NCERT Solutions Class 11 – Chapter 13 Statistics – Miscellaneous Exercise

Question 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:

Given,

We are provided with six of the observations 6, 7, 10, 12, 12 and 13.

Let us assume the missing observations to be a and b.

Now, Mean [Tex]\overline {x} = \frac{\sum{x}}{n} [/Tex] = 9 

9 = (6 + 7 + 10 + 12 + 12 + 13 + a + b)/8

But, [Tex]\overline{x}[/Tex]

Solving for a + b, we get, 

a + b = 12

Also, 

Variance [Tex]\sigma^2 = \frac{1}{n}\sum{x}^2 – (\overline{x})^2[/Tex]

Equating  [Tex]\sigma^2 = 26\space and \space \overline{x}=9, n = 8[/Tex]

We have, 

9.25 = 1/8(6² + 7² + 10² + 12² + 12² + 13² + a² + b²) –  9²

=> 9.25 + 81 = 1/8(36 + 49 + 100 + 144 + 144 + 169 + a² + b²)

=> 90.25 * 8 = 642 + a² + b²

=> a² + b² = 80

We have, b = 12 – a 

On substituting the value, 

a² + (12 – a)²= 80

=> 2a² – 24a + 64 = 0

On dividing by 2, we get 

a² – 12a + 32 = 0

Therefore, a = 4, 8 

Now, for a = 4, b = 8

And, for a = 8, b = 4

Question 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:

Given,

We are provided with five of the observations 2, 4, 10, 12, 14.

Let us assume the missing observations to be a and b.

Now, Mean

[Tex]\overline {x} = \frac{\sum{x_i}}{n}[/Tex]

But,

[Tex]\overline{x} = 8[/Tex]

8 = (2 + 4 + 10 + 12 + 14 + a + b)/(7)

Solving for a + b, we get,

a + b = 14

Also,

Variance = [Tex]\sigma^2 = \frac{1}{n}\sum{x}^2 – (\overline{x})^2[/Tex]

Equating

[Tex]\sigma^2 = 16\space and \space \overline{x}=8, n =7[/Tex]

We have, 

16 = 1/7(2² + 4² + 10² + 12² + 14² + a² + b²) – 64

=> 16 + 64 = 1/7(4 + 16 + 100 + 144 + 196 + a² + b²)

=> 560 = 460 + a² + b²

=> a²+b² = 100

We have, b = 14 – a

On substituting the value, we get

a² + (14 – a)²= 100

=> 2a² – 28a + 96 = 0

On dividing by 2, we get

a² – 14a + 48 = 0

=> (a – 8)(a – 6) = 0

Therefore, a = 6, 8

Now, for a = 6, b = 8

And, for a = 8, b = 6

Question 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:

Mean of six observations = 8

Standard deviation of six observations = 4

Let the six observations be x₁, x₂, x₃, x₄, x₅, x₆ 

Therefore, 

Mean of observations, [Tex]\overline{x} [/Tex] = (x₁ + x₂ + x₃ + x₄ + x₅ + x₆)/6 = 8

If each observation is multiplied by 3 and the resultant observations are y_i then,

y_i = 3x_i

x_i = (1/3)y_i, where i = 1….6

So, new mean 

[Tex]\bar{y} [/Tex] = (y₁ + y₂ + y₃ + y₄ + y₅ + y₆)/6

= 3(x₁ + x₂ + x₃ + x₄ + x₅ + x₆)/6

= 3 * 8

= 24

Standard Deviation [Tex](\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1 – \frac{1}{n^2}(\sum_{i=1}^{n}X)^2 } \\ (4)^2 = \frac{1}{6}\sum_{i=1}^{6}(x_i-\overline{x})^2 \\ \sum_{i=1}^{6}(x_i-\overline{x})^2 = 96 [/Tex]

[Tex]\overline{y} = 3\overline{x} \\ \overline{x} = \frac{1}{3}\overline {y}[/Tex]

Substituting values, we get, 

[Tex]\sum_{i=1}^{6}(y_i – \overline{y})^2 = 864[/Tex]

Therefore, 

Variance of new observation = 1/6 x 864

= 144

Standard Deviation = √144

= 12

Question 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, …., axⁿ are ax̅ and a²σ², respectively, (a ≠ 0).

