Question 13 Marks
Numbers $50, 42, 35, 2x + 10, 2x - 8, 12, 11, 8$ are written in descending order and their median is $25$, find $x$.
AnswerGiven the number of observation, $n = 8$
$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{\frac{\text{8}^\text{th}}{2}\text{value}+\Big(\frac{\text{8}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{4^{\text{th}}\text{value}+5^{\text{th}}\text{value}}{2}$
$=\frac{\text{2x}+10+\text{2x}-8}{2}$
$=\text{2x}+1$
Given Median $= 25$
$\therefore\text{2x}+1=25$
$\Rightarrow\text{2x}=24$
$\Rightarrow\text{x}=12$
View full question & answer→Question 23 Marks
Find the median of the data: $133, 73, 89, 108, 94, 104, 94, 85, 100, 120$
AnswerNumbers are $133, 73, 89, 108, 94, 104, 94, 85, 100, 120$
Arranging the numbers in ascending order $73, 85, 89, 94, 94, 100, 104, 108, 120, 133$ $n = 10$ (even)
$\therefore\text{Median}=\frac{\frac{\text{n}}{2}\text{th value}+\Big(\frac{\text{n}}{2}+1\Big)\text{th value}}{2}$
$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$
$=\frac{94+100}{2}$
$=\frac{194}{2}=97$
View full question & answer→Question 33 Marks
Find the median of the following observations: $46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33$. If $92$ is replaced by $99$ and $41$ by $43$ in the above data, find the new median?
AnswerGiven the numbers are $46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33$
Arranging the numbers in ascending order $33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92$
$n = 11$ (odd)
$\therefore\text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{11}+1}{2}\Big)^{\text{th}}\text{value}$
$=6^{\text{th}}\text{value}=58$
If $92$ is replaced by $99$ and $41$ by $43$ Then the new values are: $33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99$ $n = 11$ (odd)
$\therefore\text{New Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{\text{11}+1}{2}\Big)^{\text{th}}\text{value}$
$=6^{\text{th}}\text{value}=58$
View full question & answer→Question 43 Marks
Find the sum of the deviations of the variate values $3, 4, 6, 7, 8, 14$ from their mean.
AnswerValues $3, 4, 6, 7, 8, 14$
$\therefore\text{Mean}=\frac{\text{Sum of numbers}}{\text{Total numbers}}$
$\therefore\text{Mean}=\frac{3+4+6+7+8+14}{6}$
$\therefore\text{Mean}=\frac{42}{6}$
$=7$
$\therefore$ Sum of deviation of values from their mean
$= (3 - 7) + (4 - 7) + (6 - 7) + (7 - 7) + (8 - 7) + (14 - 7)$
$= - 4 - 3 - 1 + 0 + 1 + 7$
$= - 8 + 8 = 0$
View full question & answer→Question 53 Marks
Find the median of the data: $15, 6, 16, 8, 22, 21, 9, 18, 25$
AnswerNumbers are $15, 6, 16, 8, 22, 21, 9, 18, 25$
Arranging the numbers in ascending order $6, 8, 9, 15, 16, 21, 22, 25$
$n = 9$ (odd)
$\therefore\text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{value}$
$=5^\text{th}\text{value}=16$
View full question & answer→Question 63 Marks
The mean weight per student in a group of $7$ students is $55\ kg$. The individual weights of $6$ of them (in kg) are $52, 54, 55, 53, 56$ and $54$. Find the weight of the seventh student.
AnswerThe mean weight per student in a group of $7$ students $= 55\ kg$.
Weight of $6$ students (in kg) $= 52, 54, 55, 53, 56$ and $54$
Let the weight of seventh student = $x$ $kg$.
$\therefore\text{Mean Weight}=\frac{\text{Sum of weights}}{\text{Total no. of students}}$
$\Rightarrow55=\frac{52+54+55+53+56+54+\text{x}}{7}$
$\Rightarrow385=324+\text{x}$
$\Rightarrow\text{x}=385-324$
$\Rightarrow\text{x}=61\text{kg}$
$\therefore$ weight of seventh stuent $= 61\ kg$.
View full question & answer→Question 73 Marks
Find the median of the data: $92, 35, 67, 85, 72, 81, 56, 51, 42, 69$
AnswerNumbers are $92, 35, 67, 85, 72, 81, 56, 51, 42, 69$
Arranging the numbers in ascending order $35, 42, 51, 56, 67, 69, 72, 81, 85, 92$
$n = 10$(even)$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$
$=\frac{67+69}{2}$
$=\frac{136}{2}=68$
View full question & answer→Question 83 Marks
If the median of the scores $1, 2, x, 4, 5$ (where $1 < 2 < x < 4 < 5$) is $3$, then find the mean of the scores.
AnswerThe given data is $1, 2, x, 4$ and $5$.
Since $1 < 2 < x < 4 < 5$, the given data is already in ascending order.
Here, the number of observation $n = 5$, which is an odd number.
