MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 4

The value of

$\displaystyle \sum ^{n}_{i=1}\left ( x_{1} -\bar{x}\right )$ where

$\displaystyle \bar{x}$ is the arith metic mean of

$\displaystyle x_{i}$ is

0%
1
0%
0
0%
$\displaystyle n\bar{x}$
0%
none of these

Explanation

The

$\bar{x}=\frac{x_{1}}{1}$

Then

$\sum_{i=1}^{n}(x_{1}-\bar{x})=\bar{x}-\bar{x}=0$

In the data set below, what is the upper quartile?

$1, 2, 4, 2, 5, 6, 7$

0%
$2$
0%
$7$
0%
$6$
0%
$4$

Explanation

First arrange the data set in order.

$1, 2, 2, 4, 5, 6, 7$
Here the median is the middle number

$= 4$
So, the upper quartile is the median of the higher half of the data.
Then, the middle of the upper half will be

$5, 6, 7$
Therefore, the upper quartile is

$6$ .

From the following data $: 7, 3, 4, 5, 7, 7, 7, 3, 2, 1, 8, 2.$ Find the mode.

0%
$3$
0%
$7$
0%
$2$
0%
$5$

Explanation

Given data is

$7,3,4,5,7,7,7,3,2,1,8,2$

The mode is the value which appears most often. In the above data the value which appears most often is

$7$ .

Since it appears

$4$ times. So, mode is

$7$ .

Find the weighted mean of marks obtained by a student in certain examination if he has secured

$75,80,67,90$ marks in English, Tamil, Mathematics, social Science respectively, having corresponding weights

$3, 2, 2, 1.$

0%
$75.16$
0%
$76.12$
0%
$62.17$
0%
$78.12$

Explanation

Subject	Marks obtained	Weights $w_i$	$x_iw_i$
English	$75$	$3$	$225$
Tamil	$80$	$2$	$160$
Mathematics	$67$	$2$	$134$
Social science	$90$	$1$	$90$
		$\Sigma w_i=8$	$\Sigma x_iw_i=609$

Weighted Arithmetic mean $=$ $\displaystyle\dfrac {\Sigma x_iw_i}{\Sigma w_i}$ $=609/8$
Weighted Arithmetic mean $= 76.125$

The median of the following data

$36, 74, 77, 31, 68, 76, 34, 80, 64, 32, 82$ is-

0%
$68$
0%
$34$
0%
$67$
0%
$76$

Explanation

Arrange the given data in ascending order we have,

$31, 32, 34, 36, 64, 68, 74, 76, 77, 80$ and

$82$
Median

$=\cfrac{N+1}{2}=\cfrac{11+1}{2}=\cfrac{12}{2}=6$
The sixth entry is

$68$
Median is

$68$

Marks	10-20	20-30	30-40	40-50	50-60
Frequency	4	10	20	40	50

The following frequency distribution shows the marks obtained by

$124$ students in Economy at a certain school. Find the arithmetic mean using the direct method.

0%
$50$ marks
0%
$80$ marks
0%
$30$ marks
0%
$45$ marks

Explanation

Answer:- Direct Method:

Marks	Frequency	$x_i$	$f_i x_i$
$10-20$	$4$	$15$	$60$
$20-30$	$10$	$25$	$250$
$30-40$	$20$	$35$	$700$
$40-50$	$40$	$45$	$1800$
$50-60$	$50$	$55$	$2750$
	$\Sigma f_i = 124$		$\Sigma f_i x_i = 5560$

$Mean = \cfrac{\Sigma f_i x_i}{\Sigma f_i}$

$= \cfrac{5560}{124} =44.83 \simeq 45marks.$

During the first period in the class, Elsa's math quiz scores were

$90, 92, 93, 88, 95, 88, 97, 87$ and

$98$ . What was the median quiz score?

0%
$90$
0%
$91$
0%
$88$
0%
$92$

Explanation

The median of a set of data is the middlemost number in the set.
So, first arrange the data in order.

$87, 88, 88, 90, 92, 93, 95, 97, 98$
The median quiz score is

$92$ .

The daily earnings (in rupees) of

$10$ workers in a factory are

$6, 14, 17, 6, 14, 17, 14, 6, 17, 14$ .
The median wage is-

0%
Rs. $6$
0%
Rs. $14$
0%
Rs. $15$
0%
Rs. $17$

Explanation

Arranging the daily earnings in ascending order, we have

$6, 6, 6, 14, 14, 14, 14, 17, 17, 17$ the median is average of 5th and 6th
wages i.e.,

$\cfrac{14+14}{2}=14$

Find the mode of the following data

$58, 35, 54, 32, 54, 58, 34, 35, 35, 32, 58, 35, 35, 89, 58, 32, 32, 89.$

0%
$35$
0%
$58$
0%
$32$
0%
$89$

Explanation

Given,

$58, 35, 54, 32, 54, 58, 34, 35, 35, 32, 58, 35, 35, 89, 58, 32, 32, 89.$

The mode is the value which appears most often. In the above data the value which appears most often is

$35$ , since it appears

$5$ times. \

So, mode is

$35$ .

