Question 13 Marks
The following observations have been arranged in ascending order. If the median of the data is $78,$ find the value of $x.44, 47, 63, 65, x + 13, 87, 93, 99, 110.$
AnswerConsider the given data :
$44, 47, 63, 65, x + 13, 87, 93, 99, 110$
Here the number of observations is $9,$ which is odd.
Thus, the median of the given data is $\left(\frac{n+1}{2}\right)^{\text {th }}$ observation.
From the given data, $\left(\frac{9+1}{2}=5\right)^{\text {th }}$ observation is $x + 13$
Also, given that the median is $78.$
$x + 13 = 78$
$\Rightarrow x = 78 - 13$
$\Rightarrow x = 65$
View full question & answer→Question 23 Marks
Find the mean of $8, 12, 16, 22, 10$ and $4.$ Find the resulting mean, if each of the observations, given above, be: decreased by $40\%$
AnswerMean of the given data $=\frac{8+12+16+22+10+4}{6}$
$=\frac{72}{6}=12$
decreased by $40%$
New mean $=$ Original mean $- 40%$ of original mean
$\Rightarrow$ New mean $= 12 - 40%$ of $12$
$\Rightarrow$ New mean $=12-\frac{40}{100} \times 12$
$\Rightarrow$ New mean $=12-\frac{2}{5} \times 12$
$\Rightarrow$ New mean $= 12 - 0.4\times 12$
$\Rightarrow$ New mean $= 12 - 4.8$
$\Rightarrow$ New mean $= 7.2$
View full question & answer→Question 33 Marks
Find the mean of $8, 12, 16, 22, 10$ and $4$. Find the resulting mean, if the observations, given above, be: multipliedby $3$ and then divided by $2.$
AnswerMean of the given data $=\frac{8+12+16+22+10+4}{6}$
$=\frac{72}{6}=12$
multiplied by 3 and then divided by $2$
If $\bar{x}$ is the mean of n number of observation $x_1, x_2, x_3, ...., x_{n,}$
then mean of $\frac{a}{b} x_1, \frac{a}{b} x_2, \frac{a}{b} x_3, \ldots, \frac{a}{b} x_n$ is $\frac{a}{b} \bar{x}$
the mean is also multiplied by $\frac{3}{2}$.
The mean of the original data is $12.$
Hence, the new mean $=\frac{3}{2} \times 12=\frac{36}{2}=18$.
View full question & answer→Question 43 Marks
The average of n numbers $x_1, x_2, x_3…..x_n$ is $A.$ If $x_1$ is replaced by $( x+ \alpha )x_1, x_2,$ is replaced by $( x+ \alpha )x_2$ and so on.Find the new average.
AnswerMean of $n$ numbers $= A$
$\Rightarrow A =\frac{x_1+x_2+\ldots \ldots+x_n}{n}$
$\Rightarrow x_1 + x_2 + ............. + x_n$
$= n \times A ......(i)$
New Mean
$=\frac{(x+a) x_1+(x+a) x_2+\ldots \ldots .+(x+a) x_n}{n}$
$=\frac{(x+a)\left(x_1+x_2+\ldots \ldots \ldots+x_n\right)}{n}$
$=\frac{(x+a)(n \times A )}{n} .....$[ From$(i) ]$
$=\frac{(x+a) \times n \times A }{n}$
$=( x + a ) A $
View full question & answer→Question 53 Marks
Find the mean of $8, 12, 16, 22, 10$ and $4.$ Find the resulting mean, if the observations, given above, be: divided by $2.$
Answer Mean of the given data $=\frac{8+12+16+22+10+4}{6} $
$ =\frac{72}{6}=12$
divided by $2.$
If $\bar{x}$ is the mean of n number of observation $x_1, x_2, x_3, ....,x_n,$
then
mean of $\frac{x_1}{a}, \frac{x_2}{a}, \frac{x_3}{a}, \ldots,\frac{x_n}{a}$ is $\frac{\bar{x}}{a}$
Thus, when each of the given data is divided by $2,$
the mean is also divided by $3.$
The mean of the original data is $12.$
Hence, the new mean $= \frac{12}{2}=6$.
View full question & answer→Question 63 Marks
Find the mean of $8, 12, 16, 22, 10$ and $4.$ Find the resulting mean, if the observations, given above, be: multiplied by $3.$
AnswerMean of the given data $=\frac{8+12+16+22+10+4}{6}$
$=\frac{72}{6}=12$
multiplied by $3.$
If $\bar{x}$ is the mean of n number of observation $x_1, x_2, x_3, ....., x_n,$
then mean of $ax_1, ax_2, ax_3, ...., ax_n$ is $a \bar{x}$.
