NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Q: Which are the important topics one should cover in Statistics?

Standard Deviation, Mean Deviation and Central Tendency are the important topics one should cover in Statistics.

Q: Where can we download the Class 11 Maths NCERT Exemplar Solutions Chapter 15?

Students can get PDF by using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function.

Q: Are the solutions helpful for board examinations?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 15 are helpful for board examinations.

Q: Can these solutions be referred for cracking competitive examinations?

Yes, Class 11 Maths NCERT Exemplar solutions chapter 15 can be referred to when pursuing competitive examinations.

NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:21 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 15 includes solutions to questions of all types such as short answer questions, long answer questions, value-based questions, MCQ’s and HOTS. The solutions are given by experts in a simple manner and are solved step-by-step. If you who have doubts regarding the chapter Statistics, then refer to NCERT Exemplar Class 11 Maths chapter 15 solutions and follow the solutions for better understanding of the concept. Students can download the pdf for Class 11 Maths NCERT Exemplar Solutions Chapter 15 which consists of all the important questions. You can cover all the important topics and sub-topics using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function. You can also get your queries solved by referring to the solutions solved step by step in an easily comprehensible format.
Also, check - NCERT Class 11 Maths Solutions for Other Chapters

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

This Story also Contains

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:
More about NCERT Exemplar Class 11 Maths Solutions Chapter 15
NCERT Solutions for Class 11 Mathematics Chapters
Important Topic to Cover from NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

Question:1

Find the mean deviation about the mean of the distribution:
$\begin{array}{|l|l|l|l|l|l|} \hline \text { Size } & 20 & 21 & 22 & 23 & 24 \\ \hline \text { Frequency } & 6 & 4 & 5 & 1 & 4 \\ \hline \end{array}$

Answer:

We have to find the mean deviation about the mean of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

$\begin{array}{|l|l|l|} \hline \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} \\ \hline 20 & 6 & 120 \\ \hline 21 & 4 & 84 \\ \hline 22 & 5 & 110 \\ \hline 23 & 1 & 23 \\ \hline 24 & 4 & 96 \\ \hline \text { Total } & 20 & 433 \\ \hline \end{array}$
$\text{Here, mean }\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{433}{20}=21.65 \\\\ \\ \text{ So the above table can be rewritten as } \\\\$
$\begin{array}{|l|l|l|l|l|} \hline \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & d_{i}=\left|x_{i}-\bar{x}\right| & f_{i} d_{i} \\ \hline 20 & 6 & 120 & 1.65 & 9.90 \\ \hline 21 & 4 & 84 & 0.65 & 2.60 \\ \hline 22 & 5 & 110 & 0.35 & 1.75 \\ \hline 23 & 1 & 23 & 1.35 & 1.35 \\ \hline 24 & 4 & 96 & 2.35 & 9.40 \\ \hline \text { Total } & 20 & 433 & 6.35 & 25.00 \\ \hline \end{array}$

Hence, mean deviation becomes= $\frac{ \Sigma f_{i}(x_{i}-\bar{x_{i}})}{ \Sigma f_{i}}=\frac{25}{20}=1.25~ \\\\$

Question:2

Find the mean deviation about the median of the following distribution:

$\begin{array}{|l|l|l|l|l|l|} \hline \text { Marks obtained } & 10 & 11 & 12 & 14 & 15 \\ \hline \text { No. of students } & 2 & 3 & 8 & 3 & 4 \\ \hline \end{array}$

Answer:

The data distribution is given in the question.

We have to find the mean deviation about the median of the distribution in this question.

Let us make a table of the given data and fill up the other columns after calculations

$\begin{array}{|l|l|l|} \hline \text { Marks obtained }\left(x_{i}\right) & \text { Number of students }\left(f_{i}\right) & \text { Cumulative frequency } \\ \hline 10 & 2 & 2 \\ \hline 11 & 3 & 5 \\ \hline 12 & 8 & 13 \\ \hline 14 & 3 & 16 \\ \hline 15 & 4 & 20 \\ \hline \text { Total } & 20 & \\ \hline \end{array}$

$\\ \text{Now here N=20 which is even}.\\\\ Here, median \( M= \frac{1}{2} \left[ \left( \frac{N}{2} \right) ^{th}observation+ \left( \frac{N}{2}+1 \right) ^{th}~observation \right] ~ \\ \\ ~ M=\frac{1}{2} \left[ \left( \frac{20}{2} \right) ^{th} observation+ \left( \frac{20}{2}+1 \right) ^{th}~observation~ \right] ~~ \\\\ \\ ~ M=\frac{1}{2}~ \left[ 10^{th}observation+11^{th}observation~ \right] ~~ \\\\ \\ \text{ Both these observations lie in cumulative frequency 13, for which corresponding observation is 12 } \\\\ \\ M=\frac{1}{2} \left[ 12+12 \right] =12 \\\\ \\ \text{~ So the above updated table is as shown below } \\\\$

$\begin{array}{|l|l|l|l|l|} \hline \begin{array}{l} \text { Marks obtained } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Number of } \\ \text { students }\left(f_{i}\right) \end{array} & \begin{array}{l} \text { Cumulative } \\ \text { frequency } \end{array} & d_{i}=\left|x_{i}-M\right| & f_{i} d_{i} \\ \hline 10 & 2 & 2 & 2 & 4 \\ \hline 11 & 3 & 5 & 1 & 3 \\ \hline 12 & 8 & 13 & 0 & 0 \\ \hline 14 & 3 & 16 & 2 & 6 \\ \hline 15 & 4 & 20 & 3 & 12 \\ \hline \text { Total } & 20 & & 6.35 & 25 \\ \hline \end{array}$

Hence, mean deviation becomes $=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{25}{20}=1.25~ \\\\$

Question:3

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

Answer:

Set of first n natural numbers when n is an odd number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is odd number.

So, mean is

$\\ \\ \text{~~~ Or, }\frac{1-n}{2},\frac{3-n}{2}~,~\frac{5-n}{2} \ldots \ldots .\frac{n-3}{2}~,\frac{n-1}{2}~~~ \\\\ \\ ~~\frac{- \left( n-1 \right) }{2},~\frac{- \left( n-3 \right) }{2},\frac{- \left( n-5 \right) }{2} , \ldots \ldots , \frac{n-5}{2},\frac{n-3}{2},\frac{n-1}{2} \\\\ \\ \text{~ So the absolute values of deviation from the mean is } \vert x_{i}-\overline{x}~ \vert =\frac{n-1}{2},~\frac{n-3}{2}~,\frac{n-5}{2}, \ldots \ldots .\frac{n-5}{2},\frac{n-3}{2},\frac{n-1}{2}~~ \\\\ \\ \text{~ The sum of absolute values of deviations from the mean, is } \Sigma \vert x_{i}-\overline{x}~ \vert =\frac{n-1}{2}+\frac{n-3}{2}+\frac{n-5}{2}+ \ldots \ldots \frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2}~~~~~~ \\\\$

$\\ \\ \Sigma \vert x_{i}-\overline{x}~ \vert =2 \left( 1+2+3+ \ldots \ldots .+\frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2} \right) ~ \\\\ \\ \text{~ That is 2 times the sum of }\frac{n-1}{2}\text{terms,~so it can be written as } \Sigma \vert x_{i}-\overline{x}~ \vert =2 \left( \frac{\frac{n-1}{2} \left( \frac{n-1}{2}+1 \right) }{2} \right) =n \left( \frac{n-1}{2} \right) \left( \frac{n+1}{2} \right) = \left( \frac{n^{2}-1}{4} \right) ~ \\\\ \\ \text{~~ Therefore, mean deviation about the mean is }\frac{ \Sigma \vert x_{i}-\overline{x}~ \vert }{n}=\frac{ \left( \frac{n^{2}-1}{4} \right) }{n}= \left( \frac{n^{2}-1}{4n} \right) ~~ \\\\$

Question:4

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

Answer:

Set of first n natural numbers when n is an even number is given

We have to find the mean deviation about the mean

We know that first natural numbers are 1, 2, 3 ……. , n .

It is given that n is an even number.

So, mean is

$\\ \\ \overline{x}=\frac{1+2+3+ \ldots ..+n}{n}=\frac{\frac{n \left( n+1 \right) }{2}}{n}=\frac{n+1}{2}~ \\\\ \\ \text{~~The deviations of numbers from the mean are as shown below, } \\\\ \\ ~ 1-\frac{n+1}{2},~ 2-\frac{n+1}{2} , \ldots \ldots \ldots \ldots \left( n-1 \right) -\frac{n-1}{2}~,~ n-\frac{n+1}{2}~ \\\\ \\ ~Or,\frac{2- \left( n+1 \right) }{2}~,\frac{4- \left( n+1 \right) }{2} \ldots \ldots \ldots . \frac{2 \left( n-1 \right) - \left( n+1 \right) }{2},~\frac{2n- \left( n+1 \right) }{2}~ \\\\$
$\\ \\ \text{~~~ Or, }\frac{1-n}{2},\frac{3-n}{2}~,~\frac{5-n}{2} \ldots \ldots .\frac{n-3}{2}~,\frac{n-1}{2}~~~ \\\\ \\ ~~\frac{- \left( n-1 \right) }{2},~\frac{- \left( n-3 \right) }{2},\frac{- \left( n-5 \right) }{2} , \ldots \ldots , \frac{n-5}{2},\frac{n-3}{2},\frac{n-1}{2} \\\\ \\ \text{~ So the absolute values of deviation from the mean is } \vert x_{i}-\overline{x}~ \vert =\frac{n-1}{2},~\frac{n-3}{2}~,\frac{n-5}{2}, \ldots \ldots .\frac{n-5}{2},\frac{n-3}{2},\frac{n-1}{2}~~ \\\\ \\ \text{~ The sum of absolute values of deviations from the mean, is } \Sigma \vert x_{i}-\overline{x}~ \vert =\frac{n-1}{2}+\frac{n-3}{2}+\frac{n-5}{2}+ \ldots \ldots \frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2}~~~~~~ \\\\ \Sigma \vert x_{i}-\overline{x}~ \vert = \left( \frac{1}{2}+\frac{3}{2}+ \ldots \ldots .+\frac{n-5}{2}+\frac{n-3}{2}+\frac{n-1}{2} \right) ~ \left( \frac{n}{2} \right)$

$\\\text{We know that sum of first n natural numbers }= $ n^{2} $ \\ \vspace{\baselineskip} \\ \text{ Therefore, mean deviation about the mean is }\frac{ \Sigma \vert x_{i}-\overline{x}~ \vert }{n}=\frac{ \left( \frac{1}{2}+\frac{3}{2}+ \ldots ..+\frac{n-1}{2} \right) \left( \frac{n}{2} \right) }{n}~ \\\\ \\ =\frac{ \left( \frac{n}{2} \right) ^{2}}{n}=\frac{n^{2}}{4n}=\frac{n}{4} \\\\$

Question:5

Find the standard deviation of the first n natural numbers.

