Question 14 Marks
The mean of $x, x+2, x+4, x+6$ and $x+8$ is $11,$ find the mean of the first three observations.
AnswerLet $\bar{x}$ be the mean of n number of observation $x_1, x_2, x_3, ...., x_n.$
Mean $=\frac{x_1+x_2+x_3+\ldots \ldots \ldots+x_n}{n}$
Therefore,
Mean of given data
$=\frac{x+x+2+x+4+x+6+x+8}{5}$
$ =\frac{5 x+20}{5}$
$= x + 4$
Also, it's given that mean of the given data is $11.$
$\Rightarrow x + 4 = 11$
$\Rightarrow x = 7$
Hence the mean of the first three observation $= \frac{x+x+2+x+4}{3}$
$=\frac{3 x+6}{3} $
$=x+2 $
$ =7+2$
$ =9 .$
View full question & answer→Question 24 Marks
The following observations have been arranged in ascending order. If the median of these observations is $58,$ find the value of $x.24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90.$
AnswerConsider the given data :
$24, 27, 43, 48, x - 1, x + 3, 68, 73, 80, 90.$
Here the number of observations is $10,$ which is even.
Thus, the median of the given data is
$\frac{1}{2}\left[\left(\frac{n}{2}\right)^{\text {th }}\right.$ term $+\left(\frac{n}{2}+1\right)^{\text {th }}$ term $]$.
From the given data, $\left(\frac{10}{2}=5\right)^{\text {th }}$ observation is $x-1$
and $\left(\frac{10}{2}+1=6\right)^{\text {th }}$ observation is $x+3$
Also, given that the median is $58.$
Thus, we have
$\frac{1}{2}[x-1+x+3]=116$
$\Rightarrow 2 x+2=116$
$ \Rightarrow 2 x=116-2$
$ \Rightarrow 2 x=114 $
$\Rightarrow x=\frac{114}{2} $
$ \Rightarrow x=57$
View full question & answer→Question 34 Marks
Out of $10$ students, who appeared in a test, three secured less than $30$ marks and $3$ secured more than $75$ marks. The marks secured by the remaining $4$ students are $35, 48, 66$ and $40$. Find the median score of the whole group.
AnswerHere, total observations $= n = 10 ($even$)$
Thus, we have
Median $=\frac{1}{2}\left[\right.$ value of $\left(\frac{10}{2}\right)^{\text {th }}$ term + value of $\left(\frac{10}{2}+1\right)^{\text {th }}$ term $]$
$=\frac{1}{2}\left[\right.$ valueof $5^{\text {th }}$ term + value of $6^{\text {th }}$ term $]$
According to the given information, data in ascending order is as follows:
| |
$1^{st}$Term |
$2^{nd}$Term |
$3^{rd}$Term |
$4^{th}$Term |
$5^{th}$Term |
$6^{th}$Term |
$7^{th}$Term |
$8^{th}$Term |
$9^{th}$Term |
$10^{th}$Term |
| Marks |
Less than $30$ |
$35$ |
$40$ |
$48$ |
$66$ |
More than $75$ |
$\therefore$ Median $=\frac{1}{2}(40+48)=\frac{88}{2}=44$
Hence, the median score of the whole group is $44.$ View full question & answer→Question 44 Marks
In $10$ numbers, arranged in increasing order, the $7^{\text {th }}$ number is increased by $8$ , how much will the median be changed?
AnswerFor any given set of data, the median is the value of its middle term.
Here, total observations $= n = 10 ($even$)$
If n is even, we have
Median $=\frac{1}{2}\left[\right.$ value of $\left(\frac{n}{2}\right)^{\text {th }}$ term $+$ value of $\left(\frac{n}{2}+1\right)^{\text {th }}$ term $]$
Thus, for $n = 10$, we have
Median = $\frac{1}{2}\left[\right.$ value of $\left(\frac{10}{2}\right)^{\text {th }}$ term $+$ value of $\left(\frac{10}{2}+1\right)^{\text {th }}$ term $]$
$=\frac{1}{2}\left[\right.$ value of $5^{\text {th }}$ term $+$ value of $6^{\text {th }}$ term $]$
Hence, if $7^{th}$ number is diminished by $8,$
there is no change in the median value.
