MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 12

The ages of the

$115$ people who live on an apartment are grouped as follows

$:$ The mean length of the plants is

$33.43$ years using direct method. Find y in the table.

Age(years)	Number of people
$0-10$	$10$
$10-20$	$15$
$20-30$	$26$
$30-40$	Y
$40-50$	$23$
$50-60$	$16$
$60-70$	$3$
$70-80$	$1$

0%
$23$
0%
$28$
0%
$15$
0%
$21$

Explanation

Length(cm) No. of plants

$(f_i)$ Mid point

$x_i$

$f_ix_i$

$0-10$

$10$

$5$

$10\times 5=50$

$10-20$

$15$

$15\times 15=225$

$20-30$

$26$

$25$

$26\times 25=650$

$30-40$

$y$

$35$

$y\times 35=35y$

$40-50$

$23$

$45$

$23\times 45=1035$

$50-60$

$16$

$55$

$16\times 55=880$

$60-70$

$3$

$65$

$3\times 65=195$

$70-80$

$1$

$75$

$1\times 75=75$

$\Sigma f_i=115$

$\Sigma f_ix_=3110+35y$

Arithmetic mean using direct method formula is

$\overline{x}=\displaystyle\frac{\Sigma f_i x_i}{\Sigma f_i}$

$33.43 = (3110 + 35y)/115$

$433.43 \times 115 = 3110 + 35y$

$3844.45 - 3110 = 35y$

$y = 734.45/35$

$y = 20.9$

Therefore, approximately the value of

$y$ is

$21.$

Using step deviation method find the mean.

X	20-40	40-60	60-80	80-100
frequency	4	8	12	16

0%
$40$
0%
$50$
0%
$60$
0%
$70$

Explanation

X	Mid point (X)	Frequency (F)	i=class interval width	A=50 Assumed Means	u= $\dfrac{x-A}{i}$	fu
20-40	30	4	20	50	-1	-4
40-60	50=A	8	20	50	0	0
60-80	70	12	20	50	1	12
80-100	90	16	20	50	2	32
		$\Sigma f=40$				$\Sigma f u=40$

The formula used for arithmetic mean of grouped data by step deviation method is,

$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$

$A =$ Assumed mean of the given data

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

$\sum fu$ = Summation of the frequencies and deviation of a given mean data

$u = \frac{(x - A)}{i}$

$i =$ Class interval width

$\overline {X}$ = arithmetic mean

$\overline {X} =50 +\dfrac{40}{40}\times 20$

$= 50 + 20$

$= 70$

Find the measures of central tendency fro the data set

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

0%
mean $= 3.42$ , median $= 3$ and mode $= 2$
0%
mean $= 3$ , median $= 3$ and mode $= 2$
0%
mean $= 3.42$ , median $= 3$ and mode $= 3$
0%
mean $= 3$ , median $= 3$ and mode $= 3$

Explanation

Mean, median and mode of a data set are the measures of central tendency.
Step 1: Mean

$=$

$\dfrac{\sum X}{n}$

$\sum X$

$=$ sum of the data values.

$n =$ number of the data values
Mean

$=$

$\dfrac{2 + 4+5+1+7+2+3}{7}=3.42$
Step 2: The data set in the ascending order is

$1, 2, 2, 3, 4, 5, 7$
So, Median of the set is

$3$ . [Median is the middle data value of the ordered set.]
Step 3: Mode is the data value(s) that appear most often in the data set. So, the modes of the data set are

$2$ .
So, the measures of central tendency of the given set of data are mean

$= 3.42$ , median

$= 3$ and mode

$= 2$ .

The measure of central tendency of mean is affected by

0%
variables
0%
distributions
0%
outliers
0%
range

Explanation

The measure of Central tendency of mean is affected by

$outlers$ ,

because outliers are observation that lies on abnormal distance from

other values in a random sample data.

If the data set is perfectly normal, the mean, median and mode are _____to each other.

0%
Equal
0%
Unequal
0%
Similar
0%
Not comparable

Explanation

Since , when data is Normal it will be in

$bell shape$ and also

$symmetrical$ . In that case

$Median$ ,

$Mean$ &

$Mode$ of data are

$Equal$ .

When the mean is the best measure of central tendency?

0%
continuous and symmetrical data
0%
discontinuous and symmetrical data
0%
skewed data only
0%
continuous data only

Explanation

$\Rightarrow$ The mean is usually the best measure of central tendency to use when your data distribution is continuous and symmetrical, such as when your data is normally distributed.

$\Rightarrow$ However, it all depends on what you are trying to show from your data.

$\Rightarrow$ The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.

$\overline{x}=\dfrac{x_1+x_2+x_3+...x_n}{n}$

If the average (arithmetic mean) of

$t$ and

$(t+2)$ is

$x$ and if the average of

$t$ and

$(t-2)$ is

$y$ , what is the average of

$x$ and

$y$ ?