Answer:

$\\$ \text{Set~of first n natural numbers is given } $ \\ \\ \text{ We have to find the standard deviation } \\\\ \\ \text{ We can write in table the first n natutal numbers as } \\\\$
$\begin{array}{|c|l|l|l|l|l|l|l|l|} \hline x_{i} & 1 & 2 & 3 & 4 & 5 & \ldots . & \ldots . & \mathrm{n} \\ \hline x_{i}^{2} & 1 & 4 & 9 & 16 & 25 & \ldots . & \ldots \ldots & n^{2} \\ \hline \end{array}$

$\\\\ \text{The sums are } \Sigma x_{i}=1+2+3+ \ldots \ldots +n=\frac{n \left( n+1 \right) }{2}~~~~ \\\\ \\ ~~ \Sigma x_{i}^{2}=1^{2}+2^{2}+3^{2}+ \ldots \ldots +n^{2}=\frac{n \left( n+1 \right) \left( 2n+1 \right) }{6}~~~ \\\\ \\ ~ Therefore, the standard deviation can be written as, \sigma =\sqrt[]{\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~~~~ \\\\ \\ , \sigma =\sqrt[]{\frac{\frac{n \left( n+1 \right) \left( 2n+1 \right) }{6}}{n}- \left( \frac{\frac{n \left( n+1 \right) }{2}}{n} \right) ^{2}} = \sqrt[]{\frac{2n^{2}+n+2n+1}{6}-\frac{n^{2}+2n+1}{4}}~ = \sqrt[]{\frac{4n^{2}+6n+2-3n^{2}-6n-3}{12}} \\\\ \\ = \sqrt[]{\frac{n^{2}-1}{12}} \\\\$

Question:6

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 x₁, x₂, ..., x₁₅, also in seconds, is now available and we have
$\\ \sum_{i=1}^{15}{x_{i}}=279\: \: and\: \: \sum_{i=1}^{15} \mathrm{x}_{i}=5524$ .Calculate the standard derivation based on all 40 observations.

Answer:

$\\\\ Number~of observations=25, mean=18.2 seconds, standard deviation=3.25 seconds. \\\\ \\ \text{ There are another set of 15 observations }x_{1},x_{2}, \ldots ..,x_{15}\text{~ also in seconds is } \sum _{i=1}^{15}x_{i}=279~and~ \sum _{i=1}^{15}x_{i}^{2}=5524~ \\\\ \\ \text{ To find that the standard derivation based on all 40 observations } \\\\ \\ \text{ As per the given criteria in the first set Number of observations , }n_{1}=25 \\\\$

$\\ \\ Mean=18.2 \\\\ \\ \text{ And standard deviation } \sigma _{1}=3.25~ \\\\ \\ \text{ And in second set number of observations }n_{2}=15 \\\\ \\ \text{~ For the first set we have }\overline{x_{1}}=18.2=\frac{ \Sigma x_{i}}{25}~~~~~ \\\\ \\ ~ \Sigma x_{i}=25\ast18.2=455~ \\\\$
$\\ \\ \text{ Therefore the standard deviation becomes } \sigma _{1}^{2}=\frac{ \Sigma x_{i}^{2}}{25}- \left( 18.2 \right) ^{2}~ \\\\ \\ \text{ Substituting the values, we get } \left( 3.25 \right) ^{2}=\frac{ \Sigma x_{i}^{2}}{25}-331.24~~~ \\\\ \\ 10.5625+331.24=\frac{ \Sigma x_{i}^{2}}{25}=341.8025 \\\\ \\ ~~ \Sigma x_{i}^{2}=25\ast341.8025=8545.06 \\\\ \\ \text{For~the combined standard deviation of the 40 observation, n=40 and } \Sigma x_{i}^{2}=8545.06+5524=14069.69~~ \\\\ \\ ~ \Sigma x_{i}=455+279=734 \\\\ \\ \text{Therefore the standard deviation can be written as,} \sigma =\sqrt[]{\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~ \\\\$
$\\ \\ \text{~~Substituting~the values, we get Therefore the standard deviation can be written as } \\\\ \\ \sigma =\sqrt{\frac{14069.69}{40}- \left( \frac{734}{40} \right) ^{2}}= \sqrt{351.7265-336.7225}=\sqrt{15.004}=3.87 \\\\$

Question:7

The mean and standard deviation of a set of $n_1$ observations are $\bar{X_1}$ and $s_1$ , respectively while the mean and standard deviation of another set of $n_2$ observations are $\bar{X_2}$ and $s_2$ , respectively. Show that the standard deviation of the combined set of ( $n_1 + n_2$ ) observations is given by
$\mathrm{SD}=\sqrt{\frac{\mathrm{n}_{1}\left(\mathrm{s}_{1}\right)^{2}+\mathrm{n}_{2}\left(\mathrm{s}_{2}\right)^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)^{2}}}$

Answer:

$\\\\ \text{It is given that mean and Standard deviation of a set of }n_{1}\text{ observations are }\overline{x_{1}}~and~ \sigma _{1}~respectively~ \\\\ \\ \text{while the mean and standard deviation of naother set of }n_{2}\text{observations are }\overline{x_{2}}~and~ \sigma _{2}respectively \] \\ \overline{x_{1}}=\frac{1}{n_{1}} \sum _{i=1}^{n_{1}}x_{i} \\\\ \\ \overline{x_{2}}=\frac{1}{n_{2}} \sum _{j=1}^{n_{2}}y_{j} \\\\ \\ \overline{x}=\frac{1}{ \left( n_{1}+n_{2} \right) } \left[ \sum _{i=1}^{n_{1}}x_{i}+ \sum _{j=1}^{n_{2}}y_{j} \right] =\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}}{n_{1}+n_{2}} \\\\ \\ \sigma _{1}^{2}=\frac{1}{n_{1}} \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x} \right) ^{2} \\\\$

$\\ \sigma _{2}^{2}=\frac{1}{n_{2}} \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x} \right) ^{2} \\\\ \\ \sigma ^{2}= \sigma _{1}^{2}+ \sigma _{2}^{2}=\frac{1}{n_{1}+n_{2}} \left[ \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x} \right) ^{2}+\frac{1}{n_{2}} \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x} \right) ^{2} \right] \\\\ \\ \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x} \right) ^{2}= \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x_{j}}+\overline{x_{j}}-\overline{x} \right) ^{2} \\\\ \\ = \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x_{j}} \right) ^{2}+n_{1} \left( \overline{x_{j}}-\overline{x} \right) ^{2}+2 \left( \overline{x_{j}}-\overline{x} \right) \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x_{j}} \right) \\\\ \\ \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x_{j}} \right) =0 \\\\$

$\\ \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x} \right) ^{2}=n_{1}s_{1}^{2}+n_{1} \left( \overline{x_{1}}-\overline{x} \right) ^{2} \\\\ \\ d_{1}=\overline{x_{1}}-\overline{x} \\\\ \\ =\overline{x_{1}}-\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}}{n_{1}+n_{2}} \\\\ \\ =\frac{n_{2} \left( \overline{x_{1}}-\overline{x_{2}} \right) }{n_{1}+n_{2}} \\\\ \\ \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x} \right) ^{2}=n_{1}s_{1}^{2}+\frac{n_{1}n_{2}^{2} \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}} \\\\ \\ \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x} \right) ^{2}= \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x_{i}}+\overline{x_{i}}-\overline{x} \right) ^{2} \\\\$
$\\ = \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x_{i}} \right) ^{2}+n_{2} \left( \overline{x_{i}}-\overline{x} \right) ^{2}+2 \left( \overline{x_{i}}-\overline{x} \right) \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x_{i}} \right) \\\\ \\ \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x_{i}} \right) =0 \\\\ \\ \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x} \right) ^{2}=n_{2}s_{2}^{2}+n_{2} \left( \overline{x_{2}}-\overline{x} \right) ^{2} \\\\ \\ d_{2}=\overline{x_{2}}-\overline{x} \\\\ \\ =\overline{x_{2}}-\frac{n_{1}\overline{x_{1}}+n_{2}\overline{x_{2}}}{n_{1}+n_{2}} \\\\ \\ =\frac{n_{1} \left( \overline{x_{2}}-\overline{x_{1}} \right) }{n_{1}+n_{2}} \\\\$
$\\ \sum _{j=1}^{n_{2}} \left( x_{i}-\overline{x} \right) ^{2}=n_{2}s_{2}^{2}+\frac{n_{1}^{2}n_{2} \left( \overline{x_{2}}-\overline{x_{1}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}} \\\\ \\ \sigma ^{2}= \sigma _{1}^{2}+ \sigma _{2}^{2}=\frac{1}{n_{1}+n_{2}} \left[ \sum _{i=1}^{n_{1}} \left( x_{i}-\overline{x} \right) ^{2}+ \sum _{j=1}^{n_{2}} \left( y_{j}-\overline{x} \right) ^{2} \right] \\\\ \\ =\frac{1}{n_{1}+n_{2}} \left[ n_{1}s_{1}^{2}+\frac{n_{1}n_{2}^{2} \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}}+n_{2}s_{2}^{2}+\frac{n_{1}^{2}n_{2} \left( \overline{x_{2}}-\overline{x_{1}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}} \right] \\\\ \\ =\frac{1}{n_{1}+n_{2}} \left[ n_{1}s_{1}^{2}+n_{2}s_{2}^{2}+\frac{n_{1}n_{2} \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}} \left( n_{1}+n_{2} \right) \right] \\\\$

$\\ =\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1}n_{2} \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}} \\\\ \\ \sigma =\sqrt {\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1}n_{2} \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}}} \\\\$

Question:8

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

Answer:

$\\\\ \text{Given that two sets each of 20 observations, have the same standard derivation 5.} \\\\ \\ \text{ The first set has a mean 17 and the second a mean 22.} \\\\ \\ \text{ We have to show that the standard deviation of the set obtained by combining the given two sets } \\\\ \\ \text{ As per given criteria, for first set Number of observations , }n_{1}=20~ \\\\$

$\\ \text{Standard deviation }s_{1}=5~ \text{And mean } \overline{x_{1}}=17 \\\\ \\ \text{~ For second set Number of observations, }n_{2}=20~ \text{Standard deviation } s_{2}=5~ \\\\ \\ \text{and mean }\overline{x_{2}}=22 \\\\ \\ \text{We know the standard deviation for combined two series is} \sigma =\sqrt {\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+ \left( n_{1}n_{2} \right) \frac{ \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}}}~~ \\\\ \\ \text{Substituting the corresponding values we get } \sigma =\sqrt {\frac{20 \left( 5 \right) ^{2}+20 \left( 5 \right) ^{2}}{20+20}+\frac{ \left( 20\ast20 \right) \left( 17-22 \right) ^{2}}{ \left( 20+20 \right) ^{2}}}= \sqrt {\frac{1000}{40}+\frac{10000}{1600}} = \\\\ \\\\ \sqrt {25+\frac{25}{4}}=\sqrt {\frac{100+25}{4}}=\sqrt {\frac{125}{4}} = 5.59\\$

Question:9

The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{x} & \mathrm{A} & 2 \mathrm{A} & 3 \mathrm{A} & 4 \mathrm{A} & 5 \mathrm{A} & 6 \mathrm{A} \\ \hline \mathrm{f} & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where A is a positive integer, has a variance of 160. Determine the value of A.

