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Q.1: If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively. Find the mean height.
SOLUTION :
Given : the heights of 5 persons are 140 cm , 150 cm , 152 cm , 158 cm and 161 cm
Q 2 . Find the mean of 994 , 996 , 998 , 1000 , 1002.
SOLUTION :
Given :
Numbers are 994 , 996 , 998 , 1000 , 1002.
Q 3 . Find the mean of first five natural numbers.
SOLUTION :
The first five odd numbers are 1 , 2 , 3 , 4 , 5.
Q 4 . Find the mean of all factors of 10.
SOLUTION :
All factors of 6 are 1 , 2 , 5 , 10.
Mean = 4.5
Q 5 . Find the mean of first ten even natural numbers.
SOLUTION :
The first five even natural numbers are 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20
Mean = 11
Q 6 . Find the mean of x , x + 2 , x + 4 , x + 6 , x + 8.
SOLUTION :
Numbers are x , x + 2 , x + 4 , x + 6 , x + 8.
Q 7. Find the mean of first five multiples of 3.
SOLUTION :
First five multiples of 3 are 3 , 6 , 9 , 12 , 15.
= 9
Mean = 9
Q 8 . Following are the weights of 10 new born babies in a hospital on a particular day : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6 (in kg). Find the mean.
SOLUTION : The weights (in kg) of 10 new born babies are : 3.4 , 3 .6 , 4.2 , 4.5 , 3.9 , 4.1 , 3.8 , 4.5 , 4.4 , 3.6
= 4 kg
Q 9 . The percentage marks obtained by students of a class in mathematics are as follows: 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1 .Find their mean.
SOLUTION :
The percentage marks obtained by students are 64 , 36 , 47 , 23 , 0 , 19 , 81 , 93 , 72 , 35 , 3 , 1
Mean Marks = 39.5
Q 10. The numbers of children in 10 families of a locality are 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5 . Find the number of children per family.
SOLUTION :
The numbers of children in 10 families are : 2 , 4 , 3 , 4 , 2 , 3 , 5 , 1 , 1 , 5
Q 11 . If M is the mean of x_{1},x_{2},x_{3},x_{4},x_{5 }and x6_{ }, Prove that
(x_{1}−M)+(x_{2}−M)+(x_{3}−M)+(x_{4}−M)+(x_{5}−M)+(x_{6}−M) = 0.
SOLUTION :
Let M be the mean of x_{1},x_{2},x_{3},x_{4},x_{5 }and x_{6}
= x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} = 6M
To Prove : (x_{1}−M)+(x_{2}−M)+(x_{3}−M)+(x_{4}−M)+(x_{5}−M)+(x_{6}−M) = 0.
Proof : L . H . S
= (x_{1}−M)+(x_{2}−M)+(x_{3}−M)+(x_{4}−M)+(x_{5}−M)+(x_{6}−M)
= (x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6})−(M+M+M+M+M+M)
= 6M – 6M
= 0
= R . H . S
Q 12 . Duration of sunshine(in hours) in Amritsar for first 10 days of August 1997 as reported by the Meterological Department are given as follows : 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9
1. Find the mean
2.Verify that
SOLUTION :
Duration of sunshine (in hours ) for 10 days are =9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9
= 56/10 = 5.6
= (9.6−5.6)+(5.2−5.6)+(3.5−5.6)+(1.5−5.6)+(1.6−5.6)+(2.4−5.6)+(2.6−5.6)+(8.4−5.6)+(10.3−5.6)+(10.9−5.6)
= 4 – 0.4 – 2.1 – 4.1 – 4 – 3.2 – 3 + 2.8 + 4.7 + 5.3
= 16.8 – 16.8 = 0
= R.H.S
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