Question 1: Find the mean of:

(i) 11, 13, 17, 19, 23

(ii) 22, 24, 26, 28, 30, 32, 34, 36

(iii) \frac{1}{4}, 3\frac{1}{4}, 4\frac{3}{4}, 5\frac{1}{4}, 7\frac{1}{2}

(iv) 6.5, 8.2, 9.4, 4.6, 7.8, 4.9

Answer:

(i) Mean = \frac{\Sigma x_i}{n}

Mean = \frac{16+ 13+7+19+23}{5}

Mean = \frac{83}{5} = 16.6

(ii) Mean = \frac{\Sigma x_i}{n}

Mean = \frac{22+ 24+ 26+ 28+ 30+ 32+ 34+ 36}{8}

Mean = \frac{232}{8} = 29

(iii) Mean = \frac{\Sigma x_i}{n}

Mean = \frac{\frac{1}{4}+ 3\frac{1}{4}+ 4\frac{3}{4}+ 5\frac{1}{4}, 7\frac{1}{2}}{5}

Mean = \frac{20\frac{1}{2}}{5} = 4.2

(iv) Mean = \frac{\Sigma x_i}{n}

Mean = \frac{6.5+ 8.2+ 9.4+ 4.6+ 7.8+ 4.9}{6}

Mean = \frac{41.4}{6} = 6.9

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Question 2:

(i) Find the mean of all prime numbers between 20 and 40

(ii) Find the mean of first seven whole numbers

(iii) Find the mean of first five multiples of 6

Answer:

(i) Prime numbers between 20 and 40 are 23, 29, 31, 37

Mean = \frac{\Sigma x_i}{n}

Mean = \frac{23+29+31+37}{4}

Mean = \frac{120}{4} = 30

(ii) First seven whole numbers = 0, 1, 2, 3, 4, 5, 6

Mean = \frac{\Sigma x_i}{n}

Mean = \frac{0+1+2+3+4+5+6}{7}

Mean = \frac{21}{7} = 3

(iii) First five multiple of 6 = 6, 12, 18, 24, 30

Mean = \frac{\Sigma x_i}{n}

Mean = \frac{6+12+18+24+30}{5}

Mean = \frac{90}{5} = 18

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Question 3: The mean of 9, 14, x, 16, 7 and 18 is 11.5 . Find the value of x .

Answer:

Mean = \frac{\Sigma x_i}{n}

11.5 = \frac{9+14+x+16+7+18}{6}

11.5 = \frac{64+x}{6}

69 = 64+x \Rightarrow x = 5

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Question 4: The mean of 7, 9, x+3, 12, 2x-1 and 3 is 9 . Find the value of x .

Answer:

Mean = \frac{\Sigma x_i}{n}

9 = \frac{7+9+x+3+12+2x-1+3}{6}

9 = \frac{33+3x}{6}

54 = 33+x \Rightarrow x = 7

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Question 5: Find the mean of the following distribution:

Marks 12 15 20 23 25 27 30
Number of students 8 6 11 7 4 1 3

Answer:

Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

x_i f_i x_if_i
12 8 96
15 6 90
20 11 220
23 7 161
25 4 100
27 1 27
30 3 90
\Sigma(f_i)= 40 \Sigma(f_ix_i)= 784

Mean of a distribution = \frac{784}{40} = 19.6

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Question 6: The following data gives the number of boys of a particular age in a class of 40 students:

Age in Years 15 16 17 18 19 20
Number of Boys 3 8 9 11 6 3

Calculate the mean age of the students.

Answer:

Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

x_i f_i x_if_i
15 3 45
16 8 128
17 9 153
18 11 198
19 6 114
20 3 60
\Sigma(f_i)= 40 \Sigma(f_ix_i)= 698

Mean of a distribution = \frac{698}{40} = 17.45 \ years

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Question 7: The weight of 40 students in a class are given below:

Weight in Kg 30 32 33 35 36 37 38
Number of students 5 6 3 8 4 9 5

Find the mean weight.

