Q: If the mean of first \$n\$ natural numbers is \$\displaystyle\frac{5n}{9}\$, then \$n = \$

It is given that the mean of first \$n\$ natural numbers is \$\dfrac {5n}{9}\$.Therefore, \$\displaystyle\frac{1+2+3+...+n}{n} = \displaystyle\frac{5n}{9} \$\$\Rightarrow \displaystyle\frac{n+1}{2} = \displaystyle\frac{5n}{9}\$\$\Rightarrow 9n+9=10n\$\$ \Rightarrow n = 9\$

Q: Height of students of Class \$X\$ are given in the following frequency distribution:Height (in cm)\$150-155\$\$155-160\$\$160-165\$\$165-170\$\$170-175\$Number of students\$15\$\$8\$\$20\$\$12\$\$5\$Find the modal height. Also, find the mean height, Compare and interpret the two measures of central tendency.

Here:Maximum class frequency \$=20\$The class corresponding to Maximum class frequency is the modal class.So, Modal class \$=160-165\$Lower limit \$=L=160\$Modal class size \$=h=5\$Frequency of class preceding the modal class \$=f0=8\$Frequency of the modal class \$=f1=20\$Frequency of class succeeding the modal \$=f2=12\$Mode formula is given by,Mode \$=L+\left( \dfrac{f_1 -f_0}{2f_1 - f_0 -f_2}\right)\times h\$Mode \$=160+\left( \dfrac{20-8}{2(20)-8-12}\right)\times 12\$\$=160+12/20\times 5\$\$=163\$The modal height is \$163\ cm\$, which implies maximum number of students have model height \$163\ cm\$HeightMid Point \$(x_i)\$No. of students \$(f_i)\$\$f_i x_i\$\$150-155\$\$152.5\$ \$15\$\$2287.5\$\$155-160\$ \$157.5\$ \$8\$ \$1260\$\$160-165\$\$162.5\$\$20\$\$3250\$\$165-170\$\$167.5\$\$12\$\$2010\$\$170-175\$\$172.5\$\$5\$\$862.5\$\$\sum{fi}=60\$ \$\sum {fixi}=9670\$We know,\$\bar x =\dfrac{\sum_{i}f_i x_i}{\sum_{i}f_1}\$\$=9670/60\$\$=161.17\$Mean is \$161.17\$Thus, \$161.7\ cm\$ is the average height of all the students.

Q: If \$\bar{x}\$ represents the mean of n observations \$x_1, x_2, ..., x_n\$ then value of \$\sum_{i=1}^{n}(x_{i}-\bar{x})\$is :

\$\bar{x}\$ is the average of all the observations. If we draw all the numbers and the mean on a number line, then half of the numbers will lie on the right side of the mean and other half on the left side. Hence, \$(x_i - \bar{x})\$ will give equal number of positive and negative terms whose sum will be \$0\$.

Q: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

Sum of the present ages of husband, wife and child \$= (27 \times 3 + 3 \times 3)\$ years = \$90\$ years.Sum of the present ages of wife and child \$= (20 \times 2 + 5 \times 2)\$ years = 50 years.\$\therefore\$ Husband's present age \$= (90 - 50)\$ years \$= 40\$ years.

Q: Find the mean of first \$n\$ natural numbers.

\$ {\textbf{Step-1: Find sum of n natural numbers}} \$ \$ {\text{First n natural numbers = 1, 2, 3, 4,5}} \ldots \ldots \ldots \ldots \ldots {\text{. n}} \$ \$ {\text{Sum of n natural numbers = }}\left( {{\text{1 + 2 + 3 +}} \ldots \ldots + n} \right){\text { ...}}\left( {{\text{1}}} \right) \$ \$ {\text{= }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}\;{{\text{ ...}}\;{\text{(2)}}} \$ \$ {\textbf{Step-2: Find mean}} \$ \$ {\text{Total no}}{\text{. of observations = n}} \$ \$ {\text{Now, we know that}} \$ \$ {\text{Mean = }}\dfrac{{{\text{Sum of all numbers}}}}{{{\text{No}}{\text{. of observation}}}} \$ \$ {\text{Putting the values from equation (1) and (2), we get:}} \$ \$ {\text{Mean = }}\dfrac{{{\text{1 + 2 + 3 + }}.....{\text{ + n}}}}{{\text{n}}} \$ \$ = \dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{{\text{2}} \times {\text{n}}}} \$ \$ = \dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}} \$ \$ {\text{Mean of first n natural numbers = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}} \$ \$ {\textbf{Hence mean of first n natural numbers = }}\mathbf{\dfrac{{{\text{n}}\left( {{\text{n + 1}}} \right)}}{{\text{2}}}} \$

Q: If each observation of the data is increased by \$5,\$ then their mean :

