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11 months ago

Statistics Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

If for a distribution: ∑ $(x - 5) = 3$ $\sum {(x - 5)}^{2} = 43$ and the total number of items is 18, find the mean and standard deviation.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t n = 18, \sum_{}^{} (x - 5) = 3, \sum_{}^{} {(x - 5)}^{2} = 43 \\ ∴ M e a n = A + \frac{\sum_{}^{} (x - 5)}{n} = 5 + \frac{3}{18} = \frac{93}{18} = 5.166 = 5.17 \\ a n d S . D . (σ) = \sqrt[]{\frac{\sum_{}^{} {(x - 5)}^{2}}{n} - {(\frac{\sum_{}^{} (x - 5)}{n})}^{2}} = \sqrt[]{\frac{43}{18} - {(\frac{3}{18})}^{2}} \\ = \sqrt[]{2.39 - {(0.166)}^{2}} = \sqrt[]{2.39 - 0.027} = 1.54 \\ H e n c e, t h e r e q u i r e d m e a n i s 5.17 a n d S D = 1.54 \end{array}$

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11 months ago

Statistics Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the items and the sum of the squares of the items.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t \bar{x} = 50, n = 100 a n d S D (σ) = 4 \\ \bar{x} = \frac{\sum_{}^{} x_{i}}{N} \Rightarrow 50 = \frac{\sum_{}^{} x_{i}}{100} \Rightarrow \sum_{}^{} x_{i} = 5000 \\ a n d variance σ^{2} = \frac{\sum_{}^{} f_{i} x_{i}^{2}}{N} - {(\frac{\sum_{}^{} f_{i} x_{i}}{N})}^{2} \\ {(4)}^{2} = \frac{\sum_{}^{} f_{i} x_{i}^{2}}{100} - {(50)}^{2} \Rightarrow 16 = \frac{\sum_{}^{} f_{i} x_{i}^{2}}{100} - 2500 \\ ∴ \sum_{}^{} f_{i} x_{i}^{2} = (2500 + 16) * 100 \\ \Rightarrow \sum_{}^{} f_{i} x_{i}^{2} = 2516 * 100 = 251600 \\ H e n c e, t h e r e q u i r e d s u m a r e 5000 a n d 251600 . \end{array}$

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New answer posted

11 months ago

Statistics Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

The mean life of a sample of 60 bulbs was 650 hours, and the standard deviation was 8 hours. A second sample of 80 bulbs has a mean life of 660 hours and a standard deviation of 7 hours. Find the overall standard deviation.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t n_{1} = 60, \bar{x_{1}} = 650 a n d s_{1} = 8 \\ a n d n_{2} = 80, \bar{x_{2}} = 660 a n d s_{2} = 7 \\ W e k n o w t h a t f o r t h e c o m b i n e d t w o s e r i e s t h a t \\ ∴ σ = \sqrt[]{\frac{n_{1} s_{1}^{2} + n_{2} s_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1} n_{2} {(\bar{x_{1}} - \bar{x_{2}})}^{2}}{{(n_{1} + n_{2})}^{2}}} = \sqrt[]{\frac{60 * {(8)}^{2} + 80 * {(7)}^{2}}{60 + 80} + \frac{60 * 80 {(650 - 660)}^{2}}{{(60 + 80)}^{2}}} \\ = \sqrt[]{\frac{6 * 64 + 8 * 49}{14} + \frac{60 * 80 * 100}{140 * 140}} = \sqrt[]{\frac{192 + 196}{7} + \frac{1200}{49}} \\ = \sqrt[]{\frac{388}{7} + \frac{1200}{49}} = \sqrt[]{\frac{2716 + 1200}{49}} = \sqrt[]{\frac{3916}{49}} = \frac{62.58}{7} = 8.9 \\ H e n c e, t h e r e q u i r e d S D = 8.9 \end{array}$

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New answer posted

11 months ago

Statistics 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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New answer posted

11 months ago

Statistics 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

11 months ago

Statistics 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

11 months ago

Statistics 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

11 months ago

Statistics 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

11 months ago

Statistics 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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11 months ago

Statistics 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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