Answer:

A frequency distribution table is given where variance =160. We have to find the value of A, where A is a positive number. Let us make a table from the given data and fill out the other columns after calculation
$\begin{array}{|l|l|l|c|} \hline \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ \hline \text { A } & 2 & 2 \mathrm{A} & 2 A^{2} \\ \hline 2 \mathrm{A} & 1 & 2 \mathrm{A} & 4 A^{2} \\ \hline 3 \mathrm{A} & 1 & 3 \mathrm{A} & 9 A^{2} \\ \hline 4 \mathrm{A} & 1 & 4 \mathrm{A} & 16 A^{2} \\ \hline 5 \mathrm{A} & 1 & 5 \mathrm{A} & 25 \mathrm{A}^{2} \\ \hline 6 \mathrm{A} & 1 & 6 \mathrm{A} & 36 A^{2} \\ \hline \text { Total } & \mathrm{N}=7 & 22 \mathrm{A} & 92 \mathrm{A}^{2} \\ \hline \end{array}$

And we know that variance is $\\\\ ~ \sigma ^{2}= \frac{ \Sigma f_{i}x_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}x_{i}}{n} \right) ^{2}~ \\\\$

$\\ \text{ Substituting the values from the above table and also given that variance}=160 \\ we~get~ 160=\frac{92A^{2}}{7}- \left( \frac{22A}{7} \right) ^{2}= \left( \frac{92A^{2}}{7}-\frac{484A^{2}}{49} \right) =\frac{7\ast92A^{2}-484A^{2}}{49}=\frac{160A^{2}}{49}~ \\\ ~~~ 160=\frac{160A^{2}}{49}~ \\\\ \\ ~~A^{2}=49 \\\\ \\ A=7 \\\\ \\ \text{ Hence, the value of A is 7} \\\\$

Question:10

For the frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{x} & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{f} & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard distribution.

Answer:

A frequency distribution table is given and we have to find the standard deviation

Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{|l|l|l|l|} \hline \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ \hline 2 & 4 & 8 & 16 \\ \hline 3 & 9 & 27 & 81 \\ \hline 4 & 16 & 64 & 256 \\ \hline 5 & 14 & 70 & 350 \\ \hline 6 & 11 & 66 & 396 \\ \hline 7 & 6 & 42 & 294 \\ \hline \text { Total } & \mathrm{N}=60 & 277 & 1393 \\ \hline \end{array}$

$\begin{aligned} &\text { And we know that standard deviation is }\\ &\sigma=\sqrt{\frac{\Sigma f_{i} x_{i}^{2}}{n}-\left(\frac{\Sigma f_{i} x_{i}}{n}\right)^{2}} \end{aligned}$

Substituting the value from the above table

$\sigma=\sqrt{\frac{1393}{60}-\left(\frac{277}{60}\right)^{2}}=\sqrt{23.23-21.34}=1.37$

Question:11

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & \mathrm{x}-2 & \mathrm{x} & \mathrm{x}^{2} & (\mathrm{x}+1)^{2} & 2 \mathrm{x} & \mathrm{x}+1 \\ \hline \end{array}$
where x is a positive integer. Determine the mean and standard deviation of the marks.

Answer:

It is given that there are 60 students in a class. The frequency distribution of the marks obtained by the students in a test is also given.

We have to find the mean and standard deviation of the marks.

It is given that there are 60 students in the class, so

$\\ \Sigma f_{i}=60~ \\\\ \\ \left( x-2 \right) +x+x^{2}+ \left( x+1 \right) ^{2}+2x+x+1=60~ \\\\ \\ 5x-1+x^{2}+x^{2}+2x+1=60 \\\\ \\ 2x^{2}+7x=60 \\\\ \\ \text{On factorising 2}x^{2}+7x-60=0 \\\\ \\ \text{we get } \left( 2x+15 \right) \left( x-4 \right) =0 \\\\ \\ 2x=-15 or x=4 \\\\ \\ \text{Given~that x is a positive number, so x can take 4 as the only value .} \\\\ \\ \text{And let assumed mean}, a=3 \\\\ \\ \text{Now put x=4 and a=3} \text{ in the frequency distribution table and other columns after} \\\\ \\ \text{calculations, we get } \\\\$

$\begin{array}{|l|l|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & d_{i}=x_{i}-a_{i} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 0 & \mathrm{x}-2=4-2=2 & -3 & -6 & 18 \\ \hline 1 & \mathrm{X}=4 & -2 & -8 & 16 \\ \hline 2 & x^{2}=16 & -1 & -16 & 16 \\ \hline 3 & (x+1)^{2} & 0 & 0 & 0 \\ & =(4+1)^{2}=25 & & & \\ \hline 4 & 2 \mathrm{x}=2 * 4=8 & 1 & 4 & 8 \\ \hline 5 & \mathrm{X}+1=4+1=5 & 2 & 10 & 20 \\ \hline \text { Total } & \mathrm{N}=60 & 277 & =-12 & 78 \\ \hline \end{array}$

And we know that standard deviation is $\\\\ \sigma = \sqrt {\frac{ \Sigma f_{i}d_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}d_{i}}{n} \right) ^{2}}~ \\\\$

$\\ \text{Substituting the values from the above table } \sigma =\sqrt {\frac{78}{60}- \left( \frac{-12}{60} \right) ^{2}} \\\\ \\ =\sqrt {1.3- \left( 0.2 \right) ^{2}}=\sqrt {1.3-0.04}=1.12~ \\\\ \\ \text{Hence, the standard deviation is 1.12 } \\\\ \\ \text{ Now mean is }\overline{x}=A+\frac{ \Sigma f_{i}d_{i}}{N}~ =3+ \left( -\frac{12}{60} \right) =3-\frac{1}{5}=\frac{14}{5}=2.8~~ \\\\ \\ \text{Hence, the mean and standard deviation of the marks are 2.8 and 1.12 respectively} \\\\$

Question:12

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 standard deviation 7 hours. Find the overall standard deviation.

Answer:

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 standard deviation 7 hours. We have to find the overall standard deviation.

As per the given criteria. In the first set of samples number of sample bulbs , $\\\\ n_{1}=60 \\\\$

$\\ \\ \text{ Standard deviation }s_{1}=8 \text{hrs~ Mean life, }\overline{x_{1}}=650 \\\\ \\ \text{ And in second set of samples, number of sample bulbs }n_{2}=80 \text{ Standard deviation }s_{2}=7 hrs \\\\ \\ \text{~ Mean life }\overline{x_{2}}=660~ \\\\ \\ \text{We know the standard deviation for combined two series is } \\\\ \\ \sigma =\sqrt {\frac{n_{1}s_{1}^{2}+n_{2}s_{2}^{2}}{n_{1}+n_{2}}+ \left( n_{1}n_{2} \right) \frac{ \left( \overline{x_{1}}-\overline{x_{2}} \right) ^{2}}{ \left( n_{1}+n_{2} \right) ^{2}}} \\ \text{Substituting the corresponding values we get} \sigma =\sqrt {\frac{60 \left( 8 \right) ^{2}+80 \left( 7 \right) ^{2}}{60+80}+\frac{ \left( 60\ast80 \right) \left( 650-660 \right) ^{2}}{ \left( 20+80 \right) ^{2}}}= \sqrt {\frac{388}{7}+\frac{1200}{49}} = \\\\ =\sqrt {\frac{3916}{49}} = 8.9\\$

Question:13

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

Answer:

Mean and standard deviation of 100 items are 50 and 4 respectively
We have to find the sum of all items and the sum of the squares of the items.
As per the question, number of items n =100
$\\ \text{Mean of the given items, }\overline{x}=50~ \text{But we know}, \overline{x}=\frac{ \Sigma x_{i}}{n}~~ \\\\ \\ \text{Substituting the corresponding values we get} 50= \frac{ \Sigma x_{i}}{100}~~~~~ \Sigma x_{i}=50\ast100=5000 \\\\ \\ \text{Hence,~the sum of all the 100 items}=5000 \\\\ \\ \text{Also given that the standard deviation of the 100 items is 4} \\\\ \\ ~~ \sigma =4 \\\\$

$\\ \text{ But we know that } \\\\ \\ \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~~~ \\\\ \\ \text{Substituting the corresponding values, we get} 4= \sqrt {\frac{ \Sigma x_{i}^{2}}{100}- \left( \frac{5000}{100} \right) ^{2}} \\\\ \\ \text{~~Now taking square on both sides, we get }4^{2}=\frac{ \Sigma x_{i}^{2}}{100}- \left( 50 \right) ^{2}~~~ \\\\$
$\\ ~ 16=\frac{ \Sigma x_{i}^{2}}{100}-2500~ \\\\ \\ ~~ 16+2500=\frac{ \Sigma x_{i}^{2}}{100}~~ \\\\ \\ ~~\frac{ \Sigma x_{i}^{2}}{100}=2516 \\\\ \\ ~ \Sigma x_{i}^{2}=2516\ast100=251600 \\\\$

Question:14

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

Answer:

$\\\\ \text{It is given that for a distribution } \Sigma \left( x-5 \right) =3,~ \Sigma \left( x-5 \right) ^{2}=43 \text{ and the total number of item is 18} \\ \text{We have to find the mean and standard deviation} \\ \text{As per given criteria, Number of items, n=18}$

$\\ \text{~ Mean can be written as }\overline{x}=A+\frac{ \Sigma \left( x-5 \right) }{n}~~ \\ \\ \text{ Here assumed mean is 5, so substituting the corresponding values in above equation we get} \\ \\ ~\overline{x}=5+\frac{3}{18}=\frac{93}{18}=5.17~ \\ \\ \text{And we know that the standard deviation can be written as}, \sigma =\sqrt {\frac{ \Sigma \left( x-5 \right) ^{2}}{n}- \left( \frac{ \Sigma \left( x-5 \right) }{n} \right) ^{2}}~ \\$

Substituting the corresponding values we get

$\\ \sigma=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}=\sqrt{2.39-(0.166)^{2}}=\sqrt{2.39-0.027}=\sqrt{2.363}=1.54$

Thus the mean and standard deviation of given items are 5.17 and 1.54 respectively.

