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Class 7 RS Aggarwal Maths SolutionsClass7

Contents

21. Mean, Median and Mode 21A, 21B, 21C

 

EXERCISE -21A. MEAN, MEDIAN AND MODE

Mean = .

Q1. Define the terms:

I] Data : A collection of numerical figures giving some particular type of information is called data.

Eg. Height (in cm)of 5 players 155, 148, 161, 159, 163.

Ii] Raw data : obtained in a original form is called raw data. www.rsmaths99.com

Iii] Array : Arranging the data in ascending or descending order is called an array.

Iv] Tabulation of data : Arranging the data in the form of table .

V] Observation : Each numerical figure in a data is called an observation.

Vi] Frequency of an observation : the no. of times a particular observation occurs is called its frequency.

Vii] It is the subject that deals with the collection, presentation, analysis and interpretation of numerical data.

Q2. Answer given in the back page of book. Q4.

Q5. DIY

Q6. 5 natural nos. = 1, 2, 3, 4, 5, . sum of the 5 natural nos. = 1+ 2+ 3+ 4+ 5 = 15

Mean = =

Q7. First six odd natural nos. = 1,3,5,7,9,11. www.rsmaths99.com

sum of the 5 odd natural nos. = 1 + 3 + 5 + 7 + 9 + 11 = 36.

Mean =

Q8. Sum of first 7 even natural nos. = 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56.

Mean =

Q9. Sum of first 5 prime nos. = 2 + 3 + 5 + 7 + 11 = 27

Mean =

Q10. Sum of first six multiples of 5 = 5 + 10 + 15 + 20 + 25 +30 = 105.

Mean =

Q13.

Height ( in cm) (xi)

No. of plants (fi)

(fi x x i)

58

20

1160

60

25

1500

62

15

930

64

8

512

66

12

792

74

10

740

 

N = =90

x xi =5634

mean wt. = =

Q14.

Age( in years) (xi)

No. of players (fi)

(fi x x i)

14

15

210

15

14

210

16

10

160

17

8

136

18

3

54

 

N = = 50

x xi =770

mean wt. = = . www.rsmaths99.com

Q15.

Height( in cm) (xi)

No. of boys (fi)

(fi x x i)

165

9

1485

170

8

21360

175

11

1925

180

12

2160

 

N = = 40

x xi =6930

mean wt. = =

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EXERCISE 21B. MEDIAN OF UNGROUPED DATA

1.      When n is odd . Median = value of

2.      When n is even. Median =

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Q1. Find the median : www.rsmaths99.com

I] arrange in ascending order, we have: 2, 2, 3, 5, 7, 9, 9, 10, 11.

Here n = 9 , which is odd. Median = value of

= value of 5th term = 7.

Ii] arrange in ascending order, we have: 6, 8, 9, 15, 16, 18, 21, 22, 25.

Here n = 9 , which is odd. Median = value of

= value of 5th term = 16.

Iii] arrange in ascending order, we have: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25.

Here n = 11 , which is odd. Median = value of

= value of 6 th term = 16.

 

Q2. I] arrange in ascending order, we have: 9, 10, 17, 19, 21, 22, 32, 35.

Here n = 8 , which is even.

Median =

= = .

Now, 4th term = 19 & 5th term = 21.

Ii] arrange in ascending order, we have: 29, 35, 51, 55, 60, 63, 72, 82, 85, 91.

Here n = 10 , which is even. www.rsmaths99.com

Median =

= = .

Now, 5th term = 60. & 6th term = 63. = (60 + 63) =

 

Q3. First 15 odd nos. are : 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.

Here n = 15. Which is odd. Median = value of

= 15.

 

Q4. First 10 even nos. : 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

Here n = 10. Which is even. www.rsmaths99.com

Median =

= = (5 + 6)th term. = (10 + 12) = =

 

Q5. First 50 whole nos. are: 0, 1, 2, 3, .., 49.

Here n = 50. Which is even. Median =

= = .

= .

Q6. Marks of 15 students in ascending order : 17,17, 19, 19, 20, 21, 22, 23, 24, 25,

26, 29, 31, 35, 40. Here n = 15. Which is odd.

Then, median = = 23.

Q7. 31, 34, 36, 37, 40, 43, 46, 50, 52, 53. Here n = 10 which is even. www.rsmaths99.com

= = =

= www.rsmaths99.com

 

Q8.

Weight (in kg)

No. of boys

Cumulative frequency

45

8

8

46

5

13

48

6

19

50

9

28

52

7

35

54

4

39

55

2

41

Total no. of term = 31, which is odd.

Median of weight = wt. of =

Q9.

Marks

No. of students

Cumulative frequency

15

3

3

17

5

8

20

9

17

22

4

21

25

6

27

30

10

37

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Here n = 37, which is odd.

median = marks obtained by Marks obtained by 19th student.

Hence median of marks = 22.

Q10.

Height (in cm)

No. of students

Cumulative frequency

151

6

6

152

3

9

153

12

21

154

4

25

155

10

35

156

8

43

157

7

50

 

Here n = 50 , which is even. Median =

(25th + 26th term) =

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EXERCISE 21C. MODE OF UNGROUPED DATA

MODE = 3(Median) 2(mean)

Q1. Find the mode of data:

I] 4, 6, 7, 8, 8, 8, 8, 10, 11, 15

Clearly, 8 occurs maximum no. of times. Hence, mode = 8.

Ii] 18, 21, 23, 27, 27, 27, 27, 27, 36, 39, 40.

Clearly, 27 occurs maximum no. of times. Hence, mode = 27.

Q2. 28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41.

Clearly, 32 occurs maximum no. of times. Hence, mode of the age = 32 yrs.

Q3. www.rsmaths99.com

Daily wages (in Rs.) (xi)

No. of workers (fi )

Cumulative frequency

Fi x xi

100

6

6

600

125

8

14

1000

150

9

23

1350

175

12

35

2100

200

10

45

2000

 

N =

 

 

Here n = 45, which is odd.

Median = rd term. Median = Rs. 150.

Also, mean = =

Mode = 3(Median) 2(mean) = 3 x 150 2 x 1.67

= 450 313.34 = Rs. 136.66 www.rsmaths99.com