Let there be \$n\$ terms.Increasing every term by \$5\$ will increase their sum by \$5n.\$Increase in Mean= \$\cfrac{5n}{n}\$= \$5\$

Q: If the mean of \$10, 12, 18, 13, P\$ and \$17\$ is \$15\$, find the value of P

Given the mean of \$10\$ ,\$12\$ , \$18\$ , \$13\$ , \$P\$ and \$17\$ is \$15\$\$Mean\$ =\$\dfrac{\Sigma x}{n}\$\$15=\dfrac{10+12+18+13+P+17}{6}\$\$90=70+P\$\$P=90-70=20cm\$

Q: Find the mean of children per family from data given below:No. of children\$0\$\$1\$\$2\$\$3\$\$4\$\$5\$No. of families\$5\$\$11\$\$25\$\$12\$\$5\$\$2\$

No. of children \$x_i\$No. of families \$f_i\$\$f_i x_i\$\$0\$\$5\$\$0\$\$1\$\$11\$\$11\$\$2\$\$25\$\$50\$\$3\$\$12\$\$36\$\$4\$\$5\$\$20\$\$5\$\$2\$\$10\$Total\$\displaystyle \sum f_i =60\$\$\displaystyle \sum f_i x_i =127\$Mean \$=\dfrac {\displaystyle \sum f_i x_i}{\displaystyle \sum f_i} =\dfrac {127}{60}=2.12\$ (approx).

Q: An aircraft has 120 passenger seats. The number of seats occupied during 100 fights is given below :No. of seats100-104104-108108-112112-116116-120Frequency1520321815Determine the mean of seats occupied over the flights

Class intervalClass Marks \$(x_i)\$Frequency \$(f_i)\$Deviation \$(d_i=x_i-a)\$\$f_id_i\$ 100-104 102 15 -8 -120 104-108 106 20 -4 -80 108-112 110 32 0 0 112-116 114 18 4 72 116-120 118 15 8 120 \$N=\sum f_i=100\$ \$\sum f_id_i=-8\$\$\therefore\$ Assumed mean, \$a=110\$Class width \$h=4\$and total observation, \$N=100\$Hence, finding mean\$Mean(\bar{x})=a+\dfrac{\sum f_1d_1}{\sum f_1}\$\$=110+(-8/100)\$\$=110-0.08\$\$=109.92\$But we know that the seats cannot be in decimalTherefore, the no, of seats \$=109\$.

Q: The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

Required average = \$\begin{pmatrix}\dfrac{50.25\times 16+45.15\times 8}{16+8}\end{pmatrix}\$= \$\begin{pmatrix}\dfrac{804+361.20}{24}\end{pmatrix}\$= \$\dfrac{1165.20}{24}\$= 48.55

Question 1

If the mean of first $n$ natural numbers is $\displaystyle\frac{5n}{9}$, then $n = $

Accepted Answer

It is given that the mean of first $n$ natural numbers is $\dfrac {5n}{9}$.Therefore, $\displaystyle\frac{1+2+3+...+n}{n} = \displaystyle\frac{5n}{9} $$\Rightarrow \displaystyle\frac{n+1}{2} = \displaystyle\frac{5n}{9}$$\Rightarrow 9n+9=10n$$ \Rightarrow n = 9$

Question 2

Height of students of Class $X$ are given in the following frequency distribution:

Height (in cm) $150 - 155$ $155 - 160$ $160 - 165$ $165 - 170$ $170 - 175$
Number of students $15$ $8$ $20$ $12$ $5$
Find the modal height. Also, find the mean height, Compare and interpret the two measures of central tendency.

Accepted Answer

Here:Maximum class frequency $=20$The class corresponding to Maximum class frequency is the modal class.So, Modal class $=160-165$Lower limit $=L=160$Modal class size $=h=5$Frequency of class preceding the modal class $=f0=8$Frequency of the modal class $=f1=20$Frequency of class succeeding the modal $=f2=12$Mode formula is given by,Mode $=L+\left( \dfrac{f_1 -f_0}{2f_1 - f_0 -f_2}\right)\times h$Mode $=160+\left( \dfrac{20-8}{2(20)-8-12}\right)\times 12$$=160+12/20\times 5$$=163$The modal height is $163\ cm$, which implies maximum number of students have model  height $163\ cm$HeightMid Point $(x_i)$No. of students $(f_i)$$f_i x_i$$150-155$$152.5$ $15$$2287.5$$155-160$          $157.5$                   $8$                    $1260$$160-165$$162.5$$20$$3250$$165-170$$167.5$$12$$2010$$170-175$$172.5$$5$$862.5$$\sum{fi}=60$                                                     $\sum {fixi}=9670$We know,$\bar x =\dfrac{\sum_{i}f_i x_i}{\sum_{i}f_1}$$=9670/60$$=161.17$Mean is $161.17$Thus, $161.7\ cm$ is the average height of all the students.

