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Maths Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Two sets, each of 20 observations, have the same standard deviation 5. The first set has a mean of 17, and the second set has a mean of 22. Determine the standard deviation of the set obtained by combining the two given sets.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t n_{1} = 20, \bar{x_{1}} = 17 a n d σ_{1} = 5 \\ a n d n_{2} = 20, \bar{x_{2}} = 22 a n d σ_{2} = 5 \\ W e k n o w t h a t f o r t h e c o m b i n e d t w o s e r i e s t h a t \\ ∴ σ = \sqrt[]{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(\bar{x_{1}} - \bar{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}}} = \sqrt[]{\frac{20 * {(5)}^{2} + 20 * {(5)}^{2}}{20 + 20} + \frac{20 * 20 {(17 - 22)}^{2}}{{(20 + 20)}^{2}}} \\ = \sqrt[]{\frac{1000}{40} + \frac{400 * 25}{1600}} = \sqrt[]{25 + \frac{25}{4}} = \sqrt[]{\frac{125}{4}} = \sqrt[]{31.25} = 5.59 \\ H e n c e, t h e r e q u i r e d S D = 5.59 \end{array}$

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a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

The mean and standard deviation of a set of $n_{1}$ observations are ${\overset{?}{x}}_{1}$ and $s_{1}$ , respectively, while the mean and standard deviation of another set of $n_{2}$ observations are ${\overset{?}{x}}_{2}$ and $s_{2}$ , respectively. Show that the standard deviation of the combined set of $(n_{1} + n_{2})$ observations is given by:

$S . D . = \sqrt[]{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2} + n_{1} n_{2} {({\overset{?}{x}}_{1} - {\overset{?}{x}}_{2})}^{2}}{n_{1} + n_{2}}}$

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t x_{i}, i = 1, 2, 3, 4, \dots n_{1} \\ a n d y_{j}, j = 1, 2, 3, 4, \dots n_{2} \\ ∴ \bar{x_{1}} = \frac{1}{n_{1}} \sum_{i = 1}^{n} x_{i} a n d \bar{x_{2}} = \frac{1}{n_{2}} \sum_{i = 1}^{n} y_{j} \\ \Rightarrow σ_{1}^{2} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{1}})}^{2} a n d σ_{2}^{2} = \frac{1}{n_{2}} \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x_{2}})}^{2} \\ N o w m e a n o f t h e c o m b i n e d s e r i e s i s g i v e n b y \\ \bar{x} = \frac{1}{n_{1} + n_{2}} [\sum_{i = 1}^{n} x_{i} + \sum_{i = 1}^{n} y_{j}] = \frac{n_{1} \bar{x_{1}} + n_{2} \bar{x_{2}}}{n_{1} + n_{2}} \\ T h e r e f o r e, σ^{2} o f t h e c o m b i n e d s e r i e s i s \\ σ^{2} = \frac{1}{n_{1} + n_{2}} [\sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x})}^{2} + \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x})}^{2}] \\ N o w, \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x})}^{2} = \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{j}} + \bar{x_{j}} - \bar{x})}^{2} \\ = \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{j}})}^{2} + n_{1} {(\bar{x_{j}} - \bar{x})}^{2} + 2 (\bar{x_{j}} - \bar{x}) \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x_{j}})}^{2} \\ B u t \sum_{i = 1}^{n} (x_{i} - \bar{x_{i}}) = 0 [\begin{array}{l} ? T h e algebraic s u m o f t h e d e v i a t i o n o f v a l u e s o f f i r s t \\ s e r i e s f r o m t h e i r m e a n i s z e r o . \end{array}] \\ A l s o \sum_{i = 1}^{n_{1}} {(x_{i} - \bar{x})}^{2} = n_{1} s_{1}^{2} + n_{1} {(\bar{x_{1}} - \bar{x})}^{2} \\ = n_{1} s_{1}^{2} + n_{1} d_{1}^{2} \\ w h e r e d_{1} = (\bar{x_{1}} - \bar{x}) \\ S i m i l a r l y, w e h a v e \\ \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x})}^{2} = \sum_{j = 1}^{n_{2}} {(y_{j} - \bar{x_{i}} + \bar{x_{i}} - \bar{x})}^{2} = n_{2} s_{2}^{2} + n_{2} d_{2}^{2} w h e r e d_{2} = (\bar{x_{2}} - \bar{x}) \\ N o w c o m b i n e d Standard D e v i a t i o n (S D) \\ σ = \sqrt[]{\frac{n_{1} (s_{1}^{2} + d_{1}^{2}) + n_{2} (s_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}}} \\ w h e r e d_{1} = \bar{x_{1}} - \bar{x} = \bar{x_{1}} - (\frac{n_{1} \bar{x_{1}} + n_{2} \bar{x_{2}}}{n_{1} + n_{2}}) = \frac{n_{2} (\bar{x_{1}} - \bar{x_{2}})}{n_{1} + n_{2}} \\ d_{2} = \bar{x_{2}} - \bar{x} = \bar{x_{2}} - (n_{1} \bar{x_{1}} + \end{array}$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results:

Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 seconds. Further, another set of 15 observations $x_{1}, x_{2}, \dots, x_{15}$ , also in seconds, is now available, and we have: $\sum_{i = 1} x_{i} = 279$ $\sum_{i = 1} x_{i}^{2} = 5524$ Calculate the standard deviation based on all 40 observations.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t n_{1} = 25, \bar{x_{1}} = 18.2 a n d σ_{1} = 3.25 \\ a n d n_{2} = 15, \sum_{i = 1} x_{i} = 279 a n d \sum_{i = 1} x_{i}^{2} = 5524 \\ F o r t h e f i r s t s e t, w e h a v e \\ \sum_{}^{} x_{1} = 25 * 18.2 = 455 \\ ∴ σ_{1}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{25} - {(18.2)}^{2} \\ \Rightarrow {(3.25)}^{2} = \frac{\sum_{}^{} x_{i}^{2}}{25} - 331.24 \Rightarrow 10.5625 + 331.24 = \frac{\sum_{}^{} x_{i}^{2}}{25} \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 25 * (10.5625 + 331.24) \\ = 25 * 341.8025 = 8545.06 \\ F o r t h e c o m b i n e d s t a n d a r d d e v i a t i o n o f t h e 40 o b s e r v a t i o n, n = 40 \\ a n d \sum_{}^{} x_{i}^{2} = 5524 + 8545.06 = 14069.06 \\ \Rightarrow \sum_{}^{} x_{i} = 455 + 279 = 734 \\ ∴ S D = \sqrt[]{\frac{14069.06}{40} - {(\frac{734}{40})}^{2}} = \sqrt[]{351.7265 - {(18.35)}^{2}} \\ = \sqrt[]{351.7265 - 336.7225} = \sqrt[]{15.004} = 3.87 \\ H e n c e, t h e r e q u i r e d S D = 3.87 \end{array}$

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New answer posted

a year ago

Maths 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

Maths 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

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Ten

Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $B C$ and $A D$ , then the height of $P$ (in $m$ ) above the line $A C$ is:

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alok kumar singh

Contributor-Level 10

$t a n ? θ = \frac{10}{x} = \frac{h}{x_{2}} \Rightarrow x_{2} = \frac{h x}{10}$

$t a n ? ? = \frac{15}{x} = \frac{h}{x} \Rightarrow x_{1} = \frac{h x}{15}$

Now, $x_{1} + x_{2} = x = \frac{h x}{15} + \frac{h x}{10}$

$\Rightarrow 1 = \frac{h}{10} + \frac{h}{15} \Rightarrow h = 6$

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Six

If $(a + \sqrt[]{2} b c o s x) (a - \sqrt[]{2} b c o s y) = a^{2} - b^{2}$ , where $a > b > 0$ , then $\frac{d x}{d y}$ at $(\frac{π}{4}, \frac{π}{4})$ is:

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alok kumar singh

Contributor-Level 10

$(a + \sqrt[]{2} b c o s ? x) (a - \sqrt[]{2} b c o s ? y) = a^{2} - b^{2}$

$\Rightarrow a^{2} - \sqrt[]{2} a b c o s ? y + \sqrt[]{2} a b c o s ? x - 2 b^{2} c o s ? x c o s ? y = a^{2} - b^{2}$

Differentiating both sides:

$0 - \sqrt[]{2} a b (- s i n ? y \frac{d y}{d x}) + \sqrt[]{2} a b (- s i n ? x) - 2 b^{2} [c o s ? x (- s i n ? y \frac{d y}{d x}) + c o s ? y (- s i n ? x)] = 0$

At $(\frac{π}{4}, \frac{π}{4})$ :

$\begin{array}{r} a b \frac{d y}{d x} - a b - 2 b^{2} (- \frac{1}{2} \frac{d y}{d x} - \frac{1}{2}) = 0 \\ \Rightarrow \frac{d x}{d y} = \frac{a b + b^{2}}{a b - b^{2}} = \frac{a + b}{a - b}; a, b > 0 \end{array}$

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New answer posted

a year ago

Maths 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

Maths Vector Algebra

Let $x_{0}$ be the point of local maxima of $f (x) = \vec{a} \cdot (\vec{b} * \vec{c})$ where $\vec{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = - 2 \hat{i} + x \hat{j} - \hat{k}$ and $\vec{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$ . Then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ at $x = x_{0}$ is

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alok kumar singh

Contributor-Level 10

$f (x) = \overset{?}{a} \cdot (\overset{?}{b} * \overset{?}{c}) = |\begin{matrix} x & - 2 & 3 \\ - 2 & x & - 1 \\ 7 & - 2 & x \end{matrix}|$

$= x^{3} - 27 x + 26$

$f^{'} (x) = 3 x^{2} - 27 = 0 \Rightarrow x = \pm 3$ and $f^{''} (- 3) < 0$

$\Rightarrow$ local maxima at $x = x_{0} = - 3$

Thus, $\overset{?}{a} = - 3 \hat{i} - 2 \hat{j} + 3 \hat{k}, \overset{?}{b} = 2 \hat{i} - 3 \hat{j} - \hat{k}$ , and $\overset{?}{c} = 7 \hat{i} - 2 \hat{j} - 3 \hat{k}$

$\Rightarrow \overset{?}{a} \cdot \overset{?}{b} + \overset{?}{b} \cdot \overset{?}{c} + \overset{?}{c} \cdot \overset{?}{a} = 9 - 5 - 26 = - 22$

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a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are $5,7, 10,12,14,15$ , then the absolute difference of the remaining two observations is:

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alok kumar singh

Contributor-Level 10

$\overline{x} = 10$

$\Rightarrow \overset{?}{x} = \frac{63 + a + b}{8} = 10$

$\Rightarrow a + b = 17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So, $σ^{2} = 13.5 = \frac{79 + (a - 10)^{2} + (b - 10)^{2}}{8}$

$\Rightarrow a^{2} + b^{2} - 20 (a + b) = - 171$

$\Rightarrow a^{2} + b^{2} = 169$

From (1) and (2) ; $a = 12$ and $b = 5$

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