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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

In a LPP, the linear inequalities or restrictions on the variables are called __________.

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Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

In a LPP, the linear inequalities or restrictions on the variables are called __________.

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

The standard deviation of some temperature data in $?^{?} C$ is 5. If the data were converted into $?^{?} F$ , the variance would be:

(a) 81

(b) 57

(c) 36

(d) 25

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t σ_{C} = 5 \\ W e k n o w t h a t C = \frac{5}{9} (F - 32) \Rightarrow F = \frac{9 C}{5} + 32 \\ ∴ σ_{F} = \frac{9}{5} σ_{C} = \frac{9}{5} * 5 = 9 \\ ∴ σ_{F}^{2} = {(9)}^{2} = 81 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. The difference of their standard deviations is:

(a) 0

(b) 1

(c) 1.5

(d) 2.5

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e, w e h a v e C V_{1} = 50, C V_{2} = 60 \\ \bar{x_{1}} = 30 a n d \bar{x_{2}} = 25 \\ ∴ C V_{1} = \frac{σ_{1}}{\bar{x_{1}}} * 100 \Rightarrow 50 = \frac{σ_{1}}{30} * 100 \Rightarrow σ_{1} = \frac{50 * 30}{100} = 15 \\ a n d C V_{1} = \frac{σ_{2}}{\bar{x_{2}}} * 100 \Rightarrow 60 = \frac{σ_{2}}{25} * 100 \Rightarrow σ_{2} = \frac{60 * 25}{100} = 15 \\ ∴ D i f f e r e n c e σ_{1} - σ_{2} = 15 - 15 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

The following information relates to a sample of size 60: $x^{2} = 18000$ , $\sum x = 960$ . The variance is:

(a) 6.63

(b) 16

(c) 22

(d) 44

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t V a r i a n c e (σ^{2}) = \frac{\sum_{}^{} x_{i}^{2}}{N} - {(\frac{\sum_{}^{} x_{i}}{N})}^{2} \\ = \frac{18000}{60} - {(\frac{960}{60})}^{2} = 300 - 256 = 44 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Consider the first 10 positive integers. If we multiply each number by -1 and then add 1 to each number, the variance of the numbers so obtained is:

(a) 8.25

(b) 6.5

(c) 3.87

(d) 2.87

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} F i r s t 10 p o s i t i v e integers a r e 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \\ o n m u l t i p l y i n g e a c h n u m b e r b y - 1, w e g e t \\ - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9, - 10 \\ o n a d d i n g 1 t o e a c h o f t h e n u m b e r, w e g e t \\ 0, - 1, - 2, - 3, - 4, - 5, - 6, - 7, - 8, - 9 \\ ∴ \sum_{}^{} x_{i} = 0 - 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 \\ = - 45 \\ a n d \sum_{}^{} x_{i}^{2} = 0^{2} + {(- 1)}^{2} + {(- 2)}^{2} + {(- 3)}^{2} + {(- 4)}^{2} + \dots + {(- 9)}^{2} \\ = \frac{9 * 10 * 19}{6} = 285 [? \sum_{}^{} n^{2} = \frac{n (n + 1) (2 n + 1)}{6}] \\ ∴ S . D = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{N} - {(\frac{\sum_{}^{} x_{i}}{N})}^{2}} = \sqrt[]{\frac{285}{10} - {(\frac{- 45}{10})}^{2}} \\ = \sqrt[]{\frac{285}{10} - \frac{2025}{100}} = \sqrt[]{\frac{2850 - 2025}{100}} = \sqrt[]{8.25} \\ ∴ V a r i a n c e = {(S D)}^{2} = {(\sqrt[]{8.25})}^{2} = 8.25 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Consider the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ . If 1 is added to each number, the variance of the numbers so obtained is:

