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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Find the standard deviation of the first $n$ natural numbers.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} \sum_{}^{} x_{i} = 1 + 2 + 3 + 4 + \dots + n = \frac{n (n + 1)}{2} \\ \sum_{}^{} x_{i}^{2} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ ∴ S . D . (σ) = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2}} \\ = \sqrt[]{\frac{n (n + 1) (2 n + 1)}{6 n} - \frac{n^{2} {(n + 1)}^{2}}{4 n^{2}}} \\ = \sqrt[]{\frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{4}} \\ = \sqrt[]{\frac{2 n^{2} + 3 n + 1}{6} - \frac{n^{2} + 2 n + 1}{4}} \\ = \sqrt[]{\frac{4 n^{2} + 6 n + 2 - 3 n^{2} - 6 n - 3}{12}} = \sqrt[]{\frac{n^{2} - 1}{12}} \\ H e n c e, t h e r e q u i r e d S D = \sqrt[]{\frac{n^{2} - 1}{12}} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} F i r s t n n a t u r a l n u m b e r s a r e 1, 2, 3, 4, 5, 6, \dots, n . H e r e, n i s e v e n . \\ ∴ M e a n \bar{x} = \frac{1 + 2 + 3 + 4 + \dots + n}{n} = \frac{n (n + 1)}{2 n} = \frac{n + 1}{2} \\ ∴ M D = \frac{1}{n} [\begin{array}{l} | 1 - \frac{n + 1}{2} | + | 2 - \frac{n + 1}{2} | + | 3 - \frac{n + 1}{2} | + \dots + | \frac{n - 2}{2} - \frac{n + 1}{2} | + | \frac{n}{2} - \frac{n + 1}{2} | + \\ | \frac{n + 2}{2} - \frac{n + 1}{2} | + \dots + | n - \frac{n + 1}{2} | \end{array}] \\ = \frac{1}{n} [| \frac{1 - n}{2} | + | \frac{3 - n}{2} | + | \frac{5 - n}{2} | + \dots + | \frac{- 3}{2} | + | - \frac{1}{2} | + | \frac{1}{2} | + \dots + | \frac{n - 1}{2} |] \\ = \frac{1}{n} [\frac{1}{2} + \frac{3}{2} + \dots + \frac{n - 1}{2}] (\frac{n}{2}) t e r m s \\ = \frac{1}{n} {(\frac{n}{2})}^{2} = \frac{1}{n} . \frac{n^{2}}{4} = \frac{n}{4} [? S u m o f f i r s t o d d n n a t u r a l n u m b e r s = n^{2}] \\ H e n c e, t h e r e q u i r e d M D = \frac{n}{4} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.

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Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} F i r s t n n a t u r a l n u m b e r s a r e 1, 2, 3, \dots, n . H e r e, n i s o d d . \\ ∴ M e a n \bar{x} = \frac{1 + 2 + 3 + \dots + n}{n} = \frac{\frac{n (n + 1)}{2}}{n} = \frac{n + 1}{2} \\ T h e d e v i a t i o n s o f n u m b e r s f r o m m e a n (\frac{n + 1}{2}) a r e \\ 1 - \frac{n + 1}{2}, 2 - \frac{n + 1}{2}, 3 - \frac{n + 1}{2}, \dots, n - \frac{n + 1}{2} \\ i . e ., - \frac{n - 1}{2}, - \frac{n - 3}{2}, \dots, - 2, - 1, 0, 1, 2, \dots, \frac{n - 1}{2} . \\ T h e a b s o l u t e v a l u e s o f d e v i a t i o n f r o m t h e m e a n i . e ., | x_{i} - \bar{x} | a r e \\ \frac{n - 1}{2}, \frac{n - 3}{2}, \dots, 2, 1, 0, 1, 2, \dots, \frac{n - 1}{2} . \\ T h e s u m o f a b s o l u t e v a l u e s o f d e v i a t i o n f r o m t h e m e a n i . e ., | x_{i} - \bar{x} | a r e \\ = 2 (1 + 2 + 3 + \dots t o \frac{n - 1}{2} t e r m s) \\ = 2 . \frac{\frac{n - 1}{2} (\frac{n - 1}{2} + 1)}{2} = \frac{n - 1}{2} . \frac{n + 1}{2} = \frac{n^{2} - 1}{4} . \\ ∴ M e a n d e v i a t i o n a b o u t t h e m e a n = \frac{\sum_{}^{} | x_{i} - \bar{x} |}{n} = \frac{\frac{n^{2} - 1}{4}}{n} = \frac{n^{2} - 1}{4 n} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

If the midpoints of the sides of a triangle $A B, B C, C A$ are $D (1, 2, 3), E (3, 0, 1), F (1, 1, 4)$ , then the centroid of the triangle $A B C$ is ____.

