MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 10

The marks obtained by

$19$ students of a class are

$27,36,22,31,25,26,33,24,37,32,29,28,36,35,27,26,32,35$ and

$28$ . Find the lower quartile.

0%
$26$
0%
$34$
0%
$24$
0%
None of the above

Explanation

By Arrange the series in increasing order. we get

$22,24,25,26,26,27,27,28,28,29,31,32,32,33,35,35,36,36,37$

Since, number of terms

$N=19$

Lower quartile

$Q_1=\cfrac {N+1}{4}=\cfrac {19+1}{4}=5^{th} term$

$\therefore Q_1=26$

The numbers are arranged in the descending order :

$108,\,94,\,88,\,82,\,x+7,\,x-7,\,60,\,58,\,42,\,39$ . If the median is

$73$ , the value of

$x$ is:

0%
$72$
0%
$73$
0%
$76$
0%
$75$

Explanation

Numbers are

$108,94,88,82,x+7,x-7,60,58,42,39$

Median

$=73$

Median

$=\cfrac { { \left( \cfrac { n }{ 2 } \right) }+{ \left( \cfrac { n }{ 2 } +1 \right) } }{ 2 }$

$\left( \cfrac { n }{ 2 } \right)$ th obs

$=x+7$

$\left( \cfrac { n }{ 2 } +1 \right)$ th obs

$=x-7$

$73=\cfrac { (x+7)+(x-7) }{ 2 } \\ =>146=2x\\ =>x=73$

Find the median of the following data

$10,18,15,11,9,16,8$

0%
$11$
0%
$18$
0%
$9$
0%
$16$

Explanation

Arranging the data in increasing order of magnitude

$8, 9, 10, 11, 15, 16, 18$

Since there are

$7$ , i.e. an odd number of observation,

therefore, median = value of

$\left(\displaystyle\frac{7+1}{2}\right)^{th}$ observation = value of

$4^{th}$ observation =

$11$

Median of

$4,5,10,7,1,14,9$ and

$15$ will be:

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

Answer:-

Arranging the data in sequence

$1, 4, 5, 7, 9, 10, 14, 15$ .

Median =

$\cfrac{N+1}{2}$

$N$ is no. of data given

Here

$N=8$

$\Rightarrow\; \cfrac{8+1}{2}=4.5^{th}\;term$

$\cfrac{ 4\;^{th}term + \;5^{th}term}{2} = \cfrac{\left(7+9\right)}{2}=\cfrac{16}{2}=8$

$13$ consecutive days the number of person booked for violating speed limit of

$40$ km/hr. were as follows

$59, 52, 58, 61, 68, 57, 62, 50, 55, 62, 53, 54, 51$ .
The median number of speed violations per day is

0%
$61$
0%
$52$
0%
$55$
0%
$57$

Explanation

$\Rightarrow$ In the above given data

$62$ is repeated twice,

$\Rightarrow$ Now we write the data in ascending order.

$\Rightarrow$

$50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 62, 62, 68$

$\Rightarrow$ Thus, the measure of

$\dfrac{13+1}{2}=7th$ number is median value.

$\therefore$ The median number of speed violation per day is

$57$ .

The arithmetic mean of 5 numbers islf one of the number is excluded the mean of the remaining number isFind the excluded number.

0%
27
0%
25
0%
30
0%
35

Explanation

Sum of 5 numbers =27

$\times$ 5 =135
When one of the numbers is excluded.
Sum of remaining 4 numbers = 4

$\times$ 25 = 100
Excluded number = 135 -100 = 35.

Numbers

$50, 42, 35, 2x + 10, 2x - 8, 12, 11, 8$ are written in descending order and their median is

$25$ find

$x$ ?

0%
$20$
0%
$25$
0%
$12$
0%
$11$

Explanation

$\displaystyle \because$ Descending order of terms

$= 50, 42, 35, (2x + 10,), (2x - 8), 12, 11, 8$

$\displaystyle \Rightarrow$ Total number of terms

$= 8$ (even)

$\displaystyle \therefore$ Median =

$\displaystyle \frac{1}{2}\left [ \left(\frac{8}{2}\right)^{th}\text {term}+\left ( \frac{8}{2}+1 \right )^{th} \text{term}\right ]$

$= \displaystyle \frac{1}{2}\left [ \left ( 2x+10 \right )+\left ( 2x-8 \right ) \right ]$

$= \displaystyle \frac{1}{2}\left [ 4x+2 \right ]=\left ( 2x+1 \right )$
As per question--

