RS AGGARWAL CLASS 9 CHAPTER 18 MEAN, MEDIAN AND MODE OF UNGROUPED DATA MCQ

 MULTIPLE CHOICE QUESTIONS

PAGE NO-685

Question 7:

The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is
(a) 64.86
(b) 65.31
(c) 64.91
(d) 64.61

Answer 7:

(c) 64.91

Mean of 100 items = 64
Sum of 100 items = $64\times 100=6400$$64\times 100=6400$
Correct sum = (6400 + 36 + 90 $-$$-$ 26 $-$$-$ 9) = 6491
$\mathrm{Correct}\mathrm{mean}=\frac{6491}{100}=64.91$$\mathrm{Correct} \mathrm{mean}=\frac{6491}{100}=64.91$

Question 8:

The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be
(a) 50.5
(b) 51
(c) 51.5
(d) 52

Answer 8:

(b) 51

Mean of 100 observations = 50
Sum of 100 observations = $100\times 50=5000$$100\times 50=5000$
It is given that one of the observations, 50, is replaced by 150.
∴ New sum = (5000 $-$$-$ 50 + 150) = 5100
And,
$\mathrm{Resulting}\mathrm{mean}=\frac{5100}{100}=51$$\mathrm{Resulting} \mathrm{mean}=\frac{5100}{100}=51$

Question 9:

Let $x$$\overline{x}$ be the mean of $x_1,x_2,...,x_n$$x_1, x_2, ..., x_n$ and $y$$\overline{y}$ be the mean of $y_1,y_2,...,y_n$$y_1, y_2, ..., y_n$.
If $z$$\overline{z}$ is the mean of $x_1,x_2,...,x_n,y_1,y_2,...,y_n,$$x_1, x_2, ..., x_n, y_1, y_2, ... , y_n,$ then $z=?$$\overline{z} = ?$
(a) $(x+y)$$(\overline{x} + \overline{y})$
(b) $\frac{1}{2}(x+y)$$\frac{1}{2}(\overline{x} + \overline{y})$
(c) $\frac{1}{n}(x+y)$$\frac{1}{n}(\overline{x} + \overline{y})$
(d) $\frac{1}{2n}(x+y)$$\frac{1}{2n}(\overline{x} + \overline{y})$

Answer 9:

(b) $\frac{1}{2}(x+y)$$\frac{1}{2}(\overline{x} + \overline{y})$

$\overline{z}=\frac{(x_1+x_2+...+x_n)+(y_1+y_2+...+y_n)}{2n}$$\overline{z}=\frac{(x_1+x_2+...+x_n)+(y_1+y_2+...+y_n)}{2n}$

$\mathrm{Given}:\overline{x}=\frac{x_1+x_2+.....x_n}{n}\Rightarrow x_1+x_2+......+x_n=n\overline{x}\mathrm{and}\overline{y}=\frac{y_1+y_2+......+y_n}{n}\Rightarrow y_1+y_2+......+y_n=n\overline{y}\therefore \overline{z}=\frac{n\overline{x}+n\overline{y}}{2n}=\frac{1}{2}\left(\overline{x}+\overline{y}\right)$$$\begin{gathered} \mathrm{Given}: \\ \overline{x} = \frac{x_1+x_2+.....x_n}{n} \\ \Rightarrow x_1+x_2+......+x_n=n\overline{x} \\ \mathrm{and} \\ \overline{y }=\frac{y_1+y_2+......+y_n}{n} \\ \Rightarrow y_1+y_2+......+y_n=n \overline{y} \\ \therefore \overline{z} = \frac{n \overline{x} + n \overline{y}}{2n} \\ =\frac{1}{2}\left(\overline{x}+\overline{y}\right) \end{gathered}$$

Question 10:

If $x$$\overline{x}$ is the mean of $x_1,x_2,...,x_n,$$x_1, x_2, ..., x_n,$ then for $a\ne 0$$a \ne 0$, the mean of $a{x}_{1,}a{x}_2,...,a{x}_n,\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{a}$$a{x}_{1,} a{x}_2, ..., a{x}_n, \frac{x_1}{a},\frac{x_2}{a}, ...,\frac{x_n}{a}$ is
(a) $\left(a+\frac{1}{a}\right)x$$\left(a+\frac{1}{a}\right)\overline{x}$

