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Class 11 NCERT Solutions- Chapter 15 Statistics – Miscellaneous Exercise On Chapter 15

  • Last Updated : 03 Mar, 2021

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,

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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 \overline {x} = \frac{\sum{x}}{n}  = 9 

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

But, \overline{x}

Solving for a + b, we get, 

a + b = 12

Also, 

Variance \sigma^2 = \frac{1}{n}\sum{x}^2 - (\overline{x})^2



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

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

\overline {x} = \frac{\sum{x_i}}{n}

But,

\overline{x} = 8

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

Solving for a + b, we get,

a + b = 14

Also,

Variance = \sigma^2 = \frac{1}{n}\sum{x}^2 - (\overline{x})^2

Equating

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



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, x

Therefore, 

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

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

yi = 3xi

xi = (1/3)yi, where i = 1….6

So, new mean 

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

= 3(x1 + x2 + x3 + x4 + x5 + x6)/6

= 3 * 8

= 24

Standard Deviation (\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



\overline{y} = 3\overline{x} \\ \overline{x} = \frac{1}{3}\overline {y}

Substituting values, we get, 

\sum_{i=1}^{6}(y_i - \overline{y})^2 = 864

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, …,xn . 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, …xn

Mean of n observations = \overline{x}

Variance of n observations = \sigma^2

We know, 

Variance(\sigma^2) = \frac{1}{n}\sum_{i=1}^{n}y_1(x_i - \overline{x})^2

yy_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}

Now, 

Mean of the observations, ax1, ax2, …..axn\overline{x}

By substituting values, we get,

\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

Variance = a^2\sigma^2



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

\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

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 (\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}

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

Incorrect \sum_{i=1}^{n}X_1^2 = 2080 

Therefore, 

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

 = 2080 – 64 

= 2016



Calculating correct standard deviation, 

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

= √(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 (\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}

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

Incorrect \sum_{i=1}^{n}X_1^2 = 2080

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

= 2080 – 64 + 144 

= 2160

Calculating correct standard deviation,

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

= √(108 – 104.04) 



= 1.98

Question 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: 

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

SubjectMathematicsPhysics Chemistry
Mean423240.9
Standard Deviation121520

Solution:

Given values,

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

Now,

Coefficient of Variation (C.V) = \frac{Standard Deviation}{Mean} * 100

CV for Mathematics = 12/42 x 100 = 28.57

CV for Physics = 15/32 x 100 = 46.87

CV for Chemistry = 20/40.9 x 100 = 48.89

The Highest Variability of the subject is of Chemistry. 

Question 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:

Given:

n = 100

Incorrect mean, (x̅) = 20

Incorrect standard deviation (σ) = 3

Therefore, 20 = \frac{1}{100}\sum_{i=1}^{100} x_i

On solving, we get

\sum_{i=1}^{100}X_1 = 20 * 100 \\ \sum_{i=1}^{100}X_1 = 2000

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 (\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}

Incorrect \sum X_1^2 = 100(9 + 400) 

Incorrect \sum X_1^2 = 40900

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

= 40900 – 441 – 441 – 324 

= 40900 – 1206 

= 39694

Therefore, 

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

= √(409.216 – 400) 

= 3.036




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