0%
$1$
0%
$\cfrac{t}{2}$
0%
$t$
0%
$t+\cfrac{1}{2}$

Explanation

Given, the average (arithmetic mean) of

$t$ and

$t+2$ is

$x$ and the average of

$t$ and

$t−2$ is

$y$

Then

$x=\dfrac{t+(t+1)}{2}=\dfrac{2t+1}{2}$

And

$y=$

$y=\dfrac{t+(t-1)}{2}=\dfrac{2t-1}{2}$

Then average

$x$ and

$y =$

$\dfrac{x+y}{2}=\dfrac{\frac{2t+1}{2}+\frac{2t-1}{2}}{2}=\dfrac{2t+1+2t-1}{2\times 2}=\dfrac{4t}{4}=t$

What is the weighted arithmetic mean of the first

$n$ natural numbers whose weights are equal to the corresponding numbers?

0%
$\dfrac { 1 }{ 2 } \left( n+1 \right)$
0%
$\dfrac { 1 }{ 2 } \left( n+2 \right)$
0%
$\dfrac { 1 }{ 3 } \left( n+1 \right)$
0%
$\dfrac { 1 }{ 3 } \left( 2n+1 \right)$

Explanation

Weight Arithmetic mean

$= \cfrac { 1*1+2*2+3*3+....{ n }^{ 2 } }{ 1+2+3+....n }$

$=\cfrac { n(2n+1)(n+1) }{ 6*\cfrac { n(n+1) }{ 2 } } \\ =\cfrac { 2n+1 }{ 3 } \\ =\cfrac { 1 }{ 3 } (2n+1)$ .

X, Y, Z are three sets of values.
Which of the following statements is true?

0%
Mean of X is equal to Mode of Z.
0%
Mean of Z is equal to Mode of Y.
0%
Median of Y is equal to Mode of X.
0%
Mean, Median and Mode of X are same.

Explanation

Arranging the data ascending order we get

$X:1,2,2,3,3,3,7$ ;

$Y:3,5,5,7,8,9,12$ and

$Z:2,3,4,4,4,7,11$

Mean of

$X=\cfrac{1+2+2+3+3+3+7}{7}=3$

Mean of

$Y=\cfrac{3+5+5+7+8+9+12}{7}=7$

Mean of

$Z=\cfrac{2+3+4+4+47+11}{7}=5$

Mode of

$X=3$

Mode of

$Y=5$

Mode of

$Z=4$

Median of

$X=3$

Median of

$Y=7$

Median of

$Z=4$

The table below shows the number of cars Jing sold each month last year. What is the median of the data in the table?

0%
$13$
0%
$16$
0%
$19$
0%
$20.5$
0%
$23.5$

Explanation

Number of cars sold are

$25,15,22,19,16,13,19,25,26,27,28,29$

In ascending order,

$13,15,16,19,19,22,25,25,26,27,28,29$

Middle most values are

$22$ and

$25$

Median is a middle most value or average of two middle most values.

$\Rightarrow M=\dfrac {(22+25)}{2}=23.5$

The average (arithmetic mean) of

$t$ and

$y$ is

$15$ , and the average of

$w$ and

$x$ is

$15$ . What is the average of

$t, w, x$ and

$y$ ?

0%
$7.5$
0%
$15$
0%
$22.5$
0%
$30$

Explanation

Given, Average of

$t$ and

$y$

$=$

$15$

$\Rightarrow \dfrac {t \space + \space y}{2}$

$=$

$15$

$\Rightarrow t$

$+$

$y$

$=$

$15$

$\times$

$2$

$\Rightarrow t$

$+$

$y$

$=$

$30$

Average of

$w$ and

$x$

$=$

$15$

$\Rightarrow \dfrac {w \space + \space x}{2}$

$=$

$15$

$\rightarrow w$

$+$

$x$

$=$

$15$

$\times$

$2$

$\Rightarrow w$

$+$

$x$

$=$

$30$

Now, average of

$t$ ,

$y$ ,

$w$ and

$x$

$=$

$\dfrac {(t \space + \space y \space + \space w \space + \space x)}{4}$

$=$

$\dfrac {(t \space + \space y) \space + \space (w \space + \space x)}{4}$

$=$

$\dfrac {30 \space + \space 30}{4}$

$=$

$\dfrac {60}{4}$

$=$

$15$

Therefore, average of

$t$ ,

$y$ ,

$w$ and

$x$ is

$'15'$ .

The mode of the observations

$5, 4, 4, 3, 5, x, 3, 4, 3, 5, 4, 3,$ and

$5$ is

$3$ . What is the median?