(b) $\left(a+\frac{1}{a}\right)\frac{x}{2}$$\left(a+\frac{1}{a}\right)\frac{\overline{x}}{2}$

(c) $\left(a+\frac{1}{a}\right)\frac{x}{n}$$\left(a+\frac{1}{a}\right)\frac{\overline{x}}{n}$

(d) $\frac{\left(a+\frac{1}{a}\right)x}{2n}$$\frac{\left(a+{\displaystyle \frac{1}{a}}\right)\overline{x}}{2n}$

Answer 10:

(b) $\left(a+\frac{1}{a}\right)\frac{x}{2}$$\left(a+\frac{1}{a}\right)\frac{\overline{x}}{2}$
$\mathrm{Required}\mathrm{mean}=\frac{(a{x}_1+a{x}_2+...+a{x}_n)+\left(\frac{x_1}{a}+\frac{x_2}{a}+...+\frac{x_n}{a}\right)}{2n}=\frac{1}{2}\left\{\frac{a(x_1+x_2+...+x_n)}{n}+\frac{\frac{1}{a}(x_1+x_2+...+x_n)}{n}\right\}=\frac{1}{2}\left\{a\overline{x}+\frac{1}{a}\overline{x}\right\}\left(\overline{x}=\frac{x_1+x_2+...+x_n}{n}\right)=\left\{a+\frac{1}{a}\right\}\frac{\overline{x}}{2}$$$\begin{gathered} \mathrm{Required} \mathrm{mean} =\frac{(a{x}_1+a{x}_2+...+a{x}_n)+\left({\displaystyle \frac{x_1}{a}}+{\displaystyle \frac{x_2}{a}}+...+{\displaystyle \frac{x_n}{a}}\right)}{2n} \\ =\frac{1}{2}\left\{\frac{a(x_1+x_2+...+x_n)}{n}+\frac{{\displaystyle \frac{1}{a}}(x_1+x_2+...+x_n)}{n}\right\} \\ =\frac{1}{2}\left\{a\overline{x}+\frac{1}{a}\overline{x}\right\} \left(\overline{ x}=\frac{x_1+x_2+...+x_n}{n}\right) \\ =\left\{a+\frac{1}{a}\right\}\frac{\overline{x}}{2} \end{gathered}$$

Question 11:

If $x_1,x_2,...,x_n$${\overline{x}}_1, {\overline{x}}_2, ..., {\overline{x}}_n$ are the means of n groups with $n_1,n_2,...,n_n$$n_1, n_2, ..., n_n$ number of observations respectively, then the mean $x$$\overline{x}$ of all the groups taken together is
(a) ${\sum }_{i=1}^n{n}_i{x}_i$$\sum _{i=1}^n{n}_i{\overline{x}}_i$

(b) ${\sum }_{\frac{i=1}{n^2}}^n{n}_i{x}_i$$\sum _{\frac{{\displaystyle i=1}}{n^2}}^n{n}_i{\overline{x}}_i$

(c) $\frac{\sum _{i=1}^n{n}_i{x}_i}{\sum _{i=1}^n{n}_i}$$\frac{{\displaystyle \sum _{i=1}^n}{n}_i{\overline{x}}_i}{{\displaystyle \sum _{i=1}^n}{n}_i}$

(d) $\frac{\sum _{i=1}^n{n}_i{x}_i}{2n}$$\frac{{\displaystyle \sum _{i=1}^n}{n}_i{\overline{x}}_i}{2n}$

Answer 11:

$\left(c\right)\frac{\sum _{i=1}^n{n}_i{x}_i}{\sum _{i=1}^n{n}_i}$$\left(\mathrm{c}\right) \frac{{\displaystyle \sum _{i=1}^n}{n}_i{\overline{x}}_i}{{\displaystyle \sum _{i=1}^n}{n}_i}$