$\displaystyle \left ( 2x+1 \right )=25$

$\displaystyle \therefore x=\cfrac{25-1}{2}=12$

If the mean of x and

$\displaystyle \frac{1}{x}$ is M, then the mean of x

$^3$ and

$\displaystyle \frac{1}{x^3}$ is

0%
$\displaystyle \frac{M (M^2 - 3)}{2}$
0%
$M (4M^2 - 3)$
0%
$M^3$
0%
$M^3 + 3$

Explanation

Given,

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

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

$....... (1)$

On cubing both sides, we get

$(2M)^3=(x+\dfrac{1}{x})^3$

$8M^3=x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})$

$8M^3=x^3+\dfrac{1}{x^3}+3(2M)$

$8M^3=x^3+\dfrac{1}{x^3}+6M$

$x^3+\dfrac{1}{x^3}=8M^3-6M$

$\dfrac{x^3+\dfrac{1}{x^3}}{2}=M(4M^2-3)$

Hence, this is the answer.

Calculate the median for the following

Weight in kg.	20	22	24	27	28	30	31
No. of boys	8	10	11	9	7	5	3

0%
$23$
0%
$26$
0%
$23.5$
0%
$24$

Explanation

Weight (x)	No. of boys (f)	cumulative frequency(c.f)
20	8	8
22	10	18
24	11	29
27	9	38
28	7	45
30	5	50
31	3	53

Total number of boys =53

$\therefore Median=\left(\frac{n+1}{2}\right)^{th} term=\frac{53+1}{2}=\frac{54}{2}=27^{th} term$

according to the table the weight of each child from 19th boy to 29th boy is 24 kg.

$\therefore Median =24$

The mean of

$20$ observations is

$15$ . One observation

$20$ is deleted and two more observations are included to the data. If the mean of new set of observations is

$15$ , then find the sum of the two new observations included.

0%
$30$
0%
$35$
0%
$33$
0%
$32$

Explanation

Mean

$= \dfrac { \text{Sum of observations}}{\text{Total number of observations}}$
Given, mean of

$20$ observations

$= 15$ .
Thus, sum of

$20$ observations

$= 15 \times 20 = 300$
When

$20$ is deleted and two more observations are included in the data, sum of observation

$= 300 - 20 + X + Y = 280 + X + Y$ .

And number of observations is

$20 - 1 + 2 = 21$
Given, Mean of wages of new set of observations

$= 15$

$= \dfrac {280 + X + Y }{21} = 15$

$\Rightarrow 280 + X + Y= 15 \times 21 = 315$

$\Rightarrow X + Y = 315 - 280 = 35$

The mean of the following distribution is

0%
$15$
0%
$16$
0%
$17$
0%
$18$

Explanation

Class	$x_i$	$f_i$	$f_i.x_i$
0-5	2.5	4	10.0
5-10	7.5	5	37.5
10-15	12.5	7	287.5
15-20	17.5	12	210.5
20-25	22.5	7	157.5
25-30	27.5	5	137.5

Mean

$=\dfrac{\sum f_i.x_i}{\sum f_i}$

$\Rightarrow \dfrac{10+37.5+287.5+210.5+157.5+137.5}{4+5+7+12+7+5}$

$\Rightarrow \dfrac{640}{40}=16$

The weighted arithmetic mean ofthe first

$n$ natural numbers whose weights are equal to the corresponding numbers is given by,

0%
$\displaystyle \frac{1}{2}(n+1)$
0%
$\displaystyle \frac{1}{2}(n+2)$
0%
$\displaystyle \frac{1}{3}(2n+1)$
0%
$\displaystyle \frac{1}{3}n (2n+1)$

Explanation

$x_i$	$x$	Weight $(w)$
$x_1$	$1$	$1$
$x_2$	$2$	$2$
$x_3$ .... ....	$3$	$3$
$x_n$	$n$	$n$

First

$n$ natural numbers are written in the table along with their respective weight.

Weighted arithmetic mean

$=\dfrac{\sum_{n}^{r-1}w_rx_r}{\sum_{n}^{r-1}w_r}$

$=\dfrac{1\times 1+2\times 2+3\times 3+...+n\times n}{1+2+3+4+...+n}$

$=\dfrac{1^2+2^2+3^2+...+n^2}{1+2+3+4+...+n}$

Using formula of sum of series, we get

$=\dfrac{n(n+1)(2n+1)}{6}\times \dfrac{2}{n(n+1)}$

$=\dfrac{(2n+1)}{3}$

Hence, this is the required answer.