0%
$3$
0%
$4$
0%
$5$
0%
$4.5$

Explanation

Since frequency of mode

$(3)$ should be maximum

$x=3$

Now, arranging data in ascending order we get

3,3,3,3,3,4,4,4,4,,5,5,5,5

Median = Middle term

$=4$

The mean of

$48, 27, 36, 24, x$ , and

$2x$ is

$37.x =$

0%
$13\dfrac {1}{3}$
0%
$16\dfrac {2}{3}$
0%
$29$
0%
$33\dfrac {3}{4}$
0%
$40$

Explanation

Mean

$=$

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

$\Rightarrow 37=\dfrac{48+27+36+24+x+2x}{6}$

$\Rightarrow 135+3x=222$

$\Rightarrow 3x=222-135$

$\Rightarrow 3x=87$

$\Rightarrow x=\dfrac{87}{3}=29$

Given,

$10, 18, 4, 15, 3, 21, x$
If

$x$ is the median of the

$7$ numbers listed above, which of the following could be the value of

$x$ ?

0%
$5$
0%
$8$
0%
$9$
0%
$14$
0%
$16$

Explanation

Given numbers

$10,$

$18,$

$4,$

$15,$

$3,$

$21,$

$x$

Median is the middle value in the list of numbers. To find the median, the numbers have to be in the numerical order.

$x$ is the given median,

$x$ should be placed in the middle of the list in numerical order. The list contains

$7$ numbers. So,

$x$ should be placed in the

${4}^{th}$ position.

Rewriting the list in increasing order,

$3,$

$4,$

$10,$

$x,$

$15,$

$18,$

$21$

So,

$x$ should lie between

$10$ and

$15$ .

Possible values of

$x$ ,

$10$

$\leq$

$x$

$\leq$

$15$

From the given options,

Only

$14$ satisfies the above condition.

Therefore, the value of

$x$ is

$14$ .

The average of six numbers isIf the average of two of those numbers is 2, what is the average of the other four numbers?

0%
5
0%
6
0%
7
0%
8
0%
9

The median of

$2, 7, 4, 8, 9, 1$ is

0%
$4$
0%
$6$
0%
$5.5$
0%
$7$

Explanation

$1,2,4,7,8,9$

No. of terms

$=6$

Median

$=\dfrac{6+1}{2}=3.5$ term

(4)th Mean = Median

$\dfrac{4+7}{2}=5.5$

For a collection of

$11$ items, sum of all items is

$132$ , then the arithmetic mean is.

0%
$11$
0%
$12$
0%
$14$
0%
$13$

The mode of the data

$5, 5, 5, 5, 5, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4$ is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\textbf{Step 1: Find the mode.}$

$\text{The given observations are : } 5, 5, 5, 5, 5, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4$

$\text{In any series of observations, mode is the observation which appears the maximum}$

$\text{number of times.}$

$\text{Given data is 5,5,5,5,5,1,2,2,3,3,3,4,4,4,4 }$

$\text{In the given observations, 5 appears the maximum number of times. }$

$\text{So, the mode is 5.}$

$\textbf{Hence, the correct option is D.}$

A random sample of 20 people is classified in the following table according to their ages

Age	Frequency
15-25	2
25-35	4
35-45	6
45-55	5
55-65	3

What is the mean age of this group of people?

0%
$41.0$
0%
$41.5$
0%
$42.0$
0%
$42.5$

Explanation

$Mean=\ \dfrac{\sum { (frequency\times Mid-Point)_i }}{\sum{(frequency)_i}}$

$Mean=\ \dfrac{830}{20}=41.5$

The following observations have been arranged in the ascending order. If the median of the data

$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$ is

$63$ , then the value of

$x$ is ___________.

0%
$63$
0%
$62$
0%
$61$
0%
$60$

Explanation

Given data is

$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$

Also given median of data

$=63$ .

Since, total number of observation are even.

So,

$\text{median}=\dfrac{x+(x+2)}{2}$

$=\dfrac{2x+2}{2}=x+1$

Thus

$63=x+1$

$\Rightarrow x=62$

What is the mode for the data 20, 20, 20,21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 25 ?

0%
7
0%
21
0%
22
0%
25

Find the mean of the first five prime numbers.

0%
3.3
0%
7.8
0%
5.6
0%
12.1

Explanation

First five prime numbers are

$2, 3, 5, 7$ and

$11.$

Number of observations

$=5$

$Mean=\dfrac{Sum\,of\,the\,numbers}{No.\,of\,observations}$

$=\dfrac{2+3+5+7+11}{5}$

$=\dfrac{28}{5}$

$=5.6$

The mean of the values

$0, 1, 2, ..... , n$ having corresponding weight

$^nC_0, ^nC_1, ^nC_2, ..... , ^nC_n$ respectively, is?

