The marks obtained (out of 100 ) by a class of 80 students are given below: Marks Number of students 10 - 20 6 20 - 30 17 30 - 50 15 50 - 70 16 70 - 100 26 Construct a histogram to represent the data above.

Here, the class intervals are of unequal width. So, we shall first input compute adjusted frequencies of each class. The minimum class size is 20 - 10 = 10 . The adjusted frequencies are computed by using the following formula: Adjusted frequency = Minimum class size Class size × Frequency of the class. The adjusted frequencies are computed in the following table: Marks Number of Students (Frequency) Adjusted Frequency 10-20 6 10 10 × 6 = 6 20-30 7 10 10 × 7 = 7 30-50 15 10 20 × 15 = 7 . 5 50-70 16 10 20 × 16 = 8 70-100 26 10 30 × 26 = 8 . 66 Now, we construct rectangles with class-limits as bases and respective adjusted frequencies as heights.

Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way : Class interval (km/h) Frequency 30-40 3 40-50 6 50-60 25 60-70 65 70-80 50 80-90 28 90-100 14 Draw a histogram and frequency polygon representing the data above.

Let us first draw a histogram for this data and mark the mid-points of the tops of the rectangles In the figure given below, a histogram and a frequency polygon (in dotted lines) are drawn on the same scale. So, we will take x - a x i s as class interval and y - a x i s as frequency. We have to mark the middle of each class interval. So, In x - a x i s we have to take 35 , 45 , 55 , 65 , 75 , 85 , 95 . ∵ 30 + 40 2 = 35 , 40 + 50 2 = 45 , 50 + 60 2 = 55 , 60 + 70 2 = 65 , 70 + 80 2 = 75 , 80 + 90 2 = 85 , 90 + 100 2 = 95

Draw the frequency polygon representing the above data without drawing the histogram. Class Interval (km/h) Frequency 30-40 3 40-50 6 50-60 25 60-70 65 70-80 50 89-90 28 90-100 14

First, we obtain the class marks as given in the following table. Class interval (km/h) Class Marks Frequency 30 - 40 35 3 40 - 50 35 6 50 - 60 55 25 60 - 70 65 65 70 - 80 65 50 80 - 90 85 28 90 - 100 95 14 Now, we plot the points ( 35 , 3 ) , ( 45 , 6 ) , ( 55 , 25 ) , ( 65 , 65 ) , ( 75 , 50 ) , ( 85 , 28 ) and ( 95 , 14 ) . Now, we join the plotted points by line segments. The end points ( 35 , 3 ) and ( 95 , 14 ) are joined to the mid-points ( 25 , 0 ) and ( 105 , 0 ) respectively of imagined class-intervals to obtain the frequency polygon.

Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. Section A Section B Marks Frequency Marks Frequency 0 - 15 5 0 - 15 3 15 - 30 12 15 - 30 16 30 - 45 28 30 - 45 25 45 - 60 30 45 - 60 27 60 - 75 35 60 - 75 40 75 - 90 13 75 - 90 10 Represent the marks of the students of both the sections on the same by two frequency polygons. What do you observe?

First, we obtain in the class marks as given in the following tables: Section A Marks Class Marks Frequency 0 - 15 7 . 5 5 15 - 30 22 . 5 12 30 - 45 37 . 5 28 45 - 60 52 . 5 30 60 - 75 67 . 5 35 75 - 90 82 . 5 13 Section B Marks Class Marks Frequency 0 - 15 7 . 5 3 15 - 30 22 . 5 16 30 - 45 37 . 5 25 45 - 60 52 . 5 27 60 - 75 67 . 5 40 75 - 90 82 . 5 10 We represent class marks on X-axis on a suitable scale and the frequencies on Y -axis on a suitable scale. To obtain the frequency polygon for section A , we plot the points ( 7 . 5 , 5 ) ( 22 . 5 , 12 ) , ( 37 . 5 , 28 ) , ( 52 . 5 , 30 ) , ( 67 . 5 , 35 ) and ( 82 . 5 , 13 ) and join these by line segments. To obtain the frequency polygon for section B , we plot the points ( 7 . 5 , 3 ) ( 22 . 5 , 16 ) , ( 37 . 5 , 25 ) , ( 52 . 5 , 27 ) , ( 67 . 5 , 40 ) a n d ( 82 . 5 , 10 ) on the same scale and join these points by dotted line segments. The two frequency polygon are shown in the figure given below: We observe that section B has more students who scored high marks in the exam than section A . This tells us that section B has performed better in exam than section. A .