Question 3

If $\bar{x}$ represents the mean of n observations $x_1, x_2, ..., x_n$ then value of $\sum_{i=1}^{n}(x_{i}-\bar{x})$is :

Accepted Answer

$\bar{x}$ is the average of all the observations. If we draw all the numbers and the mean on a number line, then half of the numbers will lie on the right side of the mean and other half on the left side. Hence, $(x_i - \bar{x})$ will give equal number of positive and negative terms whose sum will be $0$.

Question 4

The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

Accepted Answer

Sum of the present ages of husband, wife and child $= (27 \times 3 + 3 \times 3)$ years = $90$ years.Sum of the present ages of wife and child $= (20 \times 2 + 5 \times 2)$ years = 50 years.$\therefore$ Husband's present age $= (90 - 50)$ years $= 40$ years.

Question 5

Find the mean of first $n$ natural numbers.

Accepted Answer

$  {\textbf{Step-1: Find sum of n natural&#10;numbers}} $&#10;&#10;                 $  {\text{First n natural numbers = 1, 2, 3,&#10;4,5}}  \ldots  \ldots &#10;\ldots  \ldots  \ldots {\text{. n}} $&#10;&#10;                 $  {\text{Sum of n natural numbers = }}\left(&#10;{{\text{1 + 2 + 3 +}}  \ldots  \ldots &#10;+ n} \right){\text {     ...}}\left( {{\text{1}}} \right) $&#10;&#10;                 $  {\text{= }}\dfrac{{{\text{n}}\left(&#10;{{\text{n + 1}}} \right)}}{{\text{2}}}\;{{\text{  ...}}\;{\text{(2)}}}  $&#10;&#10;$  {\textbf{Step-2: Find mean}} $&#10;&#10;                 $  {\text{Total no}}{\text{. of&#10;observations  =  n}} $&#10;&#10;                 $  {\text{Now, we know that}} $&#10;&#10;                 $  {\text{Mean = }}\dfrac{{{\text{Sum of all&#10;numbers}}}}{{{\text{No}}{\text{. of observation}}}} $&#10;&#10;                 $  {\text{Putting the values from equation (1) and (2), we get:}} $&#10;&#10;                 $  {\text{Mean = }}\dfrac{{{\text{1 + 2 + 3 +&#10;}}.....{\text{ + n}}}}{{\text{n}}} $&#10;&#10;                 $   = \dfrac{{{\text{n}}\left( {{\text{n + 1}}}&#10;\right)}}{{{\text{2}} \times {\text{n}}}} $&#10;&#10;                 $   = \dfrac{{{\text{n}}\left( {{\text{n + 1}}}&#10;\right)}}{{\text{2}}} $&#10;&#10;                 $  {\text{Mean of first n natural numbers  = }}\dfrac{{{\text{n}}\left( {{\text{n + 1}}}&#10;\right)}}{{\text{2}}} $&#10;&#10;$  {\textbf{Hence mean of first n&#10;natural numbers  = }}\mathbf{\dfrac{{{\text{n}}\left(&#10;{{\text{n + 1}}} \right)}}{{\text{2}}}} $

Question 6

If each observation of the data is increased by $5,$ then their mean :

Accepted Answer

Let there be $n$ terms.Increasing every term by $5$ will increase their sum by $5n.$Increase in Mean= $\cfrac{5n}{n}$= $5$

Question 7

If the mean of $10, 12, 18, 13, P$ and $17$ is $15$, find the value of P

Accepted Answer

Given the mean of $10$ ,$12$ , $18$ , $13$ , $P$ and $17$ is $15$$Mean$ =$\dfrac{\Sigma x}{n}$$15=\dfrac{10+12+18+13+P+17}{6}$$90=70+P$$P=90-70=20cm$

Question 8

Find the mean of children per family from data given below:No. of children$0$$1$$2$$3$$4$$5$No. of families$5$$11$$25$$12$$5$$2$

Accepted Answer

No. of children $x_i$No. of families $f_i$$f_i x_i$$0$$5$$0$$1$$11$$11$$2$$25$$50$$3$$12$$36$$4$$5$$20$$5$$2$$10$Total$\displaystyle \sum f_i =60$$\displaystyle \sum f_i x_i =127$Mean $=\dfrac {\displaystyle \sum f_i x_i}{\displaystyle \sum f_i} =\dfrac {127}{60}=2.12$ (approx).