Hence, the median is:$\Big(\frac{\text{n}+1}{2}\Big)^\text{th}\text{observation}$
$=\Big(\frac{\text{5}+1}{2}\Big)^\text{th}\text{observation}$
$=3^\text{rd}\text{observation}$
$=\text{x}$
Here, it is given that the median is $3$.
Hence, we have $x = 3$. The mean is:$\frac{1+2+3+4+5}{5}$
$=\frac{15}{5}$
$=3$
View full question & answer→Question 93 Marks
Find the median of the data: $25, 34, 31, 23, 22, 26, 35, 29, 20, 32$
AnswerNumbers are $25, 34, 31, 23, 22, 26, 35, 29, 20, 32$
Arranging the numbers in ascending order $20, 22, 23, 25, 26, 29, 31, 32, 34, 35$ $n = 10$ (even)
$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$
$=\frac{26+29}{2}$
$=\frac{55}{2}=27.5$
View full question & answer→Question 103 Marks
Find the median of the data: $83, 37, 70, 29, 45, 63, 41, 70, 34, 54$
AnswerNumbers are $83, 37, 70, 29, 45, 63, 41, 70, 34, 54$
Arranging the numbers in ascending order $29, 34, 37, 41, 45, 54, 63, 70, 70, 83$
$n = 10$(even)
$\therefore\text{Median}=\frac{\frac{\text{n}}{2}\text{th value}+\Big(\frac{\text{n}}{2}+1\Big)\text{th value}}{2}$
$=\frac{\frac{\text{10}^\text{th}}{2}\text{value}+\Big(\frac{\text{10}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{5^{\text{th}}\text{value}+6^{\text{th}}\text{value}}{2}$
$=\frac{45+54}{2}$
$=\frac{99}{2}=49.5$
View full question & answer→Question 113 Marks
Find the median of the data: $41, 43, 127, 99, 71, 92, 71, 58, 57$
AnswerNumbers are $41, 43, 127, 99, 71, 92, 71, 58, 57 $
Arranging the numbers in ascending order $41, 43, 57, 58, 71, 71, 92, 99, 127$
$n = 9$ (odd)
$\therefore\text{Median}=\Big(\frac{\text{n}+1}{2}\Big)^{\text{th}}\text{value}$
$=\Big(\frac{9+1}{2}\Big)^{\text{th}}\text{value}$
$=5^\text{th}\text{value}=71$
View full question & answer→Question 123 Marks
If $M$ is the mean of $x_1, x_2, x_3, x_4, x_5$ and $x_6$, Prove that $\left(x_1-M\right)+\left(x_2-M\right)+\left(x_3-M\right)+\left(x_4-M\right)+\left(x_5-M\right)+\left(x_6-M\right)=0$.
AnswerLet $M$ be the mean of $x_1, x_2, x_3, x_4, x_5$ and $x_6$ Then,
$\text{M}=\frac{\text{x}_1+\text{x}_2+\text{x}_3+\text{x}_4+\text{x}_5+\text{x}_6}{6}$
$= x_1 + x_2 + x_3+ x_4 + x_5+ x_6 = 6M$
To Prove:$ (x_1 - M) + (x_2 - M) + (x_3 - M) + (x_4 - M) + (x_5 - M) + (x_6 - M) = 0$.
Proof:$ L.H.S. = (x_1 - M) + (x_2 - M) + (x_3 - M) + (x_4 - M) + (x_5 - M) + (x_6 - M)$
$= (x_1 + x_2 + x_3 + x_4 + x_5 + x_6) - (M + M + M + M + M + M)$
$= 6M - 6M = 0 = R.H.S.$
View full question & answer→Question 133 Marks
If the mean of $2, 4, 6, 8, x, y$ is $5$, then find the value of $x + y.$
View full question & answer→Question 143 Marks
The following observation s have been arranged in ascending order. If the median of the data is $63$, find the value of $x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.$
AnswerTotal number of observation in the given data is $10$ (even number).
So median of this data will be mean of $\frac{10}{2}$ i.e, $5^{th}$ observation and $\frac{10}{2}+1$ i.e, $6^{th}$ observation.