Answer:

Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

x_i f_i x_if_i
30 5 150
32 6 192
33 3 99
35 8 280
36 4 144
37 9 333
38 5 190
\Sigma(f_i)= 40 \Sigma(f_ix_i)= 1388

Mean of a distribution = \frac{1388}{40} = 34.7 \ kg

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Question 8: Find the mean of the following frequency distribution:

x_i 7 8 9 10 11
f_i 19 23 31 27 20

Answer:

Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

x_i f_i x_if_i
7 19 133
8 23 184
9 31 279
10 27 270
11 20 220
\Sigma(f_i)= 120 \Sigma(f_ix_i)= 1086

Mean of a distribution = \frac{1086}{120} = 9.05

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Question 9: If the mean of the following frequency distribution is 50, find the value of p.

x_i 10 30 50 70 90
f_i 17 28 32 p 19

Answer:

Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

x_i f_i x_if_i
10 17 170
30 28 840
50 32 1600
70 p 70p
90 19 1710
\Sigma(f_i)= 96+p \Sigma(f_ix_i) = 4320+70p

Mean of a distribution:  50 = \frac{4320+70p}{96+p}

4800 +50p = 4320+70p \Rightarrow p = 24

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Question 10: Find the mean of the following grouped frequency distribution:

(i)

Class – Interval 0-10 10-20 20-30 30-40 40-50
Frequency 11 7 9 5 8

(ii)

Class – Interval 10-16 16-22 22-28 28-34 34-40
Frequency 12 8 5 9 6

Answer:

(i) Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

Class – Interval f_i x_i x_if_i
0-10 11 5 55
10-20 7 15 105
20-30 9 25 225
30-40 5 35 175
40-50 8 45 360
 40 920

Mean of a distribution = \frac{920}{40} = 23

(ii) Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

Class – Interval f_i x_i x_if_i
10-16 12 13 156
16-22 8 19 152
22-28 5 25 125
28-34 9 31 279
34-40 6 37 222
 40 934

Mean of a distribution = \frac{934}{40} = 23.35

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Question 11: The daily wages of 60 workers in a factory are given below:

Daily wages in Rs. 100-120 120-140 140-160 160-180 180-200
Number of workers 24 12 8 11 5

Find the mean daily wages.

Answer:

Mean of a distribution = \frac{\Sigma(f_ix_i)}{\Sigma f_i}

Daily wages in Rs. f_i x_i x_if_i
100-120 24 110 2640
120-140 12 130 1560
140-160 8 150 1200
160-180 11 170 1870
180-200 5 190 950
60 8220

Mean of a distribution = \frac{8220}{60} = 137

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Question 12: Find the mode of:

(i) 11, 7, 3, 7, 0, 7, 8, 10, 8, 9, 11, 7

(ii) 16, 23, 18, 20, 23, 18, 30, 25, 18, 16

Answer:

(i) In the given data, 7 occurs the maximum number of times. Hence, the mode of the given data = 7

(ii) In the given data, 18 occurs the maximum number of times. Hence, the mode of the given data = 18

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Question 13: The ages (in years) of 10 good players of a class are given below: 13, 15, 13, 14, 16, 15, 13, 16, 13 . Find the mode age.

Answer:

In the given data, 13 occurs the maximum number of times. Hence, the mode of the given data = 13

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Question 14: Daily wages of $latex 40 workers in a factory are given below.

Daily wages in Rs. 100 125 150 175 200
Number of workers 8 14 6 9 3

Find the mode of the data.

Answer:

Since the frequency of 125 is the highest, mode of the data = 125 .

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Question 15: The height of plants (in cm) in a nursery are given below.

Height in cm 28 30 32 34 36
Number of plants 36 47 80 58 72

Find the mode of the data.

Answer:

Since the frequency of 32 is the highest, mode of the data = 32 .

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Question 16: Find the median of each of the following data:

(i) 7, 11, 20, 6, 3, 16, 15, 23, 12

(ii) 9, 25, 32, 51, 7, 16, 37, 50, 0, 13, 19

(iii) 5.6, 7.2, 1.8, 4.3, 9.1, 2.6, 3.4

(iv) 122, 127, 109, 118, 125, 108

Answer:

(i) On arranging the data in ascending order we get:

3, 6, 7, 11, 12, 15, 16, 20, 23

Number of terms = 9

Therefore middle term = \frac{1}{2} (9+1)^{th} term = 5^{th} term = 12

Therefore Median = 12 

(ii) On arranging the data in ascending order we get:

0, 7, 9, 13, 16, 19, 25, 32, 37, 50, 51 

Number of terms = 11

Therefore middle term = \frac{1}{2} (11+1)^{th} term = 6^{th} term = 19

Therefore Median = 19 

(iii) On arranging the data in ascending order we get:

1.8, 2.6, 3.4, 4.3, 5.6, 7.2, 9.1

Number of terms = 7

Therefore middle term = \frac{1}{2} (7+1)^{th} term = 4^{th} term = 4.3

Therefore Median = 4.3

(iv) On arranging the data in ascending order we get:

108, 109, 118, 122, 125, 127 

Number of terms = 6

Therefore middle term = \frac{1}{2} (6)^{th} term = 3^{th} term = 118

Therefore Median = 118

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