$\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{terms}=n_1{\overline{x}}_1+n_2{\overline{x}}_2+...n_n{\overline{x}}_n\mathrm{Number}\mathrm{of}\mathrm{terms}=(n_1+n_2+...n_n)\therefore \mathrm{Mean}=\frac{\sum _{i=1}^n{n}_i{\overline{x}}_i}{\sum _{i=1}^n{n}_i}$$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{terms}=n_1{\overline{x}}_1+n_2{\overline{x}}_2+...n_n{\overline{x}}_n \\ \mathrm{Number} \mathrm{of} \mathrm{terms}=(n_1+n_2+...n_n) \\ \therefore \mathrm{Mean}=\frac{{\displaystyle \sum _{i=1}^n}{n}_i{\overline{x}}_i}{{\displaystyle \sum _{i=1}^n}{n}_i} \end{gathered}$$

Question 12:

The mean of the following data is 8.

x	3	5	7	9	11	13
y	6	8	15	p	8	4

The value of p is
(a) 23
(b) 24
(c) 25
(d) 21

Answer 12:

(c) 25
 

x	y	x$\times$$\times$y
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
Total	41 + p	303 + 9p

$\mathrm{Now},\mathrm{Mean}=\frac{303+9p}{41+p}\mathrm{Given}:\mathrm{Mean}=8\therefore \frac{303+9p}{41+p}=8\Rightarrow 303+9p=328+8p\Rightarrow p=25$$$\begin{gathered} \mathrm{Now}, \\ \mathrm{Mean}=\frac{303+9p}{41+p} \\ \mathrm{Given}: \\ \mathrm{Mean}=8 \\ \therefore \frac{303+9p}{41+p}=8 \\ \Rightarrow 303+9p=328+8p \\ \Rightarrow p=25 \end{gathered}$$

Question 13:

The runs scored by 11 members of a cricket team are
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0
The median score is
(a) 27
(b) 29
(c) 31
(d) 20

Answer 13:

(b) 29

Arranging the weight of 10 students in ascending order, we have:
0, 13, 15, 20, 27, 29, 31, 34, 43, 50, 56
Here, n is 11, which is an odd number.
Thus, we have:
$\mathrm{Median}=\mathrm{Value}\mathrm{of}\left(\frac{n+1}{2}\right)\mathrm{th}\mathrm{observation}\mathrm{Median}\mathrm{score}=\mathrm{Value}\mathrm{of}\left(\frac{11+1}{2}\right)\mathrm{th}\mathrm{term}=\mathrm{Value}\mathrm{of}6\mathrm{th}\mathrm{term}=29$$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation} \\ \mathrm{Median} \mathrm{score}=\mathrm{Value} \mathrm{of} \left(\frac{11+1}{2}\right)\mathrm{th} \mathrm{term} \\ =\mathrm{Value} \mathrm{of} 6\mathrm{th} \mathrm{term} \\ =29 \end{gathered}$$

Question 14:

The weight of 10 students (in kgs) are
55, 40, 35, 52, 60, 38, 36, 45, 31, 44
The median weight is
(a) 40 kg
(b) 41 kg
(c) 42 kg
(d) 44 kg

Answer 14:

(c) 42 kg

Arranging the numbers in ascending order, we have:
31, 35, 36, 38, 40, 44, 45, 52, 55, 60
Here, n is 10, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean}\mathrm{of}\left(\frac{n}{2}\right)\mathrm{th}\mathrm{observation}\&\left(\frac{n}{2}+1\right)\mathrm{th}\mathrm{observation}\mathrm{Median}\mathrm{weight}=\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{weights}\mathrm{of}\left(\frac{10}{2}\right)\mathrm{th}\mathrm{student}\&\left(\frac{10}{2}+1\right)\mathrm{th}\mathrm{student}=\mathrm{Mean}\mathrm{of}\mathrm{the}\mathrm{weights}\mathrm{of}5\mathrm{th}\mathrm{student}\&6\mathrm{th}\mathrm{student}=\frac{1}{2}\left(40+44\right)=42\mathrm{Hence},\mathrm{the}\mathrm{median}\mathrm{weight}\mathrm{is}42\mathrm{kg}.$$$\begin{gathered} \mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \mathrm{observation} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observation} \\ \mathrm{Median} \mathrm{weight}=\mathrm{Mean} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} \left(\frac{10}{2}\right)\mathrm{th} \mathrm{student} \& \left(\frac{10}{2}+1\right)\mathrm{th} \mathrm{student} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} \mathrm{weights} \mathrm{of} 5\mathrm{th} \mathrm{student} \& 6\mathrm{th} \mathrm{student} \\ =\frac{1}{2}\left(40+44 \right)=42 \\ \mathrm{Hence}, \mathrm{the} \mathrm{median} \mathrm{weight} \mathrm{is} 42 \mathrm{kg}. \end{gathered}$$

