NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Miscellaneous Exercise

June 11, 2022

NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Miscellaneous Exercise

Textbook	NCERT
class	Class – 11th
Subject	Mathematics
Chapter	Chapter 15
Chapter Name	Statistics
grade	Class 11th Maths solution
Medium	English
Source	last doubt

NCERT Solutions Class 11th Maths Chapter – 15 – Statistics Miscellaneous Exercise

?Chapter – 15?

✍Statistics✍

?Miscellaneous Exercise?

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

?‍♂️Solution: Let x and y be the remaining two observations.
n = 8
Variance  Variance =9.25
Mean X¯= Mean = 9
⇒ 6+7+10+12+12+13+x+y8 = 9
⇒ 60+x+y = 72
⇒ x+y=12…(1)
Variance  Variance X=1n∑i=18x_i2−(X¯)²
⇒9.25=(18×(62+72+102+122+122+132+x²+y²))−92
⇒9.25=18(642+x²+y²)−81
⇒9.25×8=642+x²+y²−648
⇒x²+y²= 80….(2)
We know  We know ,
(x+y)²+(x−y)²=2(x²+y²)
using equations (1) and (2) ⇒122+(x−y)²= 2×80[ using equations (1) and (2) ]
⇒144+(x−y)²=160
⇒(x−y)²= 16
⇒ x − y = ± 4
If x – y = 4, then x + y = 12 and x – y = 4 give x = 8 and  If x – y = 4, then x + y = 12 and x – y = 4 give x = 8 and y = 4
If x – y = – 4, then x + y = 12 and x – y = 4 give x = 4 and  If x – y = – 4, then x + y = 12 and x – y = 4 give x = 4 and y = 8
Thus, the remaining two observations are 8 and 4.

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

?‍♂️Solution: Let the remaining two observation be $x$ and $y$
The observation are $2, 4, 10, 12, 14, x, y$
Mean   $= x ˉ = 2 + 4 + 10 + 12 + 14 + x + y/7 = 8$
$\Rightarrow 56 = 42 + x + y$
$\Rightarrow x + y = 14$    ……(1)
Variance $= 16 = 1/n = l \sum 7 i=1 (X_{i} - X ˉ)^{2}$
$16 = 71 [(- 6)^{2} + (- 4)^{2} + (2)^{2} + (4)^{2} + (6)^{2} + x^{2} + y^{2} - 2 \times 8 (x + y) + 2 \times (8)^{2}]$
$16 = 71 [36 + 16 + 4 + 6 + 36 + x^{2} + y^{2} - 16 (14) + 2 (64)]$    ………..[using (1)]
$16 = 71 [108 + x^{2} + y^{2} - 224 + 128]$
$16 = 71 [12 + x^{2} + y^{2}]$
$\Rightarrow x^{2} + y^{2} = 112 - 12 = 100$
$x^{2} + y^{2} = 100$    ……….(2)
From (1), we obtain
$x^{2} + y^{2} + 2 x y = 196$    ….(3)
From (2) and (3), we obtain
$2 x y = 196 - 100$
$\Rightarrow 2 x y = 96$ ………..(4)
subtracting (4) from (2), we obtain
$x^{2} + y^{2} - 2 x y = 100 - 96$
$\Rightarrow (x - y)^{2} = 4$
$\Rightarrow x - y = \pm 2$    …………(5)
Therefore, from (1) and (5) we obtain
$x = 8$ and $y = 6$ when $x - y = 2$
$x = 6$ and $y = 8$ when $x - y = - 2$
Thus the remaining observation are $6$ and $8$

3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

?‍♂️Solution: Let the observations be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$
It is given that mean is $8$ and standard deviation is $4$ .
$\Rightarrow$ Mean $= x ˉ = x_{1}, x_{2}, x_{3}, x_{4}, x_{5} = 8$    ….(1)
If each observation is multiplied by $3$ and the resulting observation are $y_{i}$ , then
   $y_{i} = 3 x_{i}, for i = 1 to 6$
$∴$ New mean   $y ˉ = y_{1},y_{2},y_{3},y_{4},y_{5} + y ^{6/6}$
$= 3 ( x ^{1} , x ^{2} , x ^{3} , x ^{4} , x ^{5} )/6$
$= 3 \times 8$       ……..[Using (1) ]
$= 24$

We know that,

4. Given that x̅ is the mean and σ² is the variance of n observations x₁, x₂, …,x_n . Prove that the mean and variance of the observations ax₁, ax₂, ax₃, …., ax_n are ax̅ and a²σ², respectively, (a ≠ 0).

?‍♂️Solution: From the question it is given that, n observations are x₁, x₂,…..x_n

Mean of the n observation = x̅

Variance of the n observation = σ²

As we know that,

5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12

?‍♂️Solution: (i) If wrong item is omitted,

From the question it is given that,

The number of observations i.e. n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

(ii) If it is replaced by 12,

From the question it is given that,

The number of incorrect sum observations i.e. n = 200

The correct sum of observations n = 200 – 8 + 12

n = 204

Then, correct mean = correct sum/20

= 204/20

= 10.2

6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject	Mathematics	Physics	Chemistry
Mean	42	32	40.9
Standard deviation	12	15	20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

?‍♂️Solution: From the question it is given that,

Mean of Mathematics = 42

Standard deviation of Mathematics = 12

Mean of Physics = 32

Standard deviation of physics = 15

Mean of Chemistry = 40.9

Standard deviation of chemistry = 20

As we know that,

7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

?‍♂️Solution: From the question it is given that,

The total number of observations (n) = 100

Incorrect mean, (x̅) = 20

And, Incorrect standard deviation (σ) = 3