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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and variance.

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Payal Gupta | Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t n = 10, \bar{x} = 45, σ^{2} = 16 \\ ∴ \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \Rightarrow 45 = \frac{\sum_{}^{} x_{i}}{10} \Rightarrow \sum_{}^{} x_{i} = 450 \\ C o r r e c t e d \sum_{}^{} x_{i} = 450 - 52 + 25 = 423 \\ a n d C o r r e c t e d m e a n = \frac{423}{10} = 42.3 \\ N o w σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2} \Rightarrow 16 = \frac{\sum_{}^{} x_{i}^{2}}{10} - {(45)}^{2} \\ \Rightarrow 16 = \frac{\sum_{}^{} x_{i}^{2}}{10} - 2025 \Rightarrow \frac{\sum_{}^{} x_{i}^{2}}{10} = 2041 \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 2041 \times 10 \Rightarrow \sum_{}^{} x_{i}^{2} = 20410 \\ ∴ C o r r e c t e d \sum_{}^{} x_{i}^{2} = 20410 - {(52)}^{2} + {(25)}^{2} \\ = 20410 - 2704 + 625 = 18331 \\ a n d C o r r e c t e d v a r i a n c e \\ σ^{2} = \frac{18331}{10} - {(42.3)}^{2} = 1833.1 - 1789.3 = 43.8 \\ H e n c e, t h e r e q u i r e d m e a n = 42.3 a n d v a r i a n c e = 43.8 \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

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

$\begin{array}{l} G i v e n t h a t n = 100, \bar{x} = 40, σ = 10 \\ ∴ \bar{x} = \frac{\sum_{}^{} x_{i}}{n} \Rightarrow 40 = \frac{\sum_{}^{} x_{i}}{100} \Rightarrow \sum_{}^{} x_{i} = 4000 \\ C o r r e c t e d \sum_{}^{} x_{i} = 4000 - 30 - 70 + 3 + 27 = 3930 \\ a n d C o r r e c t e d m e a n = \frac{3930}{100} = 39.3 \\ N o w σ^{2} = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(40)}^{2} \Rightarrow 100 = \frac{\sum_{}^{} x_{i}^{2}}{100} - 1600 \\ \Rightarrow \sum_{}^{} x_{i}^{2} = 1700 \times 100 \Rightarrow \sum_{}^{} x_{i}^{2} = 170000 \\ ∴ C o r r e c t e d \sum_{}^{} x_{i}^{2} = 170000 - {(30)}^{2} - {(70)}^{2} + {(3)}^{2} + {(27)}^{2} \\ = 170000 - 900 - 4900 + 9 + 729 = 164938 \\ ∴ C o r r e c t S D = \sqrt[]{\frac{164938}{100} - {(39.3)}^{2}} = \sqrt[]{1649.38 - 1544.49} = \sqrt[]{104.89} = 10.24 \\ H e n c e, t h e r e q u i r e d S D = 10.24 \end{array}$

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Exemplar Solution 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

a year ago

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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Exemplar Solution 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

a year ago

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

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

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Exemplar Solution 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

a year ago

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

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

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Exemplar Solution 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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Asked by Payal Gupta

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Payal Gupta | Contributor-Level 10

a year ago

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

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

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Exemplar Solution 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

a year ago

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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Exemplar Solution 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

a year ago

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

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

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Exemplar Solution 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

a year ago

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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Exemplar Solution 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

a year ago

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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While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and variance.

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

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.

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.

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.

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.

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

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