Find the median of the following data

0%
$20$
0%
$19$
0%
$21$
0%
$22$

Explanation

$x$	$14$	$18$	$22$	$25$	$30$
$f$	$7$	$10$	$12$	$16$	$5$
C.F.	$7$	$17$	$29$	$45$	$50$

Total of frequency

$N =50$

As the total of the frequencies is even, Median will be the at the mean of observations at

$\dfrac {N}{2}$ and

$\dfrac {N}{2} + 1$ position

So, Median is mean of

$25^{th}$ and

$26^{th}$ observation , that is Mean of

$22$ and

$22 = 22$

x	5	6	10	12	13	15
f	7	8	15	13	9	8

Find the median of the following data

0%
$12$
0%
$13$
0%
$11$
0%
$15$

Explanation

$x$	$5$	$6$	$10$	$12$	$13$	$15$
$f$	$7$	$8$	$15$	$13$	$9$	$8$
C.F.	$7$	$7+8 = 15$	$15+15 = 30$	$30 + 13 = 43$	$43 + 9 = 52$	$52 + 8 = 60$

Total of frequency

$N =60$

As the total of the frequencies is even, Median will be the at the mean of observations at

$\dfrac {N}{2}$ and

$\dfrac {N}{2} + 1$ position

So, Median is mean of

$30^{th}$ and

$31^{st}$ observation, that is Mean of

$10$ and

$12 = 11$

Find the arithmetic mean for the given data:

Marks	No. of students
$0-10$	$5$
$10-20$	$10$
$20-30$	$25$
$30-40$	$30$
$40-50$	$20$
$50-60$	$10$

0%
$34$
0%
$35$
0%
$33$
0%
$38$

Explanation

Mid-values

$(x_i)$ No. of students

$(f_i)$ Product

$(f_ix_i)$

$5$

$25$

$15$

$10$

$150$

$25$

$625$

$35$

$30$

$1050$

$45$

$20$

$900$

$55$

$10$

$550$

$\Sigma f_i=100$

$\Sigma f_ix_i=3300$

$\therefore$ Arithmetic Mean

$= \displaystyle \frac{\Sigma f_ix_i}{\Sigma f_i}$

$= \displaystyle \frac{3300}{100}$

$= 33$

Hence, the arithmetic mean is

$33.$

On Monday, a person mailed

$8$ packages weighing an average (arithmetic mean) of

$\displaystyle 12\frac { 3 }{ 8 }$ pounds, and on Tuesday,

$4$ packages weighing an average of

$\displaystyle 15\frac { 1 }{ 4 }$ pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

0%
$\displaystyle 13\frac { 1 }{ 3 }$
0%
$\displaystyle 13\frac { 13 }{ 16 }$
0%
$\displaystyle 15\frac { 1 }{ 2 }$
0%
$\displaystyle 15\frac { 15 }{ 16 }$
0%
$\displaystyle 16\frac { 1 }{ 2 }$

Explanation

Mean is the sum of the values divided by the total number of values in the data set.

Given that mean of

$8$ packages is

$12\dfrac 38=\dfrac{99}{8}$

$\implies \dfrac{\text{sum of 8 packages}}{8}=\dfrac{99}{8}$

$\implies \text{sum of 8 packages}={99}$ -------(1)

Given that mean of

$4$ packages is

$15\dfrac 14=\dfrac{61}{4}$

$\implies \dfrac{\text{sum of 4 packages}}{4}=\dfrac{61}{4}$

$\implies \text{sum of 4 packages}={61}$ --------(2)

Total mean is

$\implies \dfrac{\text{sum of 8 packages}+\text{sum of 4 packages}}{8+4}$

$=\dfrac{99+61}{12}=\dfrac{160}{12}=13\dfrac 13$

Find the arithmetic mean using direct method for the following data shows distance covered by

$120$ passengers to perform their regular work.

0%
$20\ km$
0%
$25\ km$
0%
$22\ km$
0%
$15\ km$

Explanation

Answer:- Using direct method

Distance	No. of passengers $f_i$	$x_i$	$f_i x_i$
$5-15$	$30$	$10$	$300$
$15-20$	$20$	$17.5$	$350$
$20-35$	$30$	$27.5$	$825$
$35-40$	$40$	$37.5$	$1500$
	$\Sigma f_i = 120$		$\Sigma f_i x_i = 2975$

Arithmetic mean

$=\cfrac{\Sigma f_i x_i}{\Sigma f_i} =\cfrac{2975}{120} = 24.79 \simeq 25\ km$

Find the arithmetic mean of the following table using direct method.

0%
$30$ min
0%
$50$ min
0%
$35$ min
0%
$40$ min

Explanation

Arithmetic mean using direct method formula is

$\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$

$\bar { x } = \dfrac { 35,000 }{ 1,000 }$ min

$\bar { x } = 35$ min

Find the mean for the following data using shortcut method.