Choose the formula used for arithmetic mean of grouped data by Assumed mean method is

0%
$\bar { x } = A - \dfrac { \sum { fd } }{ \sum { f } }$
0%
$\bar { x } = A + \dfrac { \sum { fd } }{ \sum { f } }$
0%
$\bar { x } = A \times \dfrac { \sum { fd } }{ \sum { f } }$
0%
$\bar { x } = { A } / { \dfrac { \sum { fd } }{ \sum { f } } }$

Explanation

The formula used for arithmetic mean of grouped data by shortcut method is

$\bar { x } = A + \dfrac { \sum { fd } }{ \sum { f } }$

$A =$ Assumed mean of the given data

$\sum { f } =$ Summation of the frequencies given in the grouped data

$\sum { fd } =$ Summation of the frequencies and deviation of a given mean data

$d =$ deviation of a mean data

$\bar { x } =$ arithmetic mean

In a moderately a symmetrical distribution- The mean and mode are

$50$ and

$37.5$ respectively, find median.

0%
$45.50$
0%
$45.86$
0%
$52.5$
0%
$52.86$

Explanation

Using the empirical relation between mean, median and mode, we can write X

$-M_0=3(X-M_d)$ or

$2X=3M_d-M_0$ ;

$45.83$ .

Identify in which ideal measure of central tendency used to find the middle value, if the data are ordinal?

0%
mean
0%
median
0%
mode
0%
range

Explanation

Median is the ideal measure of central tendency used to find the middle value, if the data are ordinal.
Example: Consider the data:

$1, 2, 3, 4, 5, 5, 6, 3, 3, 2, 2, 1, 2$
First arrange the data in ordinal.
So,

$1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 5, 5, 6$
So,

$3$ is the median.
Median is the middle number.

Which ideal measure of central tendency used to find the middle value, if the data are categorical?

0%
mean
0%
median
0%
mode
0%
range

Explanation

Mode is the ideal measure of central tendency used to find the middle value, if the data are categorical.
Example: Discrete data:

$1, 2, 3, 4, 5, 5, 6, 3, 3, 2, 2, 1, 2$
Mode is the most occurring value.
Here

$2$ is the mode.

In the data set below, what is the lower quartile?

$1, 2, 4, 2, 5, 6, 7$

0%
$1$
0%
$2$
0%
$4$
0%
$5$

Explanation

First arrange the data set in order.

$1, 2, 2, 4, 5, 6, 7$
Here the median is the middle number

$= 4$
So, the lower quartile is the median of the lower half of the data.
Then, the middle of the lower half will be

$1, 2, 2$
Therefore, the lower quartile is

$2$ .

Find the first quartiles for the following data.

$8, 11, 20, 10, 2, 17, 15, 5, 16, 15, 25, 6, 5$

0%
$2$
0%
$6$
0%
$8$
0%
$5$

Explanation

First arrange the data set in order.

$2, 5, 5, 6, 8, 10, 11, 15, 15, 16, 17, 20, 25$
Here the median is the middle number

$= 11$
So, the first quartile is the median of the lower half of the data.
Therefore, the first quartile is

$5$ .

The lower quartile is the value of the middle of the first set, where

$25\%$ of the values are _____ than

$Q_1$ and

$75\%$ are larger.

0%
longer
0%
larger
0%
smaller
0%
higher

Explanation

The lower quartile is the value of the middle of the first set, where

$25\%$ of the values are smaller than

$Q_1$ and

$75\%$ are larger.
Example:

$2, 5, 5, 6, 8, 10, 11$
Here the median is the middle number

$= 6$
So, the first quartile is the median of the lower half of the data.
Then, the lower quartile of the first half will be

$2, 5, 5$ which is

$25\%$ of the given data
So, the remaining

$75\%$ numbers can be considered as larger.

The ______ is the median of of the lower half of the data set.

0%
quartile
0%
upper quartile
0%
range
0%
lower quartile

Explanation

The lower quartile is the median of of the lower half of the data set.
Example: consider the data set:

$10, 20, 30, 40, 50, 60, 70, 80, 90, 100$
The median of the lower half of the data set is

$10, 20, 30, 40, 50$
So, the lower quartile

$= 30$ .

In which quartile,

$75\%$ of observations are less than

$Q_3$ and

$25\%$ are larger?

0%
lower
0%
upper
0%
middle
0%
$Q_3$

Explanation

In upper quartile,

$75\%$ of observations are less than

$Q_3$ and

$25\%$ are larger.
Example:

$2, 5, 5, 6, 8, 10, 11$
Here the median is the middle number

$= 6$
So, the upper quartile is the median of the upper half of the data.
Then, the upper quartile of the first half will be which is

$75\%$ of the given data less than

$Q_3$ :

$8, 10, 11$
So, the remaining

$25\%$ numbers can be considered as larger.

