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NCERT Solutions for Miscellaneous Exercise Chapter 15 Class 11 - Statistics

NCERT Solutions for Miscellaneous Exercise Chapter 15 Class 11 - Statistics

Edited By Sumit Saini | Updated on Jul 18, 2022 12:11 PM IST

NCERT solutions for Class 11 Maths chapter 15 miscellaneous exercise has questions mainly on the concepts which have been discussed in the previous exercises like mean and standard deviation, variability and consistency etc. NCERT book Class 11 Maths chapter 15 miscellaneous exercise provides questions which can be helpful for the revision purpose as well. Class 11 Maths chapter 15 miscellaneous exercise is a good source to enhance one's marks in Boards as well as competitive examinations.

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  1. Statistics Class 11 Chapter 15 -Miscellaneous Exercise
  2. More About NCERT Solutions for Class 11 Maths Chapter 15 Miscellaneous Exercise
  3. Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Miscellaneous Exercises
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject Wise NCERT Exemplar Solutions

Similarly, Many concepts will be used simultaneously in a question which can help recall previously read exercises. Below provided is the Class 11 Maths chapter 15 miscellaneous exercise solutions with other exercises as well. Students can refer to these for more practice in this chapter.

Statistics Class 11 Chapter 15 -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.

Answer:

Given,

The mean and variance of 8 observations are 9 and 9.25, respectively

Let the remaining two observations be x and y,

Observations: 6, 7, 10, 12, 12, 13, x, y.

∴ Mean, \overline X = \frac{6+ 7+ 10+ 12+ 12+ 13+ x+ y}{8} = 9

60 + x + y = 72

x + y = 12 -(i)

Now, Variance

= \frac{1}{n}\sum_{i=1}^8(x_i - x)^2 = 9.25

\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2 -18(x+y)+ 2.9^2 \right ]

\implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2+ -18(12)+ 2.9^2 \right ] (Using (i))

\implies 9.25 = \frac{1}{8}\left[48+x^2+ y^2 -216+ 162 \right ] = \frac{1}{8}\left[x^2+ y^2 - 6 \right ]

\implies x^2+ y^2 = 80 -(ii)

Squaring (i), we get

x^2+ y^2 +2xy= 144 (iii)

(iii) - (ii) :

2xy = 64 (iv)

Now, (ii) - (iv):

\\ x^2+ y^2 -2xy= 80-64 \\ \implies (x-y)^2 = 16 \\ \implies x-y = \pm 4 -(v)

Hence, From (i) and (v):

x – y = 4 \implies x = 8 and y = 4

x – y = -4 \implies x = 4 and y = 8

Therefore, The remaining observations are 4 and 8. (in no order)

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

Answer:

Given,

The mean and variance of 7 observations are 8 and 16, respectively

Let the remaining two observations be x and y,

Observations: 2, 4, 10, 12, 14, x, y

∴ Mean, \overline X = \frac{2+ 4+ 10+ 12+ 14+ x+ y}{7} = 8

42 + x + y = 56

x + y = 14 -(i)

Now, Variance

= \frac{1}{n}\sum_{i=1}^8(x_i - \overline x)^2 = 16

\implies 16 = \frac{1}{7}\left[(-6)^2+ (-4)^2+ 2^2+ 4^2+ 6^2+ x^2+ y^2 -16(x+y)+ 2.8^2 \right ]

\implies 16 = \frac{1}{7}\left[36+16+4+16+36+ x^2+ y^2 -16(14)+ 2(64) \right ] (Using (i))

\implies 16 = \frac{1}{7}\left[108+x^2+ y^2 -96 \right ] = \frac{1}{7}\left[x^2+ y^2 + 12\right ]

\implies x^2+ y^2 = 112- 12 =100 -(ii)

Squaring (i), we get

x^2+ y^2 +2xy= 196 (iii)

(iii) - (ii) :

2xy = 96 (iv)

Now, (ii) - (iv):

\\ x^2+ y^2 -2xy= 100-96 \\ \implies (x-y)^2 = 4 \\ \implies x-y = \pm 2 -(v)