Question:15

Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline \mathrm{x} & 1 \leq \mathrm{x}<3 & 3 \leq \mathrm{x}<5 & 5 \leq \mathrm{x}<7 & 7 \leq \mathrm{x}<10 \\ \hline \mathrm{f} & 6 & 4 & 5 & 1 \\ \hline \end{array}$

Answer:

Frequency distribution is given. We have to find the mean and variance

Converting the ranges of x to groups the given table can be re written as

$\begin{array}{|l|l|l|l|l|} \hline \mathrm{X} \text { (Class) } & \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ \hline 1-3 & 2 & 6 & 12 & 24 \\ \hline 3-5 & 4 & 4 & 16 & 64 \\ \hline 5-7 & 6 & 5 & 30 & 180 \\ \hline 7-10 & 8.5 & 1 & 8.5 & 72.25 \\ \hline \text { Total } & & \mathrm{N}=16 & =66.5 & 340.25 \\ \hline \end{array}$

And, we know that variance is $\\\\ ~ \sigma ^{2}= \frac{ \Sigma f_{i}x_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}x_{i}}{n} \right) ^{2}~ \\\\$

$\\ \text{~ Substituting values from above table, we get } \sigma ^{2}=\frac{340.25}{16}- \left( \frac{66.5}{16} \right) ^{2}~~~ \\\\ \\ ~ \sigma ^{2}=21.265- \left( 4.16 \right) ^{2}~=21.265-17.305=3.96~ \\\\ \\ ~mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{66.5}{16}=4.16~ \\\\ \\ \text{ Hence, the mean and variance of the given frequency distribution is 4.16 and 3.96 respectively } \\\\$

Question:16

Calculate the mean deviation about the mean for the following frequency distribution:
$\begin{array}{|l|l|l|l|l|l|} \hline \text { Class interval } & 0-4 & 4-8 & 8-12 & 12-16 & 16-20 \\ \hline \text { Frequency } & 4 & 6 & 8 & 5 & 2 \\ \hline \end{array}$

Answer:

A frequency distribution table is given, and we have to find the mean deviation about the mean
Let us make a table from the given data and fill out the other columns after calculation
$\begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} \\ \hline 0-4 & 2 & 4 & 8 \\ \hline 4-8 & 6 & 6 & 36 \\ \hline 8-12 & 10 & 8 & 80 \\ \hline 12-16 & 14 & 5 & 70 \\ \hline 16-20 & 18 & 2 & 36 \\ \hline & \text { Total } & \mathrm{N}=25 & =230 \\ \hline \end{array}$
$\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{230}{25}=9.2 \\\\ \\ \text{~~ The above column can be re written as} \\\\$

$\begin{array}{|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} & \begin{array}{l} d_{i}=\mid x_{i}- \\ \bar{x} \mid \end{array} & f_{i} d_{i} \\ \hline 0-4 & 2 & 4 & 8 & 7.2 & 28.8 \\ \hline 4-8 & 6 & 6 & 36 & 3.2 & 19.2 \\ \hline 8-12 & 10 & 8 & 80 & 0.8 & 6.4 \\ \hline 12-16 & 14 & 5 & 70 & 1.8 & 24.4 \\ \hline 16-20 & 18 & 2 & 36 & 8.8 & 17.6 \\ \hline & \text { Total } & \mathrm{N}=25 & =230 & & =96.0 \\ \hline \end{array}$
$\\ mean\: \: Deviation=\frac{ \Sigma f_{i}d_{i}}{ \Sigma f_{i}}=\frac{96}{25}=3.84 \\\\$

Question:17

Calculate the mean deviation from the median of the following data:
$\begin{array}{|l|l|l|l|l|l|} \hline \text { Class interval } & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \hline \text { Frequency } & 4 & 5 & 3 & 6 & 2 \\ \hline \end{array}$

Answer:

A frequency distribution table is given and we have to find the mean deviation about the median

Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & \begin{array}{l} \text { Cumulative } \\ \text { frequency } \\ \left(\mathrm{c}_{\mathrm{f}} f\right) \end{array} \\ \hline 0-6 & 3 & 4 & 4 \\ \hline 6-12 & 9 & 5 & 9 \\ \hline 12-18 & 15 & 3 & 12 \\ \hline 18-24 & 21 & 6 & 18 \\ \hline 24-30 & 27 & 2 & 20 \\ \hline & \text { Total } & \mathrm{N}=20 & \\ \hline \end{array}$

Now here N=20, which is even

Here $\text{Median Class}= \frac{N}{2}=10^{th}~term~ \\\\$

$\\ \text{This observation lie in the class interval 12-18, so median can be written as} \\ M=l+\frac{\frac{N}{2}-cf}{f}*h \\ \text{ Here l=12, cf=9, f=3, h=6 and N=20} \\\text{Substituting these values, the above equation becomes}, M=12+\frac{\frac{20}{2}-9}{3}*6\\ M=12+\frac{10-9}{3}*6$

$\\ =12+\frac{1\ast6}{3}~~ \\\\ \\ ~ M=12+2=14 \\\\$

$\begin{array}{|l|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & d_{i}=\left|x_{i}-M\right| & f_{i} d_{i} \\ \hline 0-6 & 3 & 4 & 11 & 44 \\ \hline 6-12 & 9 & 5 & 5 & 25 \\ \hline 12-18 & 15 & 3 & 1 & 3 \\ \hline 18-24 & 21 & 6 & 7 & 42 \\ \hline 24-30 & 27 & 2 & 13 & 26 \\ \hline & \text { Total } & \mathrm{N}=20 & & =140 \\ \hline \end{array}$

$\\ mean Deviation=\frac{ \Sigma f_{i}d_{i}}{ \Sigma f_{i}}=\frac{140}{20}=7 \\\\$

Question:18

Determine the mean and standard deviation for the following distribution:
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$

Answer:

A frequency distribution table is given and we have to find the mean and standard deviation
Let us make a table from the given data and fill out the other columns after calculation

$\begin{array}{|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} \\ \hline 2 & 1 & 2 \\ \hline 3 & 6 & 18 \\ \hline 4 & 6 & 24 \\ \hline 5 & 8 & 40 \\ \hline 6 & 8 & 48 \\ \hline 7 & 2 & 14 \\ \hline 8 & 2 & 16 \\ \hline 9 & 3 & 27 \\ \hline 10 & 0 & 0 \\ \hline 11 & 2 & 22 \\ \hline 12 & 1 & 12 \\ \hline 13 & 0 & 0 \\ \hline 14 & 0 & 0 \\ \hline 15 & 0 & 0 \\ \hline 16 & 1 & 16 \\ \hline \text { Total } & \mathrm{N}=40 & =239 \\ \hline \end{array}$
$\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{239}{40}=5.975\approx6 \\\\$
$\begin{array}{|l|l|l|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} & d_{i}=x_{i}-\bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 2 & 1 & 2 & -4 & -4 & 16 \\ \hline 3 & 6 & 18 & -3 & -18 & 54 \\ \hline 4 & 6 & 24 & -2 & -12 & 24 \\ \hline 5 & 8 & 40 & -1 & -8 & 8 \\ \hline 6 & 8 & 48 & 0 & 0 & 0 \\ \hline 7 & 2 & 14 & 1 & 2 & 2 \\ \hline 8 & 2 & 16 & 2 & 4 & 8 \\ \hline 9 & 3 & 27 & 3 & 9 & 27 \\ \hline 10 & 0 & 0 & 4 & 0 & 0 \\ \hline 11 & 2 & 22 & 5 & 10 & 50 \\ \hline 12 & 1 & 12 & 6 & 6 & 36 \\ \hline 13 & 0 & 0 & 7 & 0 & 0 \\ \hline 14 & 0 & 0 & 8 & 0 & 0 \\ \hline 15 & 0 & 0 & 9 & 0 & 0 \\ \hline 16 & 1 & 16 & 10 & 10 & 100 \\ \hline \text { Total } & \mathrm{N}=40 & =239 & & =-1 & =325 \\ \hline \end{array}$
And we know that standard deviation is
$\\ \sigma = \sqrt {\frac{ \Sigma f_{i}d_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}d_{i}}{n} \right) ^{2}}= \sqrt {\frac{325}{40}- \left( \frac{-1}{40} \right) ^{2}}=\sqrt {8.125- \left( 0.025 \right) ^{2}} =2.8721\\\\$

Question:19

The weights of coffee in 70 jars are shown in the following table:
$\begin{array}{|l|l|} \hline \text { Weight (in grams) } & \text { Frequency } \\ \hline 200-201 & 13 \\ \hline 201-202 & 27 \\ \hline 202-203 & 18 \\ \hline 203-204 & 10 \\ \hline 204-205 & 1 \\ \hline 205-206 & 1 \\ \hline \end{array}$
Determine variance and standard deviation of the above distribution.

Answer:

A frequency distribution table for the weights of coffee in 70 jars is given and we have to find the variance and standard deviation of the distribution.
Let us make a table from the given data and fill out the other columns after calculation
$\begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Weight (in } \\ \text { grams) } \end{array} & \begin{array}{l} \text { Mid-value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} \\ \hline 200-201 & 200.5 & 13 & 2606.5 \\ \hline 201-202 & 201.5 & 27 & 5440.5 \\ \hline 202-203 & 202.5 & 18 & 3645 \\ \hline 203-204 & 203.5 & 10 & 2035 \\ \hline 204-205 & 204.5 & 1 & 204.5 \\ \hline 205-206 & 205.5 & 1 & 205.5 \\ \hline & \text { total } & \mathrm{N}=70 & =14137 \\ \hline & & & \\ \hline \end{array}$
$\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{N}=\frac{14137}{70}=201.9 \\\\$
$\begin{array}{|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Weight (in } \\ \text { grams) } \end{array} & \begin{array}{l} \text { Mid-value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & d_{i}=x_{i}-\bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 200-201 & 200.5 & 13 & -1.4 & -18.2 & 25.48 \\ \hline 201-202 & 201.5 & 27 & -0.4 & -10.8 & 4.32 \\ \hline 202-203 & 202.5 & 18 & 0.6 & 10.8 & 6.48 \\ \hline 203-204 & 203.5 & 10 & 1.6 & 16 & 25.6 \\ \hline 204-205 & 204.5 & 1 & 2.6 & 2.6 & 6.76 \\ \hline 205-206 & 205.5 & 1 & 3.6 & 3.6 & 12.96 \\ \hline & \text { total } & \mathrm{N}=70 & & =4 & =81.6 \\ \hline & & & & & \\ \hline \end{array}$
And we know that standard deviation is
$\sigma = \sqrt {\frac{ \Sigma f_{i}d_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}d_{i}}{n} \right) ^{2}}= \sqrt {\frac{81.6}{70}- \left( \frac{4}{70} \right) ^{2}}=\sqrt {1.17- \left( 0.057 \right) ^{2}}= \sqrt {1.17}=1.08~~ \\\\$
Variance = 1.17

Question:20

Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

Answer:

The first n terms of an A.P are given whose first term is a and common difference is d. We have to find mean and standard deviation.

