Question 1: Find the mean of:

i) \displaystyle  11, 13, 17, 19, 23   ii) \displaystyle  22, 24, 26, 28, 30, 32, 34, 36

iii) \displaystyle  \frac{1}{4}, 3\frac{1}{4}, 4\frac{3}{4}, 5\frac{1}{4}, 7\frac{1}{2}   iv) \displaystyle  6.5, 8.2, 9.4, 4.6, 7.8, 4.9

Answer:

i) \displaystyle  \text{Mean} =   \frac{\Sigma x_i}{n}   

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

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

ii) \displaystyle  \text{Mean} =   \frac{\Sigma x_i}{n}   

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

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

iii) \displaystyle  \text{Mean} =   \frac{\Sigma x_i}{n}   

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

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

iv) \displaystyle  \text{Mean} =   \frac{\Sigma x_i}{n}   

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

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

 \displaystyle  \\

Question 2:

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

ii) Find the mean of first seven whole numbers

iii) Find the mean of first five multiples of \displaystyle  6

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

\displaystyle  \\

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

Answer:

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

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

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

\displaystyle  69 = 64+x \Rightarrow x = 5

\displaystyle  \\

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

Answer:

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

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

\displaystyle  9 =  \frac{33+3x}{6}  

\displaystyle  54 = 33+x \Rightarrow x = 7

\displaystyle  \\

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:

\displaystyle \text{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

\displaystyle \text{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:

\displaystyle \text{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

\displaystyle \text{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:

\displaystyle \text{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

\displaystyle \text{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:

\displaystyle \text{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

\displaystyle \text{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:

\displaystyle \text{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

\displaystyle \text{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) \displaystyle \text{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

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

ii) \displaystyle \text{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

\displaystyle \text{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:

\displaystyle \text{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

\displaystyle \text{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 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