Question 9

An aircraft has 120 passenger seats. The number of seats occupied during 100 fights is given below :

No. of seats 100-104 104-108 108-112 112-116 116-120
Frequency 15 20 32 18 15

Determine the mean of seats occupied over the flights

Accepted Answer

Class intervalClass Marks $(x_i)$Frequency $(f_i)$Deviation $(d_i=x_i-a)$$f_id_i$  100-104 102 15 -8 -120 104-108 106 20 -4 -80 108-112 110 32 0 0 112-116 114 18 4 72 116-120 118 15 8 120   $N=\sum f_i=100$  $\sum f_id_i=-8$$\therefore$ Assumed mean, $a=110$Class width $h=4$and total observation, $N=100$Hence, finding mean$Mean(\bar{x})=a+\dfrac{\sum f_1d_1}{\sum f_1}$$=110+(-8/100)$$=110-0.08$$=109.92$But we know that the seats cannot be in decimalTherefore, the no, of seats $=109$.

Question 10

The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

Accepted Answer

Required average = $\begin{pmatrix}\dfrac{50.25\times 16+45.15\times 8}{16+8}\end{pmatrix}$= $\begin{pmatrix}\dfrac{804+361.20}{24}\end{pmatrix}$= $\dfrac{1165.20}{24}$= 48.55

Arithmetic Mean - Special Property and Weighted Mean

Language

Rate

REVISE WITH CONCEPTS

Arithmetic Mean of Infinite Series and Weighted AM

Arithmetic Mean : Grouped Data

QUICK SUMMARY WITH STORIES

Weighted Mean

Mean of grouped data by Direct method

Mean of grouped data by Assumed mean method

Step deviation method for Grouped Data

If the mean of first $n$ natural numbers is $\frac{5 n}{9}$ , then $n =$

If $\overset{x}{ˉ}$ represents the mean of n observations $x_{1}, x_{2}, . . ., x_{n}$ then value of $\sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})$ is :

The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

Find the mean of first $n$ natural numbers.

If each observation of the data is increased by $5,$ then their mean :

If the mean of $10, 12, 18, 13, P$ and $17$ is $15$ , find the value of P

Find the mean of children per family from data given below:

No. of children $0$ $1$ $2$ $3$ $4$ $5$
No. of families $5$ $11$ $25$ $12$ $5$ $2$

An aircraft has 120 passenger seats. The number of seats occupied during 100 fights is given below :

No. of seats 100-104 104-108 108-112 112-116 116-120
Frequency 15 20 32 18 15

Determine the mean of seats occupied over the flights

The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

Height (in cm)	$150 - 155$	$155 - 160$	$160 - 165$	$165 - 170$	$170 - 175$
Number of students	$15$	$8$	$20$	$12$	$5$

No. of children	$0$	$1$	$2$	$3$	$4$	$5$
No. of families	$5$	$11$	$25$	$12$	$5$	$2$

No. of seats	100-104	104-108	108-112	112-116	116-120
Frequency	15	20	32	18	15

Arithmetic Mean - Special Property and Weighted Mean

Language

Rate

REVISE WITH CONCEPTS

Arithmetic Mean of Infinite Series and Weighted AM

Arithmetic Mean : Grouped Data

QUICK SUMMARY WITH STORIES

Weighted Mean

Mean of grouped data by Direct method

Mean of grouped data by Assumed mean method

Step deviation method for Grouped Data

If the mean of first n natural numbers is 95n​, then n=

Height of students of Class X are given in the following frequency distribution: Height (in cm)150−155155−160160−165165−170170−175Number of students15820125Find the modal height. Also, find the mean height, Compare and interpret the two measures of central tendency.

If xˉ represents the mean of n observations x1​,x2​,...,xn​ then value of ∑i=1n​(xi​−xˉ)is :

The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:

Find the mean of first n natural numbers.

If each observation of the data is increased by 5, then their mean :

If the mean of 10,12,18,13,P and 17 is 15, find the value of P

Find the mean of children per family from data given below: No. of children012345No. of families511251252

An aircraft has 120 passenger seats. The number of seats occupied during 100 fights is given below : No. of seats100-104104-108108-112112-116116-120Frequency1520321815 Determine the mean of seats occupied over the flights

The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.

If the mean of first $n$ natural numbers is $\frac{5 n}{9}$ , then $n =$

If $\overset{x}{ˉ}$ represents the mean of n observations $x_{1}, x_{2}, . . ., x_{n}$ then value of $\sum_{i = 1}^{n} (x_{i} - \overset{x}{ˉ})$ is :

Find the mean of first $n$ natural numbers.

If each observation of the data is increased by $5,$ then their mean :

If the mean of $10, 12, 18, 13, P$ and $17$ is $15$ , find the value of P

Find the mean of children per family from data given below:

No. of children $0$ $1$ $2$ $3$ $4$ $5$
No. of families $5$ $11$ $25$ $12$ $5$ $2$

An aircraft has 120 passenger seats. The number of seats occupied during 100 fights is given below :

No. of seats 100-104 104-108 108-112 112-116 116-120
Frequency 15 20 32 18 15

Determine the mean of seats occupied over the flights