Thus, when each of the given data is multiplied by $3,$
the mean is also multiplied by $3.$
The mean of the original data is $12.$
Hence, the new mean $= 12 \times 3 = 36.$
View full question & answer→Question 73 Marks
Find the median of:$233, 173, 189, 208, 194, 204, 194, 185, 200$ and $220.$
AnswerFirstly arrange the numbers in ascending order
$173, 185, 189, 194, 194, 200, 204, 208, 220, 233$
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 $]$
$=\frac{1}{2}\left[\right.$ value of $\left(\frac{10}{2}\right)^{\text {th }}$ term $+$ value of $\left(\frac{10}{2}+1\right)^{\text {th }}$ term$]$
$=\frac{1}{2}\left[\right.5^{\text {th }}$ term $+6^{\text {th }}$ term $]$
$=\frac{1}{2}[200+194]$
$=\frac{1}{2}[394]$
$= 197$
Thus the Median is $197.$
View full question & answer→Question 83 Marks
Find the median of:$63, 17, 50, 9, 25, 43, 21, 50, 14$ and $34$
AnswerFirstly arrange the numbers in ascending order
$9, 14. 17, 21, 25, 34, 43, 50, 50, 63$
Now since $n = 10 ($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 $]$
$=\frac{1}{2}\left[\right.$ value of $\left(\frac{10}{2}\right)^{\text {th }}$ term + value of $\left(\frac{10}{2}+1\right)^{\text {th }}$ term $]$
$=\frac{1}{2}[25+34] $
$=\frac{1}{2}[59]$
$= 29.5$
Thus the median is $29.5.$
View full question & answer→Question 93 Marks
If the mean of observations $x, x + 2, x + 4, x + 6$ and $x + 8$ is $11,$find$:(i)$ The value of $x;(ii$) The mean of first three observations.
Answer$(i)$ Given that the mean of observations $x, x + 2, x + 4, x + 6$ and $x + 8$ is $11.$
Mean $= \frac{\text { Observations }}{n}$
$11=\frac{x+x+2+x+4+x+6+x+8}{5}$
$11=\frac{5 x+20}{5}$
$x=\frac{35}{5}$
$x=7$
$(ii)$The mean of first three observations are
$=\frac{x+x+2+x+4}{3}$
$=\frac{3 x+6}{3}$
$=\frac{3 \times 7+6}{3} .....[$ Since $x = 7 ]$
$=\frac{21+6}{3}$
$=9 .$
View full question & answer→Question 103 Marks
The mean of $5$ numbers is $18.$ If one number is excluded, the mean of the remaining number becomes $16.$ Find the excluded number.
AnswerGiven the mean of $5$ numbers is $18$
The total sum of 5 numbers
$= 18 \times 5$
$= 90$
On etimescluding an observation, the mean of remaining $4$ observation is $16$
$= 16 \times 4$
$= 64$
Therefore the sum of the remaining $4$ observations
total of 5 observations$-$total of $4$ observations
$= 90 - 64$
$= 26.$
View full question & answer→Question 113 Marks
In a series of tests, $A$ appeared for $8$ tests. Each test was marked out of $30$ and averages $25$. However, while checking his files, $A$ could only find $7$ of the $8$ tests. For these, he scored $29, 26, 18, 20, 27, 24$ and $29.$Determine how many marks he scored for the eighth test.
AnswerTotal number of tests $= 8$
The average score of $A = 25$
Let the score of the $8^{th}$ test be $x.$
Then, total score of $8$ tests $= 29 + 26 + 18 + 20 + 27 + 24 + 29 + x$
Now, we have
Mean $=\frac{\text { Total score of } 8 \text { tests }}{\text { Total number of tests }}$
$\Rightarrow 25=\frac{29+26+18+20+27+24+29+x}{8}$
$\Rightarrow 25 \times 8 = 173 + x$
$\Rightarrow x + 173 = 200$
$\Rightarrow x = 200 - 173$
$\Rightarrow x = 27$
Thus, A scored $27$ marks in the eights test.
View full question & answer→Question 123 Marks
The mean weight of $120$ students of a school is $52.75 \ kg.$ If the mean weight of $50$ of them is $51 \ kg,$
find the mean weight of the remaining students.
AnswerMean weight of $120$ students $= 52.75\ kg$
$⇒$ Sum of the weight of $120$ students $= 120 \times 52.75 = 6330 \ kg$
Mean weight of $50$ students $= 51\ kg$
$⇒$ Sum of the weight of 50 students $= 50 \times 51 = 2550\ kg$
$⇒$ Sum of the weight of remaining $( 120 - 50 ) = 70$ students
$=$ Sum of the weight of $120$ students$ - $Sum of the weight of $50$ students
$= ( 6330 - 2550 )\ kg$
$= 3780\ kg$
$⇒$ Mean weight of remaining $70$ students$ = \frac{3780}{70}=54\ kg$
View full question & answer→Question 133 Marks
The mean of $200$ items was $50.$ Later on, it was discovered that two items were misread as $92$ and $8$ instead of $192$ and $88.$Find the correct mean.
AnswerGiven that the mean of $200$ items was $50.$
Mean$ =\frac{\sum_x}{n} $
$\Rightarrow 50=\frac{\sum_x}{200}$
$\Rightarrow x=10,000$
Incorrect value of $\sum x=10,000$
correct value of
$\sum x=10,000-(92+8)+(192+88)$
$= 10,000 - 100 + 280$
$= 10,180$
Correct mean
$=\frac{\text { correct value of } \sum x}{n}$
$=\frac{10180}{200} $
$ =50.9$
View full question & answer→Question 143 Marks
The mean of $100$ observations is $40.$ It is found that an observation $53$ was misread as $83.$
Find the correct mean.
AnswerGiven the mean of $100$ observations is $40.$
$\frac{\sum x}{n}=\bar{x}$
$\Rightarrow \frac{\sum x}{n}=40$
$\Rightarrow x=40 \times 100$
$\Rightarrow x=4000$
Incorrect value of $x = 4000$
Correct value of $x =$ Incorrect value of $x -$ Incorrect observation $+$ correct observation
$= 4000 - 83 + 53$
$= 3970$
Correct mean $= \frac{\text { correct value of } \sum x}{n}$
$=\frac{3970}{100}$
$=39.7$
View full question & answer→