PAGE NO-686

Question 15:

The median of the numbers 4, 4, 5, 7, 6, 7, 7, 12, 3 is
(a) 4
(b) 5
(c) 6
(d) 7

Answer 15:

(c) 6

We will arrange the given data in ascending order as:
3, 4, 4, 5, 6, 7, 7, 7, 12
Here, n is 9, which is an odd number.
Thus, we have:
$\mathrm{Median}=\mathrm{Value}\mathrm{of}\frac{1}{2}(n+1)\mathrm{th}\mathrm{term}\mathrm{Median}\mathrm{score}=\frac{1}{2}(9+1)\mathrm{th}\mathrm{term}=5\mathrm{th}\mathrm{term}=6$$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \frac{1}{2}(n+1)\mathrm{th} \mathrm{term} \\ \mathrm{Median} \mathrm{score}=\frac{1}{2}(9+1)\mathrm{th} \mathrm{term}=5\mathrm{th} \mathrm{term}=6 \end{gathered}$$

Question 16:

The median of the numbers 84, 78, 54, 56, 68, 22, 34, 45, 39, 54 is
(a) 45
(b) 49.5
(c) 54
(d) 56

Answer 16:

(c) 54

We will arrange the data in ascending order as:
22, 34, 39, 45, 54, 54, 56, 68, 78, 84
Here, n is 10, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean}\mathrm{of}\left(\frac{n}{2}\right)\mathrm{th}\&\left(\frac{n}{2}+1\right)\mathrm{th}\mathrm{observations}=\frac{1}{2}\left(5\mathrm{th}\mathrm{observation}+6\mathrm{th}\mathrm{observation}\right)=\frac{1}{2}\left(54+54\right)=54$$$\begin{gathered} \mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(5\mathrm{th} \mathrm{observation}+6\mathrm{th} \mathrm{observation}\right) \\ =\frac{1}{2}\left(54+54\right)=54 \end{gathered}$$

Question 17:

Mode of the data 15, 17, 15, 19, 14, 18, 15, 14, 16, 15, 14, 20, 19, 14, 15 is
(a) 14
(b) 15
(c) 16
(d) 17

Answer 17:

(b) 15

Here, 14 occurs 4 times, 15 occurs 5 times, 16 occurs 1 time, 18 occurs 1 time, 19 occurs 1 time and 20 occurs 1 time. Therefore, the mode, which is the most occurring item, is 15.

Question 18:

The median of the data arranged in ascending order 8, 9, 12, 18, (x + 2), (x + 4), 30, 31, 34, 39 is 24. The value of x is
(a) 22
(b) 21
(c) 20
(d) 24

Answer 18:

(b) 21

The given data is in ascending order.
Here, n is 10, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean}\mathrm{of}\left(\frac{n}{2}\right)\mathrm{th}\&\left(\frac{n}{2}+1\right)\mathrm{th}\mathrm{observations}=\frac{1}{2}\left(5\mathrm{th}\mathrm{observation}+6\mathrm{th}\mathrm{observation}\right)=\frac{1}{2}(x+2+x+4)=(x+3)=24\mathrm{Also},x+3=24\Rightarrow x=21$$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{n}{\mathit{2}}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(5\mathrm{th} \mathrm{observation}+6\mathrm{th} \mathrm{observation}\right) \\ =\frac{1}{2}(x+2+x+4)=(x+3) \\ =24 \\ \mathrm{Also}, \\ x+3=24 \\ \Rightarrow x=21 \end{gathered}$$