Hence, From (i) and (v):

x – y = 2 \implies x = 8 and y = 6

x – y = -2 \implies x = 6 and y = 8

Therefore, The remaining observations are 6 and 8. (in no order)

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

Answer:

Given,

Mean = 8 and Standard deviation = 4

Let the observations be x_1, x_2, x_3, x_4, x_5\ and\ x_6

Mean, \overline x = \frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} = 8

Now, Let y_i be the the resulting observations if each observation is multiplied by 3:

\\ \overline y_i = 3\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{3}

New mean, \overline y = \frac{y_1+ y_2+ y_3+ y_4+ y_5+ y_6}{6}

= 3\left [\frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} \right] =3\times 8

= 24

We know that,

Standard Deviation = \sigma = \sqrt{Variance}

\dpi{100} =\sqrt{ \frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2}

\dpi{100}\\ \implies 4^2=\frac{1}{6}\sum_{i=1}^6(x_i - \overline x)^2 \\ \implies \sum_{i=1}^6(x_i - \overline x)^2 = 6\times16 = 96 -(i)

Now, Substituting the values of x_i\ and\ \overline x in (i):

\dpi{100} \\ \implies \sum_{i=1}^6(\frac{y_i}{3} - \frac{\overline y}{3})^2 = 96 \\ \implies \sum_{i=1}^6(y_i - \overline y)^2 = 96\times9 =864

Hence, the variance of the new observations = \frac{1}{6}\times864 = 144

Therefore, Standard Deviation = \sigma = \sqrt{Variance} = \sqrt{144} = 12

Question:4 Given that \small \bar {x} is the mean and \small \sigma^2 is the variance of \small n observations .Prove that the mean and variance of the observations \small ax_1,ax_2,ax_3,....,ax_n , are \small a\bar{x} and \small a^2 \sigma^2 respectively, \small (a \neq 0) .

Answer:

Given, Mean = \small \bar {x} and variance = \small \sigma^2

Now, Let y_i be the the resulting observations if each observation is multiplied by a:

\\ \overline y_i = a\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{a}

\overline y =\frac{1}{n}\sum_{i=1}^ny_i = \frac{1}{n}\sum_{i=1}^nax_i

\overline y = a\left [\frac{1}{n}\sum_{i=1}^nx_i \right] = a\overline x

Hence the mean of the new observations \small ax_1,ax_2,ax_3,....,ax_n is \small a\bar{x}

We know,

\dpi{100} \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2

Now, Substituting the values of x_i\ and\ \overline x :

\\ \implies \sigma^2= \frac{1}{n}\sum_{i=1}^n(\frac{y_i}{a} - \frac{\overline y}{a})^2 \\ \implies a^2\sigma^2= \frac{1}{n}\sum_{i=1}^n(y_i - \overline y)^2

Hence the variance of the new observations \small ax_1,ax_2,ax_3,....,ax_n is a^2\sigma^2

Hence proved.

Question:5(i) The mean and standard deviation of \small 20 observations are found to be \small 10 and \small 2 , respectively. On rechecking, it was found that an observation \small 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If wrong item is omitted.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

\overline x =\frac{1}{n}\sum_{i=1}^nx_i

\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i \implies \sum_{i=1}^{20}x_i = 200

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 = 192

Therefore, Correct Mean = (Correct Sum)/19

=\frac{192}{19}

= 10.1

Now, Standard Deviation,

\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

\\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080 ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8)

= 2080 – 64

= 2016

Hence, Correct Standard Deviation =

\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2016}{19} - (10.1)^2}

\dpi{100} = \sqrt{106.1 - 102.01} = \sqrt{4.09}

=2.02

Question:5(ii) The mean and standard deviation of \small 20 observations are found to be \small 10 and \small 2 , respectively. On rechecking, it was found that an observation \small 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If it is replaced by \small 12.