View full question & answer→Question 54 Marks
The following data have been arranged in ascending order. If their median is $63,$ find the value of $x.34, 37, 53, 55, x, x + 2, 77, 83, 89$ and $100.$
AnswerGiven numbers are $34, 37, 53, 55, x, x+2, 77, 83, 89, 100$
Here $n = 10 ($even$)$
Median = $\frac{1}{2}\left[\right.$ value of $\left(\frac{n}{2}\right)^{\text {th }}$ term $+$ value of $\left(\frac{n}{2}+1\right)^{\text {th }}$ term $]$
$=\frac{1}{2}[$ value of $\left(\frac{10}{2}\right)^{\text {th }}$ term $+$ value of $\left(\frac{10}{2}+1\right)^{\text {th }}$ term $]$
$=\frac{1}{2}[$ value of $(5)^{\text {th }}$ term $+$ value of $(5+1)^{\text {th }}$ term$]$
$=\frac{1}{2}[$ value of $(5)^{\text {th }}$ term $+$ value of $(6)^{\text {th }}$ term $]$
$63=\frac{1}{2}[x+x+2] $
$ \Rightarrow \frac{2+2 x}{2}=63$
$ \Rightarrow x+1=63 $
$ \Rightarrow x=62$
View full question & answer→Question 64 Marks
If different values of variable $x$ are $9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5$ and $11.1$;find;$(i)$ the mean $\bar{x},(ii)$ the value of $\sum\left(x_i-\bar{x}\right)$
Answer$(i)$ The given numbers are$ 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5, 11.1$
$\bar{x}=\frac{x 1+x 2+x 3+x 4+x 5+\ldots \ldots .+x n}{n}$
$=\frac{9.8+5.4+3.7+1.7+1.8+2.6+2.8+8.6+10.5+11.1}{10}$
$= 5.8$
$(ii)$ The value of $\sum_{i=1}^{10}\left(x_i-\bar{x}\right)$
We know that
$\sum_{i=1}^n\left(x_i-\bar{x}\right)=(x 1-\bar{x})+(x 2-\bar{x}) \ldots \ldots \ldots+(x n-\bar{x})=0$
$\bar{x} = 5.8$
Therefore
$\sum_{i=1}^{10}\left(x_i-\bar{x}\right)$
$= ( 9.8 - 5.8) + ( 5.4 - 5.8 ) + (3.7 - 5.8 ) + ( 1.7 - 5.8 ) + ( 1.8 - 5.8 ) +$$ (2.6 - 5.8 ) + (2.8 - 5.8 ) + ( 8.6 - 5.8 ) + ( 10.5 - 5.8 ) + ( 11.1 - 5.8 )$
$= 4 - 4 - 2.1 - 4.1 - 4 - 3.2 - 3 + 2.8 + 4.7 + 5.3$
$=0$
View full question & answer→Question 74 Marks
The mean marks $($out of $100)$ of boys and girls in an examination are $70$ and $73$ respectively. If the mean marks of all the students in that examination are $71,$find the ratio of the number of boys to the number of girls.
AnswerLet the number of boys and girls be $x$ and $y$ respectively.
Now,
Given, Mean marks of $x$ boys in the examination $= 70$
$\Rightarrow $Sum of marks of $x$ boys in the examination $= 70x$
Given, Mean marks of $y$ girls in the examination $= 73$
$\Rightarrow $ Sum of marks of $y$ girls in the examination $= 73y$
Given, Mean marks of all students $( x + y )$ in the examination $= 71$
$\Rightarrow $ Sum of marks of all students $( x + y )$students in examination $= 71( x + y )$
Now, the Sum of marks of all students $( x + y )$ students in the examination
$\Rightarrow$ Sum of marks of $x$ boys in the examination $+$ Sum of marks of $y$ girls in the examination
$\Rightarrow 71( x + y ) = 70x + 73y$
$\Rightarrow 71x + 71y = 70x + 73y$
$\Rightarrow x = 2y$
$\Rightarrow \frac{x}{y}=\frac{2}{1}$
$\Rightarrow x: y = 2 : 1$
Thus, the ratio of the number of boys to the number of girls is $2 : 1.$
View full question & answer→Question 84 Marks
Find the mean and median of the data $: 35, 48, 92, 76, 64, 52, 51, 63$ and $71.$If $51$ is replaced by $66,$ what will be the new median?
AnswerLet $\bar{x}$ be the mean of n number of observation $x_1, x_2, x_3, \dots...., x_n.$
Mean $=\frac{x_1+x_2+x_3+\ldots \ldots \ldots+x_n}{n}$
Therefore$,$
Mean of given data $=\frac{35+48+92+76+64+52+51+63+71}{9}$
$=\frac{552}{9}$
$=61.33$
Let us rewrite the given data in ascending order :
Thus, we have
$35, 48, 51, 52, 63, 64, 71, 76, 92$
There are $9$ observations, which is odd.