So,$\text{Median of data}=\frac{5^\text{th}\text{observation}+6^\text{th}\text{observation}}{2}$
$\Rightarrow63=\frac{\text{x}+\text{x}+2}{2}$
$\Rightarrow63=\frac{\text{2x}+2}{2}$
$\Rightarrow63=\text{x}+1$
$\Rightarrow\text{x}=62$
View full question & answer→Question 153 Marks
If $\overline{\text{X}}$ is the mean of the ten natural numbers $x_1, x_2, x_3, ..., x_{10}$ show that:$\Big(\text{x}_1-\overline{\text{X}}\Big)+\Big(\text{x}_2-\overline{\text{X}}\Big)+\dots+\Big(\text{x}_{10}-\overline{\text{X}}\Big)=0$
AnswerWe have,$\overline{\text{X}}=\frac{\text{x}_1+\text{x}_2+\dots+\text{x}_{10}}{10}$
$\Rightarrow\text{x}_1+\text{x}_2+\dots+\text{x}_{10}=10\bar{\text{x}}\dots(1)$
Now, $\Big(\text{x}_1-\overline{\text{X}}\Big)+\Big(\text{x}_2-\overline{\text{X}}\Big)+\dots+\Big(\text{x}_{10}-\overline{\text{X}}\Big)$
$=(\text{x}_1+\text{x}_2+\dots+\text{x}_{10})-(\bar{\text{x}}+\bar{\text{x}}+\bar{\text{x}}+\text{upto 10 terms})$
$=10\bar{\text{x}}-10\bar{\text{x}}$ [By equation $(i)$]
$\therefore\Big(\text{x}_1-\overline{\text{X}}\Big)+\Big(\text{x}_2-\overline{\text{X}}\Big)+\dots+\Big(\text{x}_{10}-\overline{\text{X}}\Big)=0...$
View full question & answer→Question 163 Marks
Find the median of the data: $12, 17, 3, 14, 5, 8, 7, 15$
AnswerNumbers are $12, 17, 3, 14, 5, 8, 7, 15$
Arranging the numbers in ascending order $3, 5, 7, 8, 12, 14, 15, 17$
$n = 8$
(even)$\therefore\text{Median}=\frac{\frac{\text{n}^\text{th}}{2}\text{value}+\Big(\frac{\text{n}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{\frac{\text{8}^\text{th}}{2}\text{value}+\Big(\frac{\text{8}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{4^{\text{th}}\text{value}+5^{\text{th}}\text{value}}{2}$
$=\frac{8+12}{2}$
$=\frac{20}{2}=10$
View full question & answer→Question 173 Marks
If the median of scores $\frac{\text{x}}2{},\frac{\text{x}}3{},\frac{\text{x}}4{},\frac{\text{x}}5{}$ and $\frac{\text{x}}6{}$ (where $x > 0$) is $6$, then find the value of $\frac{\text{x}}6{}.$
AnswerGiven that the median of the scores $\frac{\text{x}}2{},\frac{\text{x}}3{},\frac{\text{x}}4{},\frac{\text{x}}5{},\frac{\text{x}}{6},$ where $x > 0$ is $6$.
The number of scores $n$ is $5$, which is an odd number.
We have to find $\frac{\text{x}}{6}$ Note that the scores are in descending order.
Hence the median is:$\Big(\frac{\text{n}+1}{2}\Big)^\text{th}\text{score}$
$=\Big(\frac{\text{5}+1}{2}\Big)^\text{th}\text{score}$
$=3^\text{rd}\text{score}$
$=\frac{\text{x}}{4}$
But, it is given that the median is $6$.
Hence, we have:$\frac{\text{x}}{4}=6$
$\Rightarrow\text{x}=6\times4$
$\Rightarrow\frac{\text{x}}{6}=4$
View full question & answer→Question 183 Marks
If the mode of scores $3, 4, 3, 5, 4, 6, 6, x$ is $4$, find the value of $x$.
AnswerThe given data is $3, 4, 3, 5, 4, 6, 6, x.$
The mode is the value which occur maximum number of times, that is, the mode has maximum frequency.
If the maximum frequency occurs for more than $1$ value, then the number of mode is more than $1$ and is not unique.
Here it is given that the mode is $4$.
So, $x$ must be $4$, otherwise it contradicts that the mode is $4$. Hence $x = 4$
View full question & answer→Question 193 Marks
Find the median of the data: $31, 38, 27, 28, 36, 25, 35, 40$
AnswerNumbers are $31, 38, 27, 28, 36, 25, 35, 40$
Arranging the numbers in ascending order $25, 27, 28, 31, 35, 36, 38, 40$
$n = 8$ (even)
$\therefore\text{Median}=\frac{\frac{\text{n}}{2}\text{th value}+\Big(\frac{\text{n}}{2}+1\Big)\text{th value}}{2}$
$=\frac{\frac{\text{8}^\text{th}}{2}\text{value}+\Big(\frac{\text{8}}{2}+1\Big)^\text{th}\text{value}}{2}$
$=\frac{4^{\text{th}}\text{value}+5^{\text{th}}\text{value}}{2}$
$=\frac{31+35}{2}$
$=\frac{66}{2}=33$
View full question & answer→Question 203 Marks
The arithmetic mean and mode of a data are $24$ and $12$ respectively, then find the median of the data.
AnswerGiven that the arithmetic mean and mode of a data are $24$ and $12$ respectively.
That is, $MEAN = 24 MODE = 12$
We have to find median We know that
$\Rightarrow MODE = 3 \times MEDIAN - 2 \times MEAN $
$\Rightarrow 12 = 3 \times MEDIAN = 12 + (2 \times 24) $
$\Rightarrow 3 \times MEDIAN = 12 + 48 $
$\Rightarrow 3 \times MEDIAN = 60$
$\Rightarrow$ $MEDIAN$ $=\frac{60}{3}$
$\Rightarrow $ $MEDIAN$ $= 20$
View full question & answer→