Answer:

Given,

Number of observations, n = 20

Also, Incorrect mean = 10

And, Incorrect standard deviation = 2

\overline x =\frac{1}{n}\sum_{i=1}^nx_i

\implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i \implies \sum_{i=1}^{20}x_i = 200

Thus, incorrect sum = 200

Hence, correct sum of observations = 200 – 8 + 12 = 204

Therefore, Correct Mean = (Correct Sum)/20

\dpi{100} =\frac{204}{20}

= 10.2

Now, Standard Deviation,

\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

\\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080 ,which is the incorrect sum.

Thus, New sum = Old sum - (8x8) + (12x12)

= 2080 – 64 + 144

= 2160

Hence, Correct Standard Deviation =

\sigma' =\sqrt{\frac{1}{n}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2160}{20} - (10.2)^2}

= \sqrt{108 - 104.04} = \sqrt{3.96}

= 1.98

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

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

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

Answer:

Given,

Standard deviation of physics = 15

Standard deviation of chemistry = 20

Standard deviation of mathematics = 12

We know ,

C.V. = \frac{Standard\ Deviation}{Mean}\times 100

The subject with greater C.V will be more variable than others.

C.V. _P= \frac{15}{32}\times 100 = 46.87

C.V. _C= \frac{20}{40.9}\times 100 = 48.89

C.V. _M= \frac{12}{42}\times 100 = 28.57

Hence, Mathematics has lowest variability and Chemistry has highest variability.

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

Answer:

Given,

Initial Number of observations, n = 100

\overline x =\frac{1}{n}\sum_{i=1}^nx_i

\dpi{100} \implies 20 =\frac{1}{100}\sum_{i=1}^{100}x_i \implies \sum_{i=1}^{100}x_i = 2000

Thus, incorrect sum = 2000

Hence, New sum of observations = 2000 - 21-21-18 = 1940

New number of observation, n' = 100-3 =97

Therefore, New Mean = New Sum)/100

=\frac{1940}{97}

= 20

Now, Standard Deviation,

\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

\\ \implies 3^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 9 + (\overline x)^2 \\ \implies \frac{1}{100}\sum_{i=1}^nx_i ^2 = 9 + 400 = 409 \\ \implies \sum_{i=1}^nx_i ^2 = 40900 ,which is the incorrect sum.

Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)

= 40900 - 441 - 441 - 324

= 39694

Hence, Correct Standard Deviation =

\sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{39694}{97} - (20)^2}

= \sqrt{108 - 104.04} = \sqrt{3.96}

= 3.036

More About NCERT Solutions for Class 11 Maths Chapter 15 Miscellaneous Exercise

Miscellaneous exercise chapter 15 Class 11 chapter Statistics Is a good source for revision purposes specially and to check one’s level of understanding of this chapter as most of the questions are using multiple concepts. Class 11 Maths chapter 15 miscellaneous solutions mainly has 7 questions and it can take around 4-5 hours to complete the exercise. Hence, NCERT solutions for Class 11 Maths chapter 15 miscellaneous exercise is one stop solution for scoring better marks in the examination.

Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Miscellaneous Exercises

  • The Class 11 Maths chapter 15 exercise has a set of good questions which must be practiced before the examination.

  • Class 11 Maths chapter 15 miscellaneous exercises provides holistic understanding of the whole chapter as most of the concepts are used in this exercise.

  • These Class 11 Maths chapter 15 miscellaneous exercises is very rewarding from marks point of view in the examination.

Also see-

NCERT Solutions of Class 11 Subject Wise

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Frequently Asked Question (FAQs)

1. How many questions are there in Miscellaneous exercise chapter 15 ?

There are total 7 questions and some questions have subparts as well 

2. What is the use of this chapter statistics in real life ?

Statistics can be used to analyse the given data in order to understand it in a better way.

3. Are questions repeated in the examination from this chapter in the board examination ?

Yes, Sometimes in the Board exam the questions are repeated.

4. What is the level of difficulty of questions asked from this chapter ?

Moderate to difficult level of questions are provided in  this chapter. 

5. Can one skip Miscellaneous exercise if previous exercises are done properly ?

No, as it has some good questions, miscellaneous exercise must be done.. It also helps in revising the earlier concepts.  

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