Therefore, median $= \left(\frac{n+1}{2}\right)^{\text {th }}$ Observation
$\Rightarrow$ Median $=\left(\frac{9+1}{2}\right)^{\text {th }}$ Observation
$\Rightarrow$ Median $=\left(\frac{10}{2}\right)^{\text {th }}$ Observation
$\Rightarrow$ Median $= 5^{th}$ Observation
$\Rightarrow$ Median $= 63.$
If $51$ is replaced by $66,$ the new set of data in ascending order is $: 35, 48, 52, 63, 64, 66, 71, 76, 92$
Since median $= 5^{th}$ observation$,$
We have a new median $= 64.$
View full question & answer→Question 94 Marks
Find the mean of the following data $: 30, 32, 24, 34, 26, 28, 30, 35, 33, 25(i)$ Show that the sum of the deviations of all the given observations from the mean is zero.$(ii)$ Find the median of the given data.
AnswerLet $\bar{x}$ be the mean of $n$ number of observation $x_1, x_2, x_3, \dots...., x_n$
Mean $=\frac{x_1+x_2+x_3+\ldots \ldots+x_n}{n}$
Therefore,
Mean of given data $= \frac{30+32+24+34+26+28+30+35+33+25}{10}$
$=\frac{297}{10}$
$=29.7$
$(i)$ Let us tabulate the observations and their deviations from the mean
Observation
$x_i$ |
Deviation
$x_i-\bar{x}$ |
| $30$ |
$0.3$ |
| $32$ |
$2.3$ |
| $24$ |
$-5.7$ |
| $34$ |
$4.3$ |
| $26$ |
$-3.7$ |
| $28$ |
$-1.7$ |
| $30$ |
$0.3$ |
| $35$ |
$5.3$ |
| $33$ |
$3.3$ |
| $25$ |
$-4.7$ |
| Total |
$0$ |
From the table, it is clear that the sum of the deviations from the mean is Zero.
$(ii)$ Consider the given data :
$30, 32, 24, 34, 26, 28, 30, 35, 33, 25.$
Let us rewrite the above data in ascending order
$24, 25, 26, 28, 30, 30, 32, 33, 34, 35.$
There are $10$ observations, which is even.
Therefore,
Median $=\frac{1}{2}[(\frac{n}{2})^{\text {th }}$ term $+(\frac{n}{2}+1)^\text {th }$ term $]$
$=\frac{1}{2}[(\frac{10}{2})^{\text {th }}$term$+(\frac{10}{2}+1)^{\text {th }}$term$]$
$=\frac{1}{2}[(5)^{\text {th }}$term$+(\frac{10}{2}+1)^{\text {th }}$term$]$
$=\frac{1}{2}[5^{\text {th }}$term$+(5+1)^{\text {th }}$term$]$
$=\frac{1}{2}[5^{\text {th }}$term$+6^{\text {th }}$term$]$
$=\frac{1}{2}[30+30]$
$=\frac{1}{2}[60]$
$=30 .$ View full question & answer→Question 104 Marks
The mean of $10$ numbers is $24.$ If one number is included$,$ the new mean is $25.$ Find the included number.
AnswerLet $\bar{x}$ be the mean of $n$ number of observation $x_1, x_2, x_3, \dots...., x_n$
Mean of given data $=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$
Given that mean of $10$ numbers is $24.$
That is$,$
$\frac{x_1+x_2+x_3+\ldots+x_{10}}{10}=24$
$\Rightarrow x_1 + x_2 + x_3 +\dots ... + x_{10} = 10 \times 24$
$\Rightarrow x_1 + x_2 + x_3 +\dots ... + x_{10} = 240$
$\Rightarrow x_1 + x_2 + x_3 +\dots ... + x_{10} + x_{11} = 240 + x_{11}\dots ....(1)$
Also$,$ given that mean of $11$ number is $25.$
That is$,$
$\frac{x_1+x_2+x_3+\ldots+x_{10}+x_{11}}{11}=25$
$\Rightarrow x_1 + x_2 + x_3 +\dots ... + x_{10} + x_{11} = 11 \times 25$
$\Rightarrow x_1 + x_2 + x_3 +\dots ... + x_{10} + x_{11} = 275 \dots....( 2 )$
From equations $( 1 )$ and $( 2 ),$ we have :
$x_1 + x_2 + x_3 +\dots ... + x_{10} + x_{11} = 240 + x_{11} = 275$
$240 + x_{11} = 275$
$\Rightarrow x_{11} = 275 - 240 = 35$
View full question & answer→Question 114 Marks
The mean of $6$ numbers is $42.$ If one number is excluded$,$ the mean of the remaining number is $45.$ Find the excluded number.