0%
$\displaystyle\frac{2^n}{(n+1)}$
0%
$\displaystyle\frac{2^{n+1}}{n(n+1)}$
0%
$\displaystyle\frac{n+1}{2}$
0%
$\displaystyle\frac{n}{2}$

Explanation

The solution mean is

$\bar{x}=\displaystyle\frac{0\cdot 1+1\cdot \, ^nC_1+2\cdot \,^nC_2+3\cdot \,^nC_3+....+n\cdot \,^nC_n}{1+\,^nC_1+\,^nC_2+....+\,^nC_n}$

$=\frac{\displaystyle \sum_{r=0}^n\,r\cdot\, ^nC_r}{\displaystyle \sum_{r=0}^n\,^nC_r} = \frac{\displaystyle \sum_{r=1}^n \,r\cdot \frac{n}{r}\,\cdot \,^{n-1}C_{r-1}}{\displaystyle \sum_{r=0}^n\,^nC_r}$

$=\frac{n\displaystyle \sum_{r=1}^n\,^{n-1}C_{r-1}}{\displaystyle \sum_{r=0}^n \, ^nC_r}$

$=\displaystyle\frac{n\cdot 2^{n-1}}{2^n}=\frac{n}{2}$

The 'less than' ogive curve and the 'more than' ogive curve intersect at.

0%
Median
0%
Mode
0%
Arithmetic mean
0%
None of the above

Explanation

The less than ogive curve gives cumulative frequency (probability) for

$x <= a.$

The more than ogive curve gives cumulative frequency (probability) for

$x >= a.$

They intersect exactly at the median, as cumulative frequency will be % of the total at median.

The 'less than' ogive curve and the 'more than' ogive curve intersect at Median.

Option A is correct Answer.

A variate X takes values 2, 9, 3, 7, 5, 4, 3, 2,What is the median?

0%
2
0%
4
0%
7
0%
9

Explanation

The data point which is in middle of ascending or descending series is called median.

So, for data

$2,9,3,7,5,4,3,2,10$

Arranging in ascending order

$2,2,3,3,4,5,7,9,10$

So, middle value is

$4.$

Median of the given data set is

$4.$

Which of the following are positional measure of dispersion?

0%
Standard Deviation, Variance
0%
Percentile, Variance
0%
Quartile, Variance
0%
Percentile,Quartile

Which of the following is true?

0%
The arithmetic mean is affected by the presence of extreme values in the data.
0%
Median is that positional value of the variable which divides the distribution into two equal parts
0%
The median is not affected by the presence of extreme values in the data.
0%
All of these

The mean, median and mode respectively of the numbers

$7, 4, 3, 5, 6, 3, 3, 2, 4, 3, 4, 3, 3, 4, 4, 3, 2, 2, 4, 3, 5, 4, 3, 4, 3, 4, 3, 1, 2, 3$ are __________.

0%
$3.47, 3, 3$
0%
$3, 3, 3$
0%
$4, 3, 3$
0%
$5, 4, 3$

Explanation

Sum of numbers

$=7+4+3+5+6+3+3+2+4+3+4+3+3+4+4+3+2+2+4+3$

$+5+4+3+4+3+4+3+1+2+3$

$=104$

Total numbers are

$30$

$\text{mean}= \dfrac{104}{30}=3.46667=3.47$

The mode is the number which appeared most.

Therefore,

$\text{mode}=3$

For median arrange the numbers in ascending order:

$1,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,5,5,6,7$

Median

$=\dfrac{(15^{th}+16^{th}) \text{term}}{2}$

$=\dfrac{3+3}{2}=3$

In how many ways can 5 persons be seated at a round table so that two particular persons will be together?

0%
12
0%
120
0%
720
0%
180

The sum of the deviations of a set of values

$x_1, x_2$ , ......

$x_n$ measured from

$50$ is

$-10$ and the sum of deviations of the values from

$46$ is

$70$ . The mean is ___________.

0%
$49$
0%
$49.5$
0%
$49.75$
0%
$50$

Explanation

Sum of deviations from

$50$ is

$-10$

$\Rightarrow \sum(xi - 50) = -10$

$\sum x_i - 50\sum1 = -10$

$\sum x_i - 50n = -10$

$\therefore y-50n=-10.....(1)$

Sum of deviations from

$46$ is

$70$

$\Rightarrow \sum(x_i - 46) = 70$

$\sum x_i - 46\sum1 = 70$

$\sum x_i - 46n = 70$

$\therefore y-46n=70.....(2)$

Solving

$(1)$ and

$(2)$ , we get

$4n = 80$ i.e.

$n=20$

Putting value of

$n$ in

$(1)$ , we get

$y=990$

Mean

$= \dfrac{\sum x_i}{n} = \dfrac{y}{n} = \dfrac{990}{20} = 49.5$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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