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Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t t h e v e r t i c e s o f t h e Δ A B C b e \\ A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}) a n d C (x_{3}, y_{3}, z_{3}) \\ D i s t h e m i d -point o f A B \\ ∴ 1 = \frac{x_{1} + x_{2}}{2} \Rightarrow x_{1} + x_{2} = 2 \dots (i) \\ 2 = \frac{y_{1} + y_{2}}{2} \Rightarrow y_{1} + y_{2} = 4 \dots (i i) \\ a n d - 3 = \frac{z_{1} + z_{2}}{2} \Rightarrow z_{1} + z_{2} = - 6 \dots (i i i) \\ E i s t h e m i d -point o f B C \\ ∴ 3 = \frac{x_{2} + x_{3}}{2} \Rightarrow x_{2} + x_{3} = 6 \dots (i v) \\ 0 = \frac{y_{2} + y_{3}}{2} \Rightarrow y_{2} + y_{3} = 0 \dots (v) \\ a n d 1 = \frac{z_{2} + z_{3}}{2} \Rightarrow z_{2} + z_{3} = 2 \dots (v i) \\ F i s t h e m i d -point o f A C \\ ∴ - 1 = \frac{x_{1} + x_{3}}{2} \Rightarrow x_{1} + x_{3} = - 2 \dots (v i i) \\ 1 = \frac{y_{1} + y_{3}}{2} \Rightarrow y_{1} + y_{3} = 2 \dots (v i i i) \\ a n d - 4 = \frac{z_{1} + z_{3}}{2} \Rightarrow z_{1} + z_{3} = - 8 \dots (i x) \\ A d d i n g e q . (i), (i v) a n d (v i i) w e g e t \\ 2 (x_{1} + x_{2} + x_{3}) = 2 + 6 - 2 \\ ∴ x_{1} + x_{2} + x_{3} = 3 \dots (x) \\ A d d i n g e q . (i i), (v) a n d (v i i i) w e g e t \\ 2 (y_{1} + y_{2} + y_{3}) = 4 + 0 + 2 \\ ∴ y_{1} + y_{2} + y_{3} = 3 \dots (x i) \\ A d d i n g e q . (i i i), (v i) a n d (i x) w e g e t \\ 2 (z_{1} + z_{2} + z_{3}) = - 6 + 2 - 8 \\ ∴ z_{1} + z_{2} + z_{3} = - 6 \dots (x i i) \\ S u b t r a c t i n g (i) f r o m e q . (x) w e g e t \\ x_{3} = 3 - 2 = 1 \Rightarrow x_{3} = 1 \\ S u b t r a c t i n g (i v) f r o m e q . (x) w e g e t \\ x_{1} = 3 - 6 = - 3 \Rightarrow x_{1} = - 3 \\ S u b t r a c t i n g (v i i) f r o m e q . (x) w e g e t \\ x_{2} = 3 - (- 2) = 5 \Rightarrow x_{2} = 5 \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

If the distance between the points $(a, 2, 1)$ and $(1, 1, 1)$ is 5, then $a$ ____.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e g i v e n points a r e (a, 2, 1) a n d (1, - 1, 1) \\ ∴ Distance = \sqrt[]{{(a - 1)}^{2} + {(2 + 1)}^{2} + {(1 - 1)}^{2}} \\ 5 = \sqrt[]{a^{2} + 1 - 2 a + 9} \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ 25 = a^{2} + 1 - 2 a + 9 \\ \Rightarrow a^{2} - 2 a - 15 = 0 \\ \Rightarrow a^{2} - 5 a + 3 a - 15 = 0 \\ \Rightarrow a (a - 5) + 3 (a - 5) = 0 \\ \Rightarrow (a + 3) (a - 5) = 0 \\ ∴ a = - 3 o r 5 \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s 5 o r - 3 . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

The length of the longest piece of string that can be stretched straight in a rectangular room whose dimensions are 10, 13, and 8 units is ____.

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$T h e g i v e n dimensions$
$\begin{array}{l} a r e 10, 13 a n d 8 \\ L e t a = 10, b = 13 a n d c = 8 \\ ∴ Required l e n g t h = \sqrt[]{a^{2} + b^{2} + c^{2}} \\ = \sqrt[]{{(10)}^{2} + {(13)}^{2} + {(8)}^{2}} = \sqrt[]{100 + 169 + 64} = \sqrt[]{333} \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s \sqrt[]{333} . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

The plane parallel to $y z$ -plane is perpendicular to ____.

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Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e p l a n e p a r a l l e l t o y z - p l a n e i s p e r p e n d i c u l a r t o x - a x i s . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s x - a x i s . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

$x = a$ represents a plane parallel to ____.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$x = a r e p r e s e n t s a p l a n e p a r a l l e l t o y z - p l a n e .$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

A line is parallel to $x$ -axis if all the points on the line have equal ____.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} y a n d z c o o r d i n a t e s . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s y a n d z c o o r d i n a t e s . \end{array}$

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New answer posted

a year ago

Class 11th Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

A line is parallel to $x y$ -plane if all the points on the line have equal ____.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} z - c o o r d i n a t e s . \\ H e n c e, t h e v a l u e o f t h e f i l l e r i s z - c o o r d i n a t e s . \end{array}$

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