RS Aggarwal Solutions Class 9 Chapter 14 Statistics Solution

EXERCISE 14D

Question 1:
(i) first eight natural numbers are:
1,2,3,4,5,6,7and 8

(ii) First ten odd numbers are:
1,3,5,7,9,11,13,15, 17, and 19

(iii) First five prime numbers are: 2, 3, 5, 7, 11

(iv) First six even numbers are: 2,4,6,8,10,12

(v) First seven multiples of 5 are: 5,10,15, 20, 25, 30, 35

Therefore, Mean =20
(vi) Factors of 20 are: 1,2,4,5,10,20

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:
Let the given numbers be x₁,x₂……..x₂₄

Question 8:
Let the given numbers be x₁, x₂……..x₂₀
Then , the mean of these numbers =

Question 9:
Let the given numbers be x₁, x₂…………x₁₅
Then , the mean of these numbers =

Question 10:

Question 11:

Question 12:
Mean weight of the boys =48 kg
Therefore , Mean weight = \(\frac { Sum of the weight of six boys }{ 6 }=48 \)
Sum of the weight of 6 boys =(48×6)kg =288kg
Sum of the weights of 5 boys=(51+45+49+46+44)kg=235kg
Weight of the sixth boy=(sum of the weights of 6 boys ) – (sum of the weights of 5 boys)
=(288-235)=53kg.
∴ weight of the sixth boy = 53kg

Question 13:
Calculated mean marks of 50 students =39
calculated sum of these marks=(39x 50)=1950
Corrected sum of these marks
=[1950-(wrong number)+(correct number)]
=(1950-23+43) =1970
∴ correct mean = \(\frac { 1970 }{ 50}=39.4 \)

Question 14:
calculated mean of 100 items =64
sum of 100 items, as calculated = (100×64) =6400
Correct sum of these items =[6400-(wrong items )+(correct items)]
=[6400-(26+9)+(36+90)]
=[6400-35+126]=6491
∴ Correct mean = \(\frac { 6491 }{ 100}=64.91 \)

Question 15:
Mean of 6 numbers = 23
Sum of 6 numbers =(23×6 )=138
Again , mean of 5 numbers =20
Sum of 5 numbers=(20x 5 ) =100
The excluded number= (sum of 6 numbers )-(sum of 5 numbers)
=(138-100) =38
∴ The excluded number=38.