AnswerLet $\bar{x}$ be the mean of $n$ number of observation $x_1, x_2, x_3,\dots ...., x_n$
Mean of given data $=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$
Given that mean of $6$ number is $42.$
That is$,$
$\frac{x_1+x_2+x_3+\ldots+x_6}{6}=42$
$\Rightarrow x_1 + x_2 + x_3 +\dots ... + x_6 = 6 \times 42$
$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 252 - x_6\dots...( 1 )$
Also$,$ given that the mean of $5$ number is $45.$
That is$,$
$\frac{x_1+x_2+x_3+x_4+x_5}{5}=45$
$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 5 \times 45$
$\Rightarrow x_1 + x_2 + x_3 + x_4 + x_5 = 225\dots ....( 2 )$
From equation $( 1 )$ and $( 2 ),$ we have
$x_1 + x_2 + x_3 + x_4 + x_5 = 252 - x_6 = x_1 + x_2 +x_3 + x_4 + x_5 = 225$
$252 - x_6 = 225$
$\Rightarrow x_6 = 252 - 225 = 27$
View full question & answer→Question 124 Marks
The heights $($in $\ cm)$ of the volleyball players from team $A$ and team $B$ were recorded as:Team $A :180, 178, 176, 181, 190, 175, 187$Team $B :174, 175, 190, 179, 178, 185, 177$Which team had a greater average height?Find the median of team $A$ and team $B$.
AnswerTotal number of players in each team $= 7$
Mean height of team $A =\frac{\text { Sum of the height of players of the team A }}{\text { Total number of team A players }}$
$=\frac{180+178+176+181+190+175+187}{7}$
$=\frac{1267}{7}$
$= 181 \ cm$
Mean height of team $B =\frac{\text { Sum of the height of players of the team B }}{\text { Total number of team B players }}$
$=\frac{174+175+190+179+178+185+177}{7}$
$=\frac{1258}{7}$
$= 179.7 \ cm$
Thus, team $A$ has greater average height.
Median of team $A$:
Arranging heights in ascending order, we get
$175, 176, 178, 180, 181, 187, 190$
Total number of observations $= n = 7 ($odd$)$
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$
Observation $=\left(\frac{7+1}{2}\right)^{\text {th }}$
Observation $=4^{\text {th }}$
Observation $=180 \ cm$
Median of team $B$ :
Arranging heights in ascending order, we get
$174, 175, 177, 178, 179, 185, 190$
Total number of observations $= n = 7 ($odd$)$
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$
Observation $=\left(\frac{7+1}{2}\right)^{\text {th }}$
Observation $=4^{\text {th }}$
Observation $=178 \ cm$
View full question & answer→Question 134 Marks
The mean weight of $60$ students in a class is $40 \ kg$. The mean weight of boys is $50 \ kg$ while that of girls is $30 \ kg$. Find the number of boys and girls in the class.
AnswerTotal number of students $= 60$
Mean weight of 60 students = 40
Let the number of boys$ = x$
Then, the number of girls $= 60 - x$
Mean weight of boys $=\frac{\text { Total weight of boys }}{\text { Total number of boys }}$
$\Rightarrow 50=\frac{\text { Total weight of boys }}{x}$
$\Rightarrow $ Total weight of boys $= 50x$
Mean weight of girls $=\frac{\text { Total weight of girls }}{\text { Total number of girls }}$
$\Rightarrow 30=\frac{\text { Total weight of girls }}{60-x}$
$\Rightarrow $ Total weight of girls $= 30( 60 - x )$
Now,
Mean weight of $60$ students $=\frac{\text { Total weight of boys }+ \text { Total weight of girls }}{\text { Total number of students }}$
$\Rightarrow 40=\frac{50 x+30(60-x)}{60}$
$\Rightarrow 2400 = 50x + 1800 - 30x$
$\Rightarrow 20x = 600$
$\Rightarrow x = 30$
$\Rightarrow 60 - x = 60 - 30 = 30$
Hence, the number of boys is $30$ and the number of girls is also $30.$
View full question & answer→Question 144 Marks
Find the mean and median of all the positive factors of $72.$
AnswerLet us find the factors of $72:$
$72 = 1 \times 72$
$= 2 \times 36$
$= 3 \times 24$
$= 4 \times 18$
$= 6 \times 12$
$= 8 \times 9$
$= 9 \times 8$
$= 12 \times 6$
$= 18 \times 4$
$= 24 \times 3$
$= 36 \times 2$
$= 72 \times 1$
Therefore, the data set is: $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$
Mean of the above data set $=\frac{1+2+3+4+6+8+9+12+18+24+36+72}{12}$
$=\frac{195}{12}$
$=16.25$
Since the number of observation is $12,$ which is even,
median is given by
Median$=\frac{1}{2}[\left(\frac{n}{2}\right)^{\text {th }}$ term$+\left(\frac{n}{2}+1\right)^{\text {th }}$ term$]$
$=\frac{1}{2}[\left(\frac{12}{2}\right)^{\text {th }}$ term$+\left(\frac{12}{2}+1\right)^{\text {th }}$ term$]$
$=\frac{1}{2}[6^{\text {th }}$ term$+7^{\text {th }}$ term$]$
$=\frac{1}{2}[8+9]$
$=\frac{1}{2} \times 17$
$=8.5$
View full question & answer→