Solution:

Let us assume the observations to be x₁, …x_n

Mean of n observations = [Tex]\overline{x}[/Tex]

Variance of n observations = [Tex]\sigma^2[/Tex]

We know, 

[Tex]Variance(\sigma^2) = \frac{1}{n}\sum_{i=1}^{n}y_1(x_i – \overline{x})^2 [/Tex]

y[Tex]y_i = ax_i \\ x_i = \frac{1}{a}y_i \\ \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i \\ \overline{y} = \frac{1}{n}\sum_{i=1}^{n}ax_i \\ \overline{y} = a\overline{x}[/Tex]

Now, 

Mean of the observations, ax¹, ax², …..axⁿ = [Tex]\overline{x}[/Tex]

By substituting values, we get,

[Tex]\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(\frac{1}{a}y_1 – \frac{1}{a}\overline{y})^2 \\ a^2\sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(y_1-\overline{y})^2[/Tex]

Variance = [Tex]a^2\sigma^2[/Tex]

Question 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) On omission of wrong item 

n = 20

Incorrect mean = 20

Incorrect SD = 2

[Tex]\overline{X} = \frac{1}{n}\sum_{i=1}^{20}X_1 \\ 10 = \frac{1}{20}\sum_{i=1}^{20}X_1 \\ \sum_{i=1}^{20}X_1 = 200[/Tex]

Now, 

Incorrect sum of observations = 200 

Correct sum of observations = 200 – 8 = 192

Therefore, 

Correct mean = Correct sum/19

= 192/19

= 10.1

Standard Deviation [Tex](\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1 – \frac{1}{n^2}(\sum_{i=1}^{n}X)^2 } \\ 2 = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1^2 – (\overline{X})^2}[/Tex]

 4 = 1/20 Incorrect [Tex]\sum_{i=1}^{n}X_1^2[/Tex] – 100 

Incorrect [Tex]\sum_{i=1}^{n}X_1^2[/Tex] = 2080 

Therefore, 

Correct [Tex]\sum_{i=1}^{n}X_1^2[/Tex] = Incorrect [Tex]\sum_{i=1}^{n}X_1^2[/Tex] – (8)²

 = 2080 – 64 

= 2016

Calculating correct standard deviation, 

Correct SD = [Tex]\sqrt{\frac{Correct \sum X_1^2}{n}-(Correct Mean)^2} \\ = \sqrt{\frac{2016}{19}-(10.1)^2}  [/Tex]

= √(1061.1 – 102.1)

= 2.02

(ii) If it is replaced by 12, 

Incorrect sum of observations, n = 200

Correct sum of observations n = 200 – 8 + 12

n = 204

Correct Mean = Correct Sum / 20

= 204/20

= 10.2

Standard Deviation [Tex](\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1 – \frac{1}{n^2}(\sum_{i=1}^{n}X)^2 } \\ 2 = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1^2 – (\overline{X})^2}[/Tex]

4 = 1/20 Incorrect [Tex]\sum_{i=1}^{n}X_1^2 [/Tex]– 100 

Incorrect [Tex]\sum_{i=1}^{n}X_1^2[/Tex] = 2080

Therefore, Correct [Tex]\sum_{i=1}^{n}X_1^2[/Tex] = Incorrect [Tex]\sum_{i=1}^{n}X_1^2  [/Tex]– (8)² + (12)² 

= 2080 – 64 + 144 

= 2160

Calculating correct standard deviation,

[Tex]Correct SD = \sqrt{\frac{Correct \sum X_1^2}{n}-(Correct Mean)^2} \\ = \sqrt{\frac{2160}{20}-(10.2)^2} [/Tex]

= √(108 – 104.04) 

= 1.98

Question 6. 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:

Given:

n = 100

Incorrect mean, (x̅) = 20

Incorrect standard deviation (σ) = 3

Therefore, [Tex]20 = \frac{1}{100}\sum_{i=1}^{100} x_i[/Tex]

On solving, we get

[Tex]\sum_{i=1}^{100}X_1 = 20 * 100 \\ \sum_{i=1}^{100}X_1 = 2000[/Tex]

Incorrect sum of observations = 2000

Now, Correct sum of observations = 2000 – 21- 21 – 18

= 1940

Correct mean = Correct Sum / 97

= 1940/97

= 20

Also,

Standard Deviation [Tex](\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1 – \frac{1}{n^2}(\sum_{i=1}^{n}X)^2 } \\ 3 = \sqrt{\frac{1}{n}\sum_{i=1}^{n}X_1^2 – (\overline{X})^2} \\ 3 = \sqrt{\frac{1}{100}* Incorrect \sum X_1^2-(20)^2} [/Tex]

Incorrect [Tex]\sum X_1^2[/Tex] = 100(9 + 400) 

Incorrect [Tex]\sum X_1^2[/Tex] = 40900

Correct [Tex]\sum_{i=1}^{n}X_1^2 [/Tex]= Incorrect [Tex]\sum_{i=1}^{n}X_1^2[/Tex] – (21)² – (21)² – (18)² 

= 40900 – 441 – 441 – 324 

= 40900 – 1206 

= 39694

Therefore, 

Correct S.D = [Tex]\sqrt{\frac{Correct\sum X_1^2}{n}-(Correct mean)^2} \\ = \sqrt{\frac{39694}{97}-(20)^2} \\ [/Tex]

= √(409.216 – 400) 

= 3.036