The given AP in tabular form is as shown below,

$\begin{array}{|l|l|l|} \hline \multicolumn{1}{|c|} {x_{i}} & \multicolumn{1}{|c|} {d_{i}=x_{i}-a} & d_{i}^{2} \\ \hline a & 0 & 0 \\ \hline a+d & d & d^{2} \\ \hline a+2 d & 2 d & 4 d^{2} \\ \hline a+3 d & 3 d & 9 d^{2} \\ \hline- & - & - \\ \hline a+(n-1) d & (n-1) d & (n-1)^{2} d^{2} \\ \hline \end{array}$
$\\ \text{Here we have assumed a as mean } \\ \text{~ Given that AP has n terms.} \\ \text{ And we know the sum of all terms of AP can be written as }$

$\\ \sum x_{i}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\ \overline{x}=\frac{ \sum x_{i}}{n}=\frac{2a+ \left( n-1 \right) d}{2}=a+\frac{ \left( n-1 \right) }{2}d \\ \sum d_{i}= \sum \left( x_{i}-a \right) =d \left[ 1+2+3+ \ldots + \left( n-1 \right) \right] =d \left( \frac{n \left( n-1 \right) }{2} \right)$

$\\ \sum d_{i}^{2}= \sum \left( x_{i}-a \right) ^{2}=d^{2} \left[ 1^{2}+2^{2}+3^{2}+ \ldots + \left( n-1 \right) ^{2} \right] =d^{2} \left( \frac{n \left( n-1 \right) \left( 2n-1 \right) }{6} \right) \\\\ \\ \sigma =\sqrt {\frac{ \sum \left( x_{i}-a \right) ^{2}~}{n}- \left( \frac{ \sum \left( x_{i}-a \right) }{n} \right) ^{2}} \\\\ \\ =\sqrt {\frac{d^{2} \left( \frac{n \left( n-1 \right) \left( 2n-1 \right) }{6} \right) }{n}-\frac{d^{2}n^{2} \left( n-1 \right) ^{2}}{4n^{2}}} \\\\ \\ =\sqrt {\frac{d^{2} \left( n-1 \right) \left( 2n-1 \right) }{6}-\frac{d^{2} \left( n-1 \right) ^{2}}{4}} \\\\ \\ =\sqrt {\frac{d^{2} \left( n-1 \right) }{2} \left( \frac{2n-1}{3}-\frac{n-1}{2} \right) } \\\\ \\ =\sqrt {\frac{d^{2} \left( n-1 \right) }{2} \left( \frac{4n-2-3n+3}{6} \right) }=\sqrt {\frac{d^{2} \left( n-1 \right) }{2} \left( \frac{n+1}{6} \right) }=d\sqrt {\frac{n^{2}-1}{12}} \\\\$

Question:21

Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Ravi } & 25 & 50 & 45 & 30 & 70 & 42 & 36 & 48 & 35 & 60 \\ \hline \text { Hashina } & 10 & 70 & 50 & 20 & 95 & 55 & 42 & 60 & 48 & 80 \\ \hline \end{array}$
Who is more intelligent and who is more consistent?

Answer:

The marks obtained, out of 100, by 2 students Ravi and Hashina in 10 tests are given
We have to find who is more intelligent and who is more consistent
The marks of Ravi taken separately as follows
$\begin{array}{|l|l|l|} \hline \multicolumn{1}{|c|} {x_{i}} & \multicolumn{1}{|c|} {d_{i}=x_{i}-45} & \multicolumn{1}{|c|} {d_{i}^{2}} \\ \hline 25 & -20 & 400 \\ \hline 50 & 5 & 25 \\ \hline 45 & 0 & 0 \\ \hline 30 & -15 & 225 \\ \hline 70 & 25 & 625 \\ \hline 42 & -3 & 9 \\ \hline 36 & -9 & 81 \\ \hline 48 & 3 & 9 \\ \hline 35 & -10 & 100 \\ \hline 60 & 15 & 225 \\ \hline \text { TOTAL }=441 & =-9 & =1699 \\ \hline \end{array}$
Here we have assumed 45 as mean
And we know that the standard deviation can be written as,
$\\ \sigma =\sqrt {\frac{ \Sigma \left( x-a \right) ^{2}}{n}- \left( \frac{ \Sigma \left( x-a \right) }{n} \right) ^{2}}~~ \\\\ \\ \sigma =\sqrt {\frac{1699}{10}- \left( -\frac{9}{10} \right) ^{2}} =\sqrt {169.9-0.81}= \sqrt {169.09}~=13 \\\\ \\ \text{Now mean is }\overline{x}=A+\frac{ \Sigma f_{i}d_{i}}{N}~ =45- \left( \frac{9}{10} \right) =44.1~~ \\\\ \\ \text{ For Hashina} \\\\$
$\begin{array}{|l|l|l|} \hline \multicolumn{1}{|c|} {x_{i}} & \multicolumn{1}{|c|} {d_{i}=x_{i}-53} & \multicolumn{1}{|c|} {d_{i}^{2}} \\ \hline 10 & -43 & 1849 \\ \hline 70 & 17 & 289 \\ \hline 50 & -3 & 9 \\ \hline 20 & -33 & 1089 \\ \hline 95 & 42 & 1764 \\ \hline 55 & 2 & 4 \\ \hline 42 & -11 & 121 \\ \hline 60 & 7 & 49 \\ \hline 48 & -5 & 25 \\ \hline 80 & 27 & 729 \\ \hline \text { TOTAL }=530 & =0 & =5928 \\ \hline \end{array}$
Here as $\\\\ \frac{530}{10}=53~~So, 53 is mean \\\\$

And we know that the standard deviation can be written as,
$\sigma =\sqrt {\frac{ \Sigma \left( x-a \right) ^{2}}{n}- \left( \frac{ \Sigma \left( x-a \right) }{n} \right) ^{2}}~~ \\\\ \\ \sigma =\sqrt {\frac{5928}{10}- \left( -\frac{0}{10} \right) ^{2}} =\sqrt {592.8}=24.35 \\\\ \\ \text{ We have the mean and S.D of the Hashina and Ravi . } \\\\ \\ \text{For Ravi , C.V}=\frac{ \sigma }{\overline{x~}} * 100=\frac{13}{44.1} * 100 =29.48 \\\\$
$\\ For Hasina , C.V=\frac{ \sigma }{\overline{x~}} * 100=\frac{24.35}{53} * 100~=45.94~ \\\\ \\ \text{ Now as CV } \left( \text{of Ravi } \right) <\text{CV of Hasina} \\\\ \\ \text{~Hence, Ravi is more consistent } \\\\ \\ \text{ Mean of Hasina}>\text{Mean of Ravi } \\\\ \\ \text{~ Hence, Hasina is more intelligent} \\\\$

Question:22

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.
We have to find the correct standard deviation
As per the given criteria, Number of observations, n=100
Mean of the given observations before correction, $\\\\ \overline{X~}=40 \\\\$

But we know $\\\overline{X~}=\frac{ \Sigma x_{i}}{n}~~ \\\\$

Substituting the corresponding values, we get $40=\frac{ \Sigma x_{i}}{100}~~ \\\\$

$\\ \\ ~~ \Sigma x_{i}=40 * 100=4000~ \\\\ \\ \text{It is given in the question that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27} \\\\ \\ ~respectively~~~~ \\\\ \\ So~~ \Sigma x_{i}=4000-30-70+3+27=3930~ \\\\ \\ \text{ So the correct mean after correction is }\overline{X~}=\frac{ \Sigma x_{i}}{n}=\frac{3930}{100}=39.3~ \\\\$

Also given that standard deviation of the 100 observations is 10 before correction $\sigma =10 \\\\$

$\\ \text{But~we know that } \\ \sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~~ \\ ~\text{ Substituting the corresponding values we get }10=\sqrt {\frac{ \Sigma x_{i}^{2}}{100}- \left( \frac{4000}{100} \right) ^{2}}~~ \\ \text{~Now taking square on both sides, we get }10^{2}=\frac{ \Sigma x_{i}^{2}}{100}- \left( 40 \right) ^{2}~~ \\ ~ 100=\frac{ \Sigma x_{i}^{2}}{100}-1600~ \\ ~ 100+1600=\frac{ \Sigma x_{i}^{2}}{100}~~ \\ ~~ \Sigma x_{i}^{2}=170000~ \\ \text{It is said that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively. }$

$\\ \text{So,~correction is } \Sigma x_{i}^{2}=170000- \left( 30 \right) ^{2}- \left( 70 \right) ^{2}+ \left( 3 \right) ^{3}+ \left( 27 \right) ^{2}~~ \\\\ \\ ~~~ \Sigma x_{i}^{2}=170000-900-4900+9+729=164938 \\\\ \\ \text{~~So, the correct standard deviation after correction is } \\\\ \\ \sigma = \sqrt {\frac{164938}{100}- \left( \frac{3930}{100} \right) ^{2}} =\sqrt {1649.38- \left( 39.3 \right) ^{2}} = \sqrt {104.89}=10.24 \\\\$

Question:23

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 the variance.