The mean of the following distribution is 50 x f 10 17 30 5 a+3 50 32 70 7 a-11 90 19 Find the value of a and hence the frequencies of 30 and 70 .

x i f i f i x i 10 17 170 30 5 a + 3 150 a + 90 50 32 1600 70 7 a - 11 490 a - 770 90 19 1710 Total Σ f i = 60 + 12 a Σ f i x i = 2800 + 640 a We know that Mean ( x ¯ ) = ∑ f i x i Σ f i = 2800 + 640 a 60 + 12 a ⇒ 2800 + 640 a 60 + 12 a = 50 ⇒ 2800 + 640 a = 3000 + 600 a ⇒ 40 a = 200 ⇒ a = 200 40 ⇒ a = 5 So, frequency of 30 = 5 a + 3 = 5 ( 5 ) + 3 = 25 + 3 = 28 and frequency of 70 = 7 a - 11 = 7 ( 5 ) - 11 = 35 - 11 = 24

A total of 25 patients admitted to a hospital are tested for levels of blood sugar, ( mg / dl ) and the results obtained were as follows: 87 71 83 67 85 77 69 76 65 85 85 54 70 68 80 73 78 68 85 73 81 78 81 77 75 Find mean, median and mode ( mg / dl ) of the above data.

Mean : Sum of all observations = 1891 Number of observations, n = 25 Mean ( x ¯ ) = x 1 + x 2 … … + x n n = 1891 25 = 75 . 64 Median: Arranging the observations in ascending order, we get 54 , 65 , 67 , 68 , 68 , 69 , 70 , 71 , 73 , 73 , 75 , 76 , 77 , 77 , 78 , 78 , 80 , 81 , 81 , 83 , 85 , 85 , 85 , 85 , 87 Here, Number of observations ( n ) = 25 (odd) ∴ Median = n + 1 2 t h value = 25 + 1 2 t h value = 13 t h value = 77 Mode: We see from the given data that the observation 85 occurs maximum number of times ( 4 times). ∴ Mode = 85 . Therefore, Mean= 75 . 64 , Median= 77 , Mode= 85 .

NCERT Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

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NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

Attempt the free practice questions on Chapter 14: Statistics and Probability, Exercise 4: Exercise with hints and solutions to strengthen your understanding. NCERT Exemplar Mathematics - Class 9 solutions are prepared by Experienced Embibe Experts.

Questions from NCERT Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise with Hints & Solutions

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 2

The following are the marks (out of $100$ ) of $60$ students in mathematics.
$16, 13, 5, 80, 86, 7, 51, 48, 24, 56, 70, 19, 61, 17, 16, 36, 34, 42, 34, 35, 72, 55, 75, 31, 52, 28, 72, 97, 74, 45, 62, 68, 86, 35, 85, 36, 81, 75, 55, 26, 95, 31, 7, 78, 92, 62, 52, 56, 15, 63, 25, 36, 54, 44, 47, 27, 72, 17, 4, 30 .$

Construct a grouped frequency distribution table with width $10$ of each class in such a way that one of the classes is $3$ ( $20$ not included).

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 6

The marks obtained (out of $100$ ) by a class of $80$ students are given below:

Marks	Number of students
$10 - 20$	$6$
$20 - 30$	$17$
$30 - 50$	$15$
$50 - 70$	$16$
$70 - 100$	$26$

Construct a histogram to represent the data above.

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 7

Following table shows a frequency distribution for the speed of cars passing through at a particular spot on a high way :

Class interval (km/h)	Frequency
30-40	3
40-50	6
50-60	25
60-70	65
70-80	50
80-90	28
90-100	14

Draw a histogram and frequency polygon representing the data above.

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 8

Draw the frequency polygon representing the above data without drawing the histogram.

Class Interval (km/h)	Frequency
30-40	3
40-50	6
50-60	25
60-70	65
70-80	50
89-90	28
90-100	14

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 9

Following table gives the distribution of students of sections $A$ and $B$ of a class according to the marks obtained by them.

Section $A$		Section $B$
Marks	Frequency	Marks	Frequency
$0 - 15$	$5$	$0 - 15$	$3$
$15 - 30$	$12$	$15 - 30$	$16$
$30 - 45$	$28$	$30 - 45$	$25$
$45 - 60$	$30$	$45 - 60$	$27$
$60 - 75$	$35$	$60 - 75$	$40$
$75 - 90$	$13$	$75 - 90$	$10$

Represent the marks of the students of both the sections on the same by two frequency polygons. What do you observe?

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 10

The mean of the following distribution is $50$

$x$	$f$
10	17
30	5 a+3
50	32
70	7 a-11
90	19

Find the value of $a$ and hence the frequencies of $30$ and $70$ .

HARD

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 11

The mean marks (out of $100$ ) of boys and girls in an examination are $70$ and $73$ , respectively. If the mean marks of all the students in that examination is $71$ , find the ratio of the number of boys to the number of girls.

MEDIUM

9th CBSE

IMPORTANT

NCERT Exemplar Mathematics - Class 9>Chapter 14 - Statistics and Probability>Exercise>Q 12

A total of $25$ patients admitted to a hospital are tested for levels of blood sugar, $(mg / dl)$ and the results obtained were as follows:

$87$	$71$	$83$	$67$	$85$
$77$	$69$	$76$	$65$	$85$
$85$	$54$	$70$	$68$	$80$
$73$	$78$	$68$	$85$	$73$
$81$	$78$	$81$	$77$	$75$

Find mean, median and mode ( $mg / dl$ ) of the above data.

NCERT Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise

Questions from NCERT Solutions for Chapter: Statistics and Probability, Exercise 4: Exercise with Hints & Solutions

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