______ is called as half of the difference between upper and lower quartiles.

0%
Standard deviation
0%
Mean deviation
0%
Quartile deviation
0%
Interquartile range

The notation used for second quartile is

0%
$Q_3$
0%
$Q_1$
0%
$Q_2$
0%
$Q_4$

Explanation

The notation used for second quartile is

$Q_2$

Which type of variable is best to measure the central tendency of median?

0%
nominal
0%
ratio
0%
interval
0%
ordinal

Explanation

Ordinal is best to measure the central tendency of median.
Example: data set

$= {2, 3, 4, 5, 6, 7, 8}$
In the ordinal number of data set, median is the middle number, which is

$5$ .

What is the best measure of central tendency for the given diagram?

0%
mode
0%
mean
0%
median
0%
range

In which distribution, measure of central tendency is equal?

0%
left skew
0%
right skew
0%
normal
0%
all the above

Find the average of the expressions

$2x+4, 5x-1$ and

$-x+3$ .

0%
$x+2$
0%
$x-2$
0%
$2x+2$
0%
$2x-2$

Explanation

Average of the expression

$=$

$\dfrac{sum \quad of \quad expression }{total \quad no. \quad of \quad expressions}$

$\Rightarrow \dfrac{2x+4+5x-1-x+3}{3}$

$\Rightarrow \dfrac{6x+6}{3}$

$\Rightarrow \dfrac{3(2x+2)}{3}=2x+2$

$12-n, 12, 12+n$
What is the average (arithmetic mean) of the

$3$ quantities in the list above?

0%
$4$
0%
$12$
0%
$18$
0%
$4+\dfrac{n}{3}$
0%
$12+\dfrac{n}{3}$

Explanation

Given

$3$ Quantities:

$12$

$-$

$n$ ,

$12$ ,

$12$

$+$

$n$

Average of the above

$3$ Quantities

$=$

$\dfrac {\text{Sum of the above}\: 3\: \text{Quantities}} {\text{Number of Quantities}}$

$=$

$\dfrac { 12 \space - \space n \space + \space 12 \space + \space 12 \space + \space n }{3}$

$=$

$\dfrac {12 \times 3}{3}$

$=$

$12$

Therefore, average of the Quantities listed above is

$12$ .

The mean of

$a, b, c, d$ and

$e$ is

$28$ . If the mean of

$a, c$ and

$e$ is

$24$ , what is the mean of

$b$ and

$d$ ?

0%
$31$
0%
$32$
0%
$33$
0%
$34$

Explanation

Formula used:

$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$

Given

$\Rightarrow$

$\dfrac{a+b+c+d+e}{5}=28$ and

$\dfrac{a+c+e}{3} =24$

To find

$\Rightarrow$

$\dfrac{b+d}{2}$ which is nothing but the mean of

$b$ and

$d$ .

$\begin{aligned}{}\dfrac{{a + b + c + d + e}}{5} &= 28\\a + b + c + d + e& = 140\quad \dots (i)\end{aligned}$

Also,

$\begin{aligned}{}\dfrac{{a + c + e}}{3} &= 24\\a + c + e& = 72\quad \dots (ii)\end{aligned}$

By using

$(i)$ and

$(ii)$ ,

$\begin{aligned}{}a + b + c + d + e &= 140\\b + d + 72 &= 140\\b + d &= 68\\\dfrac{{b + d}}{2} &= 34\end{aligned}$

Hence, option

$D$ is the correct answer.

If the median of

$\dfrac {x}{3}, \dfrac {x}{2}, \dfrac {x}{4}, \dfrac {2x}{9}$ and

$x$ is

$5$ , find the value of

$x$ .

0%
$10$
0%
$15$
0%
$20$
0%
$25$

Explanation

$\Rightarrow$ The given numbers are

$\dfrac{x}{3},\dfrac{x}{2},\dfrac{x}{4},\dfrac{2x}{9},x$

Arranging given numbers in ascending order,

$\dfrac{2x}{9},\dfrac{x}{4},\dfrac{x}{3},\dfrac{x}{2},x$

$\Rightarrow$ Now, here we can see

$\dfrac{x}{3}$ is middle number.

According to the question,

$\therefore$

$\dfrac{x}{3}=5$

$\therefore$

$x=15$

In which two measure of central tendency do not affect the outliers?

0%
median and mean
0%
mean and mode
0%
median and mode
0%
mode and range

Determine which type of variable is best to measure the central tendency of mode?

0%
nominal
0%
ratio
0%
interval
0%
ordinal

Explanation

Nominal is best to measure the central tendency of mode.
The mode can be very useful for dealing with nominal data. For example, if a sandwich shop sells

$10$ different types of sandwiches, the mode would represent the most popular sandwich.

What is the best measure of central tendency for the following diagram?

0%
mode
0%
mean
0%
median
0%
range

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Economics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.