(a) 6.5

(b) 2.87

(c) 3.87

(d) 8.25

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n n u m b e r s a r e 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \\ N u m b e r s o b t a i n e d w h e n 1 i s a d d e d t o t h e a b o v e n u m b e r s i s 2, 3, 4, 5, 6, 7, 8, 9, 10 a n d 11 . \\ ∴ \sum_{}^{} x_{i} = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 \\ = \frac{10}{2} [2 * 2 + (10 - 1) . 1] = 5 [4 + 9] = 5 * 13 = 65 \\ N o w \sum_{}^{} x_{i}^{2} = 2^{2} + 3^{2} + 4^{2} + \dots + 1 1^{2} \\ = (1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + 1 1^{2}) - {(1)}^{2} \\ = \frac{n (n + 1) (2 n + 1)}{6} - 1 = \frac{11 * 12 * 23}{6} - 1 \\ = 22 * 23 - 1 = 506 - 1 = 505 \\ ∴ V a r i a n c e (σ^{2}) = \frac{\sum_{}^{} x_{i}^{2}}{n} - {(\frac{\sum_{}^{} x_{i}}{n})}^{2} = \frac{505}{10} - {(\frac{65}{10})}^{2} \\ = 50.5 - {(6.5)}^{2} = 50.5 - 42.25 = 8.25 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Standard deviations for the first 10 natural numbers is:

(a) 5.5

(b) 3.87

(c) 2.97

(d) 2.87

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t S D o f f i r s t n n a t u r a l n u m b e r s \sqrt[]{\frac{n^{2} - 1}{12}} \\ H e r e n = 10 \\ ∴ S D = \sqrt[]{\frac{{(10)}^{2} - 1}{12}} = \sqrt[]{\frac{99}{12}} = \sqrt[]{8.25} = 2.87 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Let $x_{1}, x_{2}, \dots, x_{n}$ be $n$ observations. Let $w_{i} = l x_{i} + k$ for $i = 1, 2, \dots, n$ , where $l$ and $k$ are constants. If the mean of $x_{i}$ â€™s is 48 and their standard deviation is 12, the mean of $w_{i}$ â€™s is 55 and the standard deviation of $w_{i}$ â€™s is 15, the values of $l$ and $k$ should be:

(a) $l = 1.25, k = - 5$

(b) $l = 2.5, k = - 5$

(c) $l = - 1.25, k = 5$

(d) $l = 2.5, k = 5$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t w_{i} = x_{i} + k, \bar{x_{i}} = 48, S D (x_{i}) = 12, \\ w_{i} = 55 a n d S D (w_{i}) = 15 \\ t h e n \bar{w_{i}} = \bar{x_{i}} + k (\bar{w_{i}} = m e a n o f w_{i}' s a n d \bar{x_{i}} i s t h e m e a n o f x_{i}' s) \\ \Rightarrow 55 = 48 + k \dots (i) \\ S D o f w_{i} = S D o f x_{i} \\ 15 = l * 12 \\ \Rightarrow l = \frac{15}{12} = 1.25 \dots (i i) \\ f r o m e q n . (i) a n d (i i) w e h a v e \\ k = \bar{w_{i}} - \bar{x_{i}} = 55 - 1.25 * 48 = 55 - 60 = - 5 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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3 months ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the observations with mean $m$ and standard deviation $s$ . The standard deviation of the observations $k x_{1}, k x_{2}, k x_{3}, k x_{4}, k x_{5}$ is:

(a) $k + s$

(b) s/k

(c) $k \cdot s$

(d) $s$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} H e r e m = \frac{\sum_{}^{} x_{i}}{N}, S = \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{5} - {(\frac{\sum_{}^{} x_{i}}{5})}^{2}} \\ ∴ S D = \sqrt[]{\frac{K^{2} \sum_{}^{} x_{i}^{2}}{5} - {(\frac{K \sum_{}^{} x_{i}}{5})}^{2}} = \sqrt[]{\frac{K^{2} \sum_{}^{} x_{i}^{2}}{5} - K^{2} {(\frac{\sum_{}^{} x_{i}}{5})}^{2}} \\ = K \sqrt[]{\frac{\sum_{}^{} x_{i}^{2}}{5} - {(\frac{\sum_{}^{} x_{i}}{5})}^{2}} = K . S \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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