Answer:

$\\ \text{If at the time of calculation of mean and variance of 10 readings, a student wrongly used the reading} \\\\ \\ \text{ 52 instead of the correct reading 25. The mean and variance were obtained as 45 and 16 respectively.} \\ \text{ We have to find the correct mean and the variance.} \\ ~~ \text{As per the given criteria, Number of observations}, n=10 \\\\ \text{Mean of the given observations before correction,} \\\\ \overline{X~}=45 \\ \text{~ But we know, }\overline{X~}=\frac{ \Sigma x_{i}}{n}~~ \\\\$
$\\ ~~\text{Substituting the corresponding values, we get }45=\frac{ \Sigma x_{i}}{10}~~ \\\\ \\ ~~ \Sigma x_{i}=45 * 10=450~ \\\\ \\ \text{It~is~given in the question that one reading 25 was taken wrongly as 52 } \\\\ \\ So~~ \Sigma x_{i}=450-52+25=423~ \\\\ \\ \text{ So the correct mean after correction is }\overline{X~}=\frac{ \Sigma x_{i}}{n}=\frac{423}{10}=42.3~~ \\\\$
$\\ \text{~Also given that variance of the 10 observations is 16 before correction } \sigma ^{2}=16~ \\\\ \\ \text{But~we know that } \\\\ \\ \sigma ^{2}=\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}~~ \\\\ \\ ~ \text{Substituting the corresponding values we get} 16=\frac{ \Sigma x_{i}^{2}}{10}- \left( 45 \right) ^{2}~~ \\\\ \\ ~\text{Now taking square on both sides, we get } 16=\frac{ \Sigma x_{i}^{2}}{10}-2025~~ \\\\ \\ ~ 16+2025=\frac{ \Sigma x_{i}^{2}}{10}~~ \\\\ \\ ~~ \Sigma x_{i}^{2}=20410~ \\\\$
$\\ \text{It is said that one reading 25 was wrongly taken as 52. } \\\\ \\ \text{So,~correction is } \Sigma x_{i}^{2}=20410- \left( 52 \right) ^{2}+ \left( 25 \right) ^{2}~~=20410-2704+625=18331 \\\\ \\ \text{~~So, the variance after correction is } \\\\ \\ \sigma ^{2}=\frac{18331}{10}- \left( \frac{423}{10} \right) ^{2} =1833.1- \left( 42.3 \right) ^{2}=1833.1-1789.29~=43.81 \\\\$

Question:24

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is
A. 2
B. 2.57
C. 3
D. 3.75

Answer:

Given data is 3,10, 10, 4, 7 ,10 ,5 . They are total 7.

$\\ \text{Here mean, }\overline{X~}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7 \\\\ \text{ This can be written in table form as,} \\\\$

$\begin{array}{|l|l|} \hline \text { Data }\left(x_{i}\right) & \multicolumn{1}{|c|} {d_{i}=\left|x_{i}-\bar{x}\right|} \\ \hline 3 & 4 \\ \hline 10 & 3 \\ \hline 10 & 3 \\ \hline 4 & 3 \\ \hline 7 & 0 \\ \hline 10 & 3 \\ \hline 5 & 2 \\ \hline \text { Total } & =18 \\ \hline \end{array}$

$\\ \text{Hence Mean Deviation becomes,} M.D=\frac{ \Sigma d_{i}}{7}=\frac{18}{7}=2.57 \\\\$

Question:25

Mean deviation for n observations x₁, x₂, ..., x_n from their mean $\bar{X}$ is given by

$\\A. \sum _{i=1}^{n}\left(x_{1}-\bar{x}\right)\\\\ B.\frac{1}{n} \sum _{i=1}^{n} \left |\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}} \right |\\\\C. \sum _{i=1}^{n} {\left(x_{i}-\bar{x}\right)^{2}}\\\\D \frac{1}{n}\sum _{i=1}^{n}(x-\bar{x})^{2}$

Answer:

$\\ \text{We know for n observations }x_{1},~x_{2}, \ldots \ldots ., x_{n}~having~\overline{X}\text{ is given by } \\ \\ M.D=\frac{ \Sigma d_{i}}{n}~ \\ \\ \text{ But we know that }d_{i}= \vert x_{i}-\overline{x} \vert ~~ \\ \\ \text{So,~mean~deviation becomes M.D}=\frac{ \Sigma \vert x_{i}-\overline{x} \vert }{n}~~ \\ \\ Or\: \: M.D=\frac{1}{n} \sum _{i=0}^{n} \vert x_{i}-\overline{x~} \vert \\$

Question:26

When tested, the lives (in hours) of 5 bulbs were noted as follows:
1357, 1090, 1666, 1494, 1623
The mean deviations (in hours) from their mean is
A. 178
B. 179
C. 220
D. 356

Answer:

$\\ \text{Given~that the lives of 5 bulbs in hours is 1357, 1090, 1666, 1494 , 1623 } \\ \\ \text{ Here mean }\overline{X}=\frac{1357+1090+1666+1494+1623}{5}=\frac{7230}{7}=1446~ \\ \\ \text{ This can be written in table form as, } \\$
$\begin{array}{|l|l|} \hline \text { Lives (in hours }) x_{i} & \multicolumn{1}{|c|} {d_{i}=x_{i}-\bar{x}} \\ \hline 1357 & 89 \\ \hline 1090 & 356 \\ \hline 1666 & 220 \\ \hline 1494 & 48 \\ \hline 1623 & 177 \\ \hline \text { Total } & =890 \\ \hline \end{array}$
Hence, mean deviation becomes $\\\\ M.D=\frac{ \Sigma d_{i}}{n}=\frac{890}{5}=178 \\$
$\text{M.D}=\frac{ \Sigma d_{i}}{n}=\frac{890}{5}=178 \\$

Question:27

Following are the marks obtained by 9 students in a mathematics test:
50, 69, 20, 33, 53, 39, 40, 65, 59
The mean deviation from the median is:
A. 9
B. 10.5
C. 12.67
D. 14.76

Answer:

Given that the marks obtained by 9 students in a mathematics test are 50, 69 , 20 ,33, 53, 39 ,40, 65, 59 As number of students =9 which is odd
$\\ \text{So median will be}\frac{9+1}{2}=5^{th}~term~ \\ \\ \text{~ Arranging these in ascending order, we get 20,33,39,40,50,53,59,65,69 } \\ \\ \text{~ So the }5^{th}\text{ term after arranging is 50 } \\ \\ \text{~ So, median is 50 } \\ \\ \text{This~can~be written in table form as, } \\$
$\begin{array}{|l|l|} \hline \text { Marks }\left(x_{i}\right) & \multicolumn{1}{|c|} {d_{i}=\mid x_{i}-\text {Median} \mid} \\ \hline 20 & 30 \\ \hline 33 & 17 \\ \hline 39 & 11 \\ \hline 40 & 10 \\ \hline 50 & 0 \\ \hline 53 & 3 \\ \hline 59 & 9 \\ \hline 65 & 15 \\ \hline 69 & 19 \\ \hline \text { Total } & =114 \\ \hline \end{array}$
Hence Mean deviation becomes $\\ M.D=\frac{ \Sigma d_{i}}{n}=\frac{114}{9}=12.67 \\$

Question:28

The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is

A. $\sqrt{\frac{52}{7}}$
B. $\frac{52}{7}$
C. $\sqrt{6}$
D. $6$

Answer:

Given data can be written in table form as,
$\begin{array}{|l|l|} \hline \text { Data }\left(x_{i}\right) & x_{i}^{2} \\ \hline 6 & 36 \\ \hline 5 & 25 \\ \hline 9 & 81 \\ \hline 13 & 169 \\ \hline 12 & 144 \\ \hline 8 & 64 \\ \hline 10 & 100 \\ \hline \text { Total }=63 & 619 \\ \hline \end{array}$
$\\ \text{But~we know that standard deviation can be written as } \\ \\ \sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}= \sqrt {\frac{619}{7}- \left( \frac{63}{7} \right) ^{2}}~~~ \\ \\ \sigma = \sqrt {\frac{619}{7}- \left( 9 \right) ^{2}}=\sqrt {\frac{619}{7}-81}~ =\sqrt {\frac{52}{7}} \\$

Question:29

Let x₁, x₂, ..., x_n be n observations and $\bar{X}$ be their arithmetic mean. The formula for the standard deviation is given by

$\\A. \left (\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}} \right )^{2} \\B. \frac{\left(x_{i}-\bar{x}\right)^{2}}{n} \\C. \sqrt{\frac{\left(\mathrm{X}_{\mathrm{i}}-\overline{\mathrm{X}}\right)^{2}}{\mathrm{n}}} \\D \sqrt{\frac{x_{i}^{2}}{n}+\bar{x}^{2}}$

Answer:

$. \\ \text{We know standard deviation for }x_{1},~x_{2}, \ldots .. , x_{n}\text{~ observations can be written as } \\ \\ \sigma =\sqrt {\frac{1}{n}~~ \sum _{i=0}^{n} \left( x_{i}-\overline{x} \right) ^{2}~} \\$

Question:30

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is
A. 50000
B. 250000
C. 252500
D. 255000

Answer:

As per the question

Number of observation, n =100

$\\ \text{Mean of the given observations, }\overline{x}=50~ \\ \\ \text{But~we know, }\overline{X~}=\frac{ \Sigma x_{i}}{n}~ \\ \\ \text{Substituting the corresponding values, we get }50=\frac{ \Sigma x_{i}}{100}~~ \\ \\ \Sigma x_{i}=50 * 100=5000 \\ \\ \text{It is also given that the standard deviation of the 100 observations is 5 } \\\\ \sigma =5 \\ \\ \sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}} \\ \\ \text{Substituting the corresponding values, we get }5=\sqrt {\frac{ \Sigma x_{i}^{2}}{100}- \left( \frac{5000}{100} \right) ^{2}}~ \\$

$\\ \text{ Now taking square on both sides we get }5^{2}=\frac{ \Sigma x_{i}}{100}- \left( 50 \right) ^{2}~ \\ \\ 25=\frac{ \Sigma x_{i}}{100}-2500 \\ \\ ~ 25+2500=\frac{ \Sigma x_{i}^{2}}{100}~~~~ \\ \\ ~~ \Sigma x_{i}^{2}=2525 * 100 \\ \\ ~ \Sigma x_{i}^{2}=252500 \\$

Question:31

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. ks
C. s + k
D. $\frac{s}{k}$

Answer:

Given observations are a, b, c, d, e

So, the mean of the 5 observations is

$\\ m=\frac{a+b+c+d+e}{5}=\frac{ \sum x_{i}}{5} \\\sum x_{i}=5m=a+b+c+d+e \\$

And the standard deviation of the 5 observations is

$\\ \sigma =\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}- \left( \frac{ \sum x_{i}}{5} \right) ^{2}}=\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\$

Now we will find the mean and standard deviation of the observations a+k , b+k, c+k, d+k, e+k we get

So, the mean of these 5 observations is

$\\ m_{1}=\frac{ \left( a+k \right) + \left( b+k \right) + \left( c+k \right) + \left( d+k \right) + \left( e+k \right) }{5} \\ \\ =\frac{a+b+c+d+e}{5}+k=m+k \\ \\ \sigma _{1}=\sqrt {\frac{ \sum \left( x_{i}+k \right) ^{2}~}{5}- \left( \frac{ \sum \left( x_{i}+k \right) }{5} \right) ^{2}}=\sqrt {\frac{ \sum \left( x_{i}^{2}+k^{2}+2kx_{i} \right) }{5}- \left( m+k \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum \left( x_{i}^{2} \right) }{5}+\frac{5k^{2}}{5}+\frac{2k \sum \left( x_{i} \right) }{5}- \left( m+k \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum \left( x_{i}^{2} \right) }{5}+k^{2}+2mk-m^{2}-k^{2}-2mk} \\ \\ =\sqrt {\frac{ \sum \left( x_{i}^{2} \right) }{5}-m^{2}} \\ \\ \sigma _{1}= \sigma \\$

