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New question posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

<p><strong>In a LPP, the linear inequalities or restrictions on the variables are called __________.</strong></p>

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Asked by Payal Gupta

New answer posted•a year ago

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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago

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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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year 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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Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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New answer posted•a year ago

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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year 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

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year 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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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New answer posted•a year ago

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) $l = 2.5, k = 5$

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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New answer posted•a year 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) $s$

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Fifteen

Let $a, b, c, d, e$ be the observations with mean $m$ and standard deviation $s$ . The standard deviation of the observations $a + k, b + k, c + k, d + k, e + k$ is:

(a) $s$

(b) $k \cdot s$

(c) $s + k$

(d) $\frac{S}{K}$

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

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

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