3$	$1$

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 8

Find the value of P when the mean of the following data is

21.6 .

$x_{i}$	$6$	$12$	$18$	$24$	$30$	$36$
$f_{i}$	$5$	$4$	$P$	$6$	$4$	$6$

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 9

The heights (in

c m

) of

9

students of a class are as follows:

155, 160, 145, 149, 150, 147, 152, 144, 148

Find the median of this data.

MEDIUM

9th CBSE

IMPORTANT

Full Marks Mathematics 9>Chapter 14 - Statistics>TEST YOUR SKILLS>Q 10

The following table gives a description of the marks obtained by

41

students of a class:

Marks obtained	$25$	$15$	$20$	$30$	$17$	$22$
Number of students	$8$	$2$	$10$	$4$	$5$	$12$

Find the median of marks obtained.

Weight (in $kg$ )	Number of Students
$30.5 - 35.5$	$9$
$35.5 - 40.5$	$6$
$40.5 - 45.5$	$15$
$45.5 - 50.5$	$3$
$50.5 - 55.5$	$1$
$55.5 - 60.5$	$2$
Total	$36$

Following frequency distribution table is representing the weights of 36 students of a class: Weight (in kg) Number of Students 30.5-35.5 9 35.5-40.5 6 40.5-45.5 15 45.5-50.5 3 50.5-55.5 1 55.5-60.5 2 Total 36 Represent this data by a histogram.

Important Questions on Statistics

Learn and Prepare for any exam you want !

Following frequency distribution table is representing the weights of $36$ students of a class:

Weight (in $kg$ ) Number of Students

$30.5 - 35.5$ $9$

$35.5 - 40.5$ $6$

$40.5 - 45.5$ $15$

$45.5 - 50.5$ $3$

$50.5 - 55.5$ $1$

$55.5 - 60.5$ $2$

Total $36$

Represent this data by a histogram.