Question:32

Let x₁, x₂, x₃, x₄, x₅ be the observations with mean m and standard deviation s.
The standard deviation of the observations kx₁, kx₂, kx₃, kx₄, kx₅ is
A. k + s
B. $\frac{s}{k}$
C. ks
D. s

Answer:

$\\ \text{Given observations are }x_{1},~x_{2},~x_{3},~x_{4},~x_{5}~~~ \\ \\ \text{ So the mean of the 5 observations is} \\ \\ m=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}=\frac{ \sum x_{i}}{5} \\ \\ \sum x_{i}=5m=x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \\ \\ \sigma =\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}- \left( \frac{ \sum x_{i}}{5} \right) ^{2}}=\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\ \\ m_{1}=\frac{kx_{1}+kx_{2}+kx_{3}+kx_{4}+kx_{5}}{5}=\frac{k \left( x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \right) }{5}=km \\$
$\\ \sigma _{1}=\sqrt {\frac{ \sum \left( kx_{i} \right) ^{2}~}{5}- \left( \frac{ \sum kx_{i}}{5} \right) ^{2}} \\ \\ =\sqrt {\frac{ \sum k^{2} \left( x_{i} \right) ^{2}~}{5}- \left( m_{1} \right) ^{2}} \\ \\ =\sqrt {\frac{k^{2} \sum \left( x_{i} \right) ^{2}~}{5}-k^{2}m^{2}} \\ \\ =\sqrt {k^{2} \left( \frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2} \right) } \\ \\ =k\sqrt {\frac{ \sum \left( x_{i} \right) ^{2}~}{5}-m^{2}} \\ \\ \sigma _{1}=k \sigma ~ \\ \\ \text{ Hence, the standard deviation of new set of observations k}x_{1}\text{, k}x_{2}\text{, k}x_{3}\text{, k}x_{4}\text{, k}x_{5} \: \: \:\ \ is\: \: \: \ \ k \sigma \\$

Question:33

Let $x\textsubscript{1}, x\textsubscript{2}, ... x\textsubscript{n}$ be n observations. $w\textsubscript{i} = lx\textsubscript{i} + k for i = 1, 2, ...n,$ where l and k are constants. If the mean of $x\textsubscript{i}$ 's is 48 and their standard deviation is 12, the mean of $w_i$ ’s is 55 and standard deviation of $w_i$ ’s is 15, the values of l and k should be
A. l = 1.25, k = – 5
B. l = – 1.25, k = 5
C. l = 2.5, k = – 5
D. l = 2.5, k = 5

Answer:

$\\ Given~x_{1},~x_{2} \ldots ..x_{n}\text{ be n observations } \\ \\ \text{And mean of these n observations, }\overline{x}=48~ \\ \\ \text{And their standard deviation S}D_{x}=12~ \\ \\ \text{Another series of n observations is given such that }w_{i}=lx_{i}+k~~for i=1,2 , \ldots .n \text{where L and k are constants} \\ \\ \text{ And mean of these n observations, }\overline{w}=55~ \text{And their standard deviation} SD_{w}=15 \\ \\ \text{ Applying the given condition for mean we get }w_{i}=lx_{i}+k~ \\ \\ \text{Substituting~the corresponding given values of means, we get }55=l \left( 48 \right) +k \ldots \ldots \left( i \right) \\$
$\\ \text{ We know that if standard deviation of x series is s, then standard deviation of kx series is ks } \\ \\ \text{ So standard deviation of }x_{1},~x_{2}, \ldots .x_{n}\text{is S}D_{x}\text{~ And hence the SD of l}x_{1}\text{, l}x_{2} \ldots ..lx_{n}\text{ is lS}D_{x}~ \\ \\ \text{Similarly, If standard deviation of x series is s, then standard deviation of k+x series is s,} \\ \\ \text{ So S.D of l}x_{1}\text{, l}x_{2} \ldots . lx_{n}\text{is lS}D_{x} \\ \\ \text{And hence the SD of l}x_{1}+k, lx_{2}+k \ldots ..lx_{n}+k is lSD_{x}~ \\ \\ \text{ So applying the given condition for standard deviation we get S}D_{w}=lSD_{x}~ \\ \\ \text{Substituting the given values we get }15=l \left( 12 \right) ~ l=\frac{15}{12}=1.25 \\ \\ \text{ Now substituting the value of l in equation } \left( i \right) ,~\text{we get} 55= \left( 1.25 \right) \left( 48 \right) +k \\ \\ ~ 55=60+k \\ \\ ~ k=-5 \\$

Question:34

Standard deviations for first 10 natural numbers is
A. 5.5
B. 3.87
C. 2.97
D. 2.87

Answer:

$\\ \text{We know that the standard deviation of the first n natural numbers is }\sqrt {\frac{n^{2}-1}{12}}~ \\ \\ \text{Now for first 10 natural numbers} n=10 \\ \\ \text{ substituting this in the equation of standard deviation we get }\sigma =\sqrt {\frac{ \left( 10 \right) ^{2}-1}{12}}=\sqrt {\frac{100-1}{12}}= \sqrt {\frac{99}{12}}= \sqrt {8.25} =2.87 \\$

Question:35

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

Answer:

We know that the standard deviation of the first n natural numbers is $\sqrt {\frac{n^{2}-1}{12}}~ \\$
Now for first 10 natural numbers n=10
substituting this in the equation of standard deviation we get $\sigma =\sqrt {\frac{ \left( 10 \right) ^{2}-1}{12}}=\sqrt {\frac{100-1}{12}}= \sqrt {\frac{99}{12}}= \sqrt {8.25} =2.87 \\$
Now when 1 is added to each numbers of 1 , 2, 3, 4, 5, 6,7 ,8 ,9, 10 ; we get new series as 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1
Now we know if standard deviation of x series is s, then standard deviation of k+x series is s,
So the S.D of 1+1, 2+1, 3+1, 4+1, 5+1, 6+1, 7+1, 8+1, 9+1, 10+1 series is also same as the S.D of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 series.
$\\ \sigma =\sqrt {8.25}~ \\$
Now for variance we will square on both the sides $\sigma ^{2}=8.25 \\$

Question:36

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

Answer:

First 10 positive integers are 1,2, 3, 4, 5,6 ,7, 8,9, 10 On multiplying each number by
-1 we get-1, -2, -3, -4 , -5, -6, -7, -8, -9 , -10
On adding 1 to each of the number, we get 0, -1, -2, -3, -4, -5, -6, -7, -8, -9
$\\ \Sigma x_{i}=0-1-2-3-4-5-6-7-8-9=-45~ and \\ \\ ~ \Sigma x_{i}^{2}=0^{2}+ \left( -1 \right) ^{2}+ \left( -2 \right) ^{2}+ \ldots \ldots + \left( -9 \right) ^{2}~ \\ \\ \text{ But we know that } \Sigma x_{i}^{2}=\frac{9 \left( 9+1 \right) \left( 2 \left( 9 \right) +1 \right) }{6}=285 \\ \\ \sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~ \\ \\ \text{Substituting the corresponding values, we get }\sigma =\sqrt {28.5- \left( -4.5 \right) ^{2}}=\sqrt {28.5-20.25}= \sqrt {8.25}~ \\ \\ \text{Now for variance both the sides are squared } \sigma ^{2}=8.25 \\$

Question:37

The following information relates to a sample of size 60:

$$ \sum $ x\textsuperscript{2}=18000, $ \sum $ x = 960.\\$
The variance is
A. 6.63
B. 16
C. 22
D. 44

Answer:

$\\ \text{We know that S.D can be written as }\sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~ \\ \\ \text{But given } \Sigma x^{2}=18000, \Sigma x=960,~N=60, \\ \\ \text{substituting these corresponding values, we get }\sigma =\sqrt {\frac{18000}{60}- \left( \frac{960}{60} \right) ^{2}}= \sqrt {300- \left( 16 \right) ^{2}}=\sqrt {300-256}= \sqrt {44}~ \\ \\ \text{Now for variance we square both the sides } \sigma ^{2}=44 \\$

Question:38

Coefficient of variation of two distributions are 50 and 60, and their arithmetic means are 30 and 25 respectively. Difference of their standard deviation is
A. 0
B. 1
C. 1.5
D. 2.5

Answer:

$\\ \text{Given coefficient of variation of two distributions are C}V_{1}=50 and CV_{2}=60 \\ \\ \text{ And the arithmetic means are }\overline{x_{1}}=30, \overline{x_{2}}=25~ \\ \\ \text{We know that coefficient of variation can be written as CV}= \frac{ \sigma }{\overline{x~}} * 100~ \\ \\ \text{Now for first distribution, we have C}V_{1}= \frac{ \sigma _{1}}{\overline{x_{1}~}} * 100 \\ \\ \text{Substituting corresponding values, we get 50}= \frac{ \sigma _{1}}{30} * 100 \\ \\ ~ \sigma _{1}=15 \ldots \ldots . \left( i \right) ~~ \\ \\ \text{ Now for second distribution, we have C}V_{2}= \frac{ \sigma _{2}}{\overline{x_{2}~}} * 100~ \\$
$\\ \text{Substituting corresponding values, we get 60}= \frac{ \sigma _{2}}{25} * 100 \\ \\ ~ \sigma _{2}=15 \ldots \ldots . \left( ii \right) ~~~ \\ \\ \text{So from both the equations we get that the difference of their standard deviation is 0 } \\$

Question:39

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

Answer:

$\\ \text{Given that the standard deviation of some temperature data in degree celcius, } \sigma _{c}=5 \\ \\ \text{We know that C}=\frac{5}{9} \left( F-32 \right) ~ F=\frac{9C}{5}+32~ \\ \\ \text{ Now we know that If S.D of x series is s, then S.D of kx series is ks. } \\ \\ \text{ So if SD of some temperature data is } \sigma _{c}=5 \\ \\ \text{Then SD of some temperature data in}\frac{9C}{5}~,\frac{9C}{5} \sigma _{c}=5 * \frac{9}{5}=9~ \\ \\ \text{Similarly, If SD of x series is s, then SD of k+x series is s}, \\$
$\\ \text{ So SD of some temperature data in}\frac{9C}{5}~ \sigma _{c}=9 \\ \\ \text{ And thus the SD of some temperature data in}\frac{9C}{5}+32 will be \sigma _{F}=\frac{9C}{5}~ \\ \\ ~ \sigma _{C}=9 \\ \\ \text{Now for variance, we will square on both sides, we get } \sigma _{F}=9^{2}=81 \\$

