The document RD Sharma Solutions Ex-24.1, (Part - 1), Measures Of Central Tendency, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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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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