Selina Concise Class 9 Maths Chapter 19 Mean and Median Exercise 19A Solutions
EXERCISE – 19A
(1) Find the mean of 43, 51, 50, 57 and 54.
Solution:
Given Number are = 43, 51, 50, 57 and 54.
Mean x̄ = 43 + 51 + 50 + 57 + 54/ 5
= 255/5
x̄ = 51
(2) Find the mean of first Six natural numbers.
Solution:
The natural numbers are –
1, 2, 3, 4, 5, 6.
Mean x̄ = 1+ 2+ 3+ 4+ 5+ 6/6
= 21/6
x̄ = 3.5
(3) Find the mean of first ten odd natural number.
Solution:
The odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
Mean x̄ = 1+3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19/ 10
x̄ = 100/10
x̄ = 10
(4) Find the mean of all factors of 10.
Solution:
The Factors of 10 are,
1, 2, 5 and 10.
Mean x̄ = 1 + 2 + 5 + 10/4
= 18/4
X̄ = 4.5
(5) Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.
Solution:
The given numbers are –
X + 3, x + 5, x + 7, x + 9 and x + 11/ 5
X̄ = X + 3 + x + 5+ x + 7+ x + 9+ x + 11/ 5
X̄ = 5x + 35/5
X̄ = 5 (x + 7)/5
X̄ = x + 7
(6) If the different values of variable x are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1;
Find
Solution:
The given numbers are –
9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1.
Mean X̄ = 9.8 + 5.4 + 3.7 + 1.7 + 1.8+ 2.6 + 2.8 + 8.6 + 10.5 + 11.1/ 10
X̄ = 58.9/10
X̄ = 5.8
(ii)
Xi | Xi – X̄ | |
9.8 | 9.8 – 5.8 | 4 |
5.4 | 5.4 – 5.8 | -0.4 |
3.7 | 3.7 – 5.8 | -2.1 |
1.7 | 1.7 – 5.8 | -4.1 |
1.8 | 1.8 – 5.8 | -4 |
2.6 | 2.6 – 5.8 | -3.2 |
2.8 | 2.8 – 5.8 | -3 |
8.6 | 8.6 – 5.8 | 2.8 |
10.5 | 10.5 – 5.8 | 4.7 |
11.1 | 11.1 – 5.8 | 5.3 |
(7) The mean of 15 observations is 32 find the resulting mean, if each observation is:
(i) Increased by 3
(ii) Decreased by 3
(iii) Multiplied by 2
(iv) Divided by 0.5
(v) Increased by 60%
(vi) Decreased by 20%
Solution:
(i) mean x̄ = 32
Now,
New mean x̄ = 32 + 3
x̄ = 35
(ii) mean x̄ = 32
New mean x̄ = 32 – 7
x̄ = 25
(iii) mean x̄ = 32
New mean x̄ = 32 × 2
= 64
(iv) mean x̄ = 32
New mean = 32/0.5
x̄ = 32 × 10/ 5
= 32 × 2
x̄ = 64
(v) mean x = 32
New mean = 32 + 60/100 × 100
= 32 + .2
= 32 + .2
= 51.2
x̄ = 51.2
(vi) mean x̄ = 32
New mean x̄ = 32 – × 20/100
= 32 – (32 × 0.2)
= 32 – 6.4
x̄ = 25.6
(8) The mean of 5 numbers is 18. If one number is excluded, the mean of remaining number becomes 16. Find the excluded number.
Solution:
Mean x̄ = 18
Total Sum of 5 numbers = x1 +x2 + x3 ———+ x5 / 5 = 18
X1 + x2 + x3 + x4 + x5 = 18 × 5
X1 + x2 + x3 + x4 + x5 = 90 —– (i)
x̄N = 16
If one number is excluded, the mean of remaining number becomes 16.
X1 + x2 + x3 + x4/ 4 = 16
X1 + x2 + x3 + x4 = 16 × 4
X1 + x2 + x3 + x4 = 64 —- (ii)
Subtracting equation (i) – (ii),
X1 + x2 + x3 + x4 + x5 = 90
X1 + x2 + x3 + x4 = 64
————————————-
X5 = 26
∴ The excluded number is 26
(9) If the mean of observations x, x + 2, x + 4, x + 6 and x + 8, 11, find:
(i) The value of x:
(ii) The mean of first three observations
Solution:
Mean x̄ = 11
X + x + 2 + x + 4 + x + 6 + x + 8/5 = 11
5x + 20/5 = 11
5x + 20 = 11 × 5
5x + 20 = 55
5x = 55 – 20
5x = 35
X = 35/5
X = 7
∴ The value of x = 7
(ii) The first three numbers are –
X, x + 2 and x + 4
X = 7
X + 2 = 7 + 2 = 9
X + 4 = 7 + 4 = 11
∴ mean x̄ = 7 + 9 + 11/3
= 27/3
x̄ = 9
(10) The mean of 100 observations is 40. It is found that an observation 53 was misread as 83. Find the correct mean.
Solution:
Mean x̄ = 40
X1 + x2 +…… x100/100 = 40
X1 + x2 + …… x100 = 40 × 100
X1 + x2 +….. x100 = 4000
X1 + x2 …… + 83 = 4000
X1 + x2+…… + 83 – 30 = 4000
X1 + x2 +……+ 53 = 3970
Original mean x̄ = 3970/100
x̄ = 39.7
∴ correct mean = x̄ = 39.7
Here is your solution of Selina Concise Class 9 Maths Chapter 19 Mean and Median Exercise 19A
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