Question:40

Fill in the blanks
Coefficient of variation = $\frac{...}{mean}*100$

Answer:

Standard Deviation
We know that the Coefficient of variation can be written as
$\\ C.V=\frac{ \sigma }{\overline{x~}} * 100 where\: \: \sigma = \text {standard deviation }$

Question:41

Fill in the blanks
If $\bar{X}$ is the mean of n values of x, then $\sum _{i=1}^{n} \left( x_{i}-\overline{x} \right)$ is always equal to _______.
If a has any value other than $\bar{X}$ , then $\sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) ^{2}$ is _________ than $\sum \left (X_i-a \right )^2$

Answer:

Zero, less than Given x is mean of n values, then the sum of all n terms is denoted by $\sum _{i=1}^{n}x_{i}, \\$
so difference of both these is laways equal to zero i.e. $} \sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) =0 \\$
And square of the above equation is also equal to zero, so $} \sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) ^{2}=0 \\$
Now if "a" has the value other than x then $~ \sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) ^{2}>0~ \\$
$\\ So,~ \sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) ^{2}<~ \sum _{i=1}^{n} \left( x_{i}-a \right) ^{2}~~ \\ \\ \text{So, if }\mathrm{a}\text{~ has any value other than }\overline{x},~then~~ \sum _{i=1}^{n} \left( x_{i}-\overline{x} \right) ^{2}\text{~is less than } \sum _{i=1}^{n} \left( x_{i}-a \right) ^{2} \\$

Question:42

Fill in the blanks:
If the variance of a data is 121, then the standard deviation of the data is _______.

Answer:

11
We know the square root of variance is standard deviation $\\ \sigma ^{2}=121$
Taking square root on both sides, we get $\sigma =\sqrt {121}=11~~$
So, if the variance of a data is 121, then the Standard deviation of the data is 11.

Question:43

Fill in the blanks
The standard deviation of a data is ___________ of any change in origin but is _____ on the change of scale.

Answer:

Independent, dependent
Change of origin means some value has been added or subtracted in the observation. And we know the S.D does not change if any value is added or subtracted from the observations, So SD is independent of change in origin.
However, Standard deviation is only affected by a change in scale, that is, when some value is multiplied or divided to observations. Hence, the SD of any data is independent of any change in origin but dependent on any change of scale.

Question:44

The sum of the squares of the deviations of the values of the variable is _______ when taken about their arithmetic mean.

Answer:

Minimum
The sum of the squares of the deviations of the values of the variable is minimum or least when taken about their arithmetic mean.

Question:45

Fill in the blanks
The mean deviation of the data is _______ when measured from the median.

Answer:

Least
Mean deviation is sum of all deviations of a set of a data about the data’s mean. In addition, it is widely believed that the median is usually between the mean and the mode. So, the mean deviation of the data is least when measured from the median.

Question:46

Fill in the blanks
The standard deviation is _______ to the mean deviation taken from the arithmetic mean.

Answer:

Greater than or equal to
SD is the difference between square of deviation of data about mean and square of mean. In addition, mean deviation is sum of all deviations of a set of data about the data’s mean. Hence, the SD is greater than or equal to the mean deviation taken from the arithmetic mean.

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

· Central Tendency

· Mean

· Median

· Mode

· Measure of Dispersion

· Range

· Quartile deviation

· Mean deviation

· Standard deviation

· Mean deviation for grouped data

· Mean deviation for ungrouped data

· Variance

· Analysis of Frequency Deviation

More about NCERT Exemplar Class 11 Maths Solutions Chapter 15

By referring to NCERT Exemplar Class 11 Maths chapter 15 solutions students can easily gather the various concepts of Statistics. Statistics carries a significant number of marks according to the CBSE paper pattern and the students should concentrate on important concepts like standard deviation and mean deviation, as they are frequently asked in the examination.

The step-wise and straightforward format of the NCERT exemplar solutions for Class 11 Maths chapter 15 makes it easy for the students to understand the most complex concepts of Statistics in a quick and easygoing manner. The pdf is exclusively created by our experts for students who find the concepts of Statistics difficult.

Students can easily score marks in this chapter by thoroughly studying a few topics which are commonly asked in the final examinations. The students will understand the concept in the simplest way possible by going through the NCERT Exemplar Class 11 Maths solutions chapter 15 Statistics.

As most students tend to skip the entire chapter due to doubts in Statistics, our experts have formulated the NCERT Exemplar Class 11 Maths chapter 15 solutions in the most student-friendly manner.

NCERT Solutions for Class 11 Mathematics Chapters

Chapter 1 Sets

Chapter 2 Relations and Functions

Chapter 3 Trigonometric Functions

Chapter 4 Principle of Mathematical Induction

Chapter 5 Complex Numbers and Quadratic Equations

Chapter 6 Linear Inequalities

Chapter 7 Permutations and Combinations

Chapter 8 Binomial Theorem

Chapter 9 Sequences and Series

Chapter 10 Straight lines

Chapter 11 Conic Sections

Chapter 12 Introduction to Three Dimensional Geometry

Chapter 13 Limits and Derivatives

Chapter 14 Mathematical Reasoning

Chapter 16 Probability

Important Topic to Cover from NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

Standard Deviation: Questions on Standard Deviation are frequently asked by the examiners. The students should go through the NCERT Exemplar Class 11 Maths chapter 15 solutions to solve all their doubts regarding standard deviation. The student should just follow the simple steps given in the solution and practice.

Mean Deviation: The students should not ignore solving the question of mean deviation as they are commonly asked in the exam. The students should memorize the formula for mean and standard deviation upon which the problem is based and solve the entire question by referring to the steps given in the Class 11 Maths NCERT Exemplar solutions chapter 15.

Mean, Median, Mode, and Range: These are the fundamental concepts of Statistics and can be asked in the examination for MCQ or short answer type questions.

Check Chapter-Wise NCERT Solutions of Book

Chapter-1	Sets
Chapter-2	Relations and Functions
Chapter-3	Trigonometric Functions
Chapter-4	Principle of Mathematical Induction
Chapter-5	Complex Numbers and Quadratic equations
Chapter-6	Linear Inequalities
Chapter-7	Permutation and Combinations
Chapter-8	Binomial Theorem
Chapter-9	Sequences and Series
Chapter-10	Straight Lines
Chapter-11	Conic Section
Chapter-12	Introduction to Three Dimensional Geometry
Chapter-13	Limits and Derivatives
Chapter-14	Mathematical Reasoning
Chapter-15	Statistics
Chapter-16	Probability

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Physics Solutions

Read more NCERT Solution subject wise -

Also, read NCERT Notes subject wise -

Also Check NCERT Books and NCERT Syllabus here:

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

Frequently Asked Questions (FAQs)

1. Which are the important topics one should cover in Statistics?

Standard Deviation, Mean Deviation and Central Tendency are the important topics one should cover in Statistics.

2. Where can we download the Class 11 Maths NCERT Exemplar Solutions Chapter 15?

Students can get PDF by using the NCERT Exemplar Class 11 Maths solutions chapter 15 PDF Download function.

3. Are the solutions helpful for board examinations?

Yes, NCERT Exemplar Class 11 Maths solutions chapter 15 are helpful for board examinations.

4. Can these solutions be referred for cracking competitive examinations?

Yes, Class 11 Maths NCERT Exemplar solutions chapter 15 can be referred to when pursuing competitive examinations.

Also Read

NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

Hydrogen Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 9 Notes - Download PDF

The S-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 10 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Exercise 10.1 Class 11 Maths Chapter 10 - Straight Lines

NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

Articles

IGNOU Assignment Result 2024

Dec 21, 2024

DBSE Time Table 2025, Check Delhi Board Exam Schedule for Class 10 & 12

Dec 20, 2024

How to Make APAAR ID via DigiLocker and School: A Complete Guide

Dec 17, 2024

Sainik School Entrance Exam 2025: Application, Exam Date, How to Apply, Syllabus

Dec 17, 2024

MVPP Scholarship Exam 2024: Mukhyamantri Vigyan Pratibha Pariksha Registration, Exam Date

Dec 16, 2024

Silverzone ISSO Answer Key 2024-25: Download PDF

Dec 16, 2024

Silverzone IOEL Answer Key 2024 – Download International Olympiad of English Language Solutions

Dec 16, 2024

Silverzone Olympiad IOM Answer Key 2024-25: Qualifying Marks, Score

Dec 16, 2024

Upcoming School Exams

Bihar Open Board 10th Examination

Application Date:11 November,2024 - 10 January,2025

Admit Card

Result

Bihar Open Board 12th Examination

Application Date:11 November,2024 - 10 January,2025

Admit Card

Result

National Means Cum-Merit Scholarship

Admit Card Date:27 November,2024 - 22 December,2024

Eligibility Criteria

Application Process

Exam Pattern

Gujarat Board Higher Secondary School Certificate

Late Fee Application Date:13 December,2024 - 22 December,2024

Exam Pattern

Result

Admit Card

Madhya Pradesh State Open School 10th Examination

Admit Card Date:13 December,2024 - 31 December,2024

Admit Card

Result

View All School Exams

Get answers from students and experts

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

More about NCERT Exemplar Class 11 Maths Solutions Chapter 15

NCERT Solutions for Class 11 Mathematics Chapters

Important Topic to Cover from NCERT Exemplar Class 11 Maths Solutions Chapter 15 Statistics:

NCERT Exemplar Class 11 Solutions

Frequently Asked Questions (FAQs)

Also Read

Articles

Upcoming School Exams

Bihar Open Board 10th Examination

Bihar Open Board 12th Examination

National Means Cum-Merit Scholarship

Gujarat Board Higher Secondary School Certificate

Madhya Pradesh State Open School 10th Examination

Popular Questions

Colleges After 12th

Popular Course After 12th

Applications for Admissions are open.

JEE Main Important Physics formulas

JEE Main Important Chemistry formulas

TOEFL ® Registrations 2024

Pearson | PTE

JEE Main high scoring chapters and topics

JEE Main Important Mathematics Formulas

Explore on Careers360

NCERT Solutions

NCERT Exemplars

NCERT Books

NCERT Notes

CBSE

CBSE Class 10

CBSE Class 12th

CISCE Board 10th

IMO

IEO

NSO

NMMS

Schools By Medium

By Ownership

By City

By State

NVS

JNVST

Download Careers360 App

All this at the convenience of your phone

Scan and download the app