0%
$6.23$
0%
$7.33$
0%
$8.13$
0%
$9.43$

Explanation

Answer:- By shortcut Method

Children (x)	Pocket Allowance (f)	d=x-A	fd
2	10	-4	-40
4	20	-2	-40
6=A	30	0	0
8	40	2	80
10	50	4	200
	$\Sigma f = 150$		$\Sigma fd = 200$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f}=6+\cfrac{200}{150} ={6+1.33}=7.33$

B)7.33

A survey was conducted by a group of students in which they collected the following data regarding the number of plants in

$30$ houses in a locality. Find the mean number of plants per house.

0%
$23.4$
0%
$25.4$
0%
$22.4$
0%
$26.4$

Explanation

Given:-

Total no. of houses

$,\Sigma f_i = 30$

Representing the data in frequency distribution table

No. of plants	No. of houses $(f_i)$	Classmark $(x_i)$	$f_i x_i$
$0-6$	$4$	$3$	$12$
$6-12$	$6$	$9$	$54$
$12-18$	$1$	$15$	$15$
$18-24$	$3$	$21$	$63$
$24-30$	$5$	$27$	$135$
$30-36$	$6$	$33$	$198$
$36-42$	$5$	$39$	$195$
	$\Sigma f_i=30$		$\Sigma f_i x_i = 672$

$\therefore \; mean=\cfrac{\Sigma f_i x_i}{\Sigma f_i}=\cfrac{672}{30} = 22.4$

Hence, the mean no. of plants per house

$= 22.4$

Find the mean of the following frequency distribution using direct method.

0%
$16.20$
0%
$14.16$
0%
$16.17$
0%
$14.28$

Explanation

Answer:- Using direct method

Representing the data into frequency distribution table

$x_i =\text{mid point of each class interval}$

Class interval	No. of workers $(f_i)$	$x_i$	$f_i x_i$
0-5	4	2.5	10
5-10	5	7.5	37.5
10-15	6	12.5	75
15-20	7	17.5	122.5
20-25	8	22.5	180
	$\Sigma f_i=30$		$\Sigma f_ix_i=425$

$\therefore \; mean = \cfrac{\Sigma f_ix_i}{\Sigma f_i} =\cfrac{425}{30}=14.16$

Calculate arithmetic mean using direct method for the following data shows distance covered by

$50$ persons to perform their work inside the factory.

0%
$20.5 km$
0%
$32.8 km$
0%
$25.2 km$
0%
$27.6 km$

Explanation

Answer:- Using direct method

Representing the data into frequency distribution table

$x_i =\text{mid point of each class interval}$

Class interval	No. of workers $(f_i)$	$x_i$	$f_i x_i$
1-11	4	6	24
11-21	6	16	96
21-31	10	26	260
31-41	12	36	432
41-51	18	46	828
	$\Sigma f_i=50$		$\Sigma f_ix_i=1640$

$\therefore \; mean = \cfrac{\Sigma f_ix_i}{\Sigma f_i} =\cfrac{1640}{50}=32.8$

Average distance covered by 50 persons = 32.8

B)32.8

Find the mean of the above data.

0%
$90$
0%
$91$
0%
$94$
0%
$92$

Explanation

Answer:- Using the direct method

Marks obtained $x_i$	No. of students $f_i$	$f_i x_i$
100	4	400
89	5	445
98	8	784
87	2	174
75	1	75
95	3	285
	$\sum f_i = 23$	$\sum f_i x_i = 2167$

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

$= \cfrac{2167}{23}$

$= 94.21$

$\approx 94$

Number of dog's	2-4	4-6	6-8	8-10
weight (kg)	20	10	12	15

Find the arithmetic mean of dog's weight using direct method.

0%
$5.17$ kg
0%
$4.56$ kg
0%
$5.77$ kg
0%
$4.58$ kg

Explanation

Number of dog's	Weight (kg) $(f_i)$	Midpoint $(X_i)$	$f_iX_i$
2-4	20	3	$20\times 3=60$
4-6	10	5	$10\times 5=50$
6-8	12	7	$12\times 7=84$
8-10	15	9	$15\times 9=135$
	$\sum f_i = 57$		$\sum f_ix_i = 329$

Arithmetic mean using direct method formula is

$\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$

$\bar { x } = \dfrac { 329 }{ 57 }$

$\bar { x } = 5.77$
The average mean weight of the dog is

$5.77 kg$ .

Walmart supermarket recorded the length of time, to the nearest hour, that a sample of

$100$ cars was parked in their lot. Estimate the average hour using direct method.

0%
$20$ hours
0%
$31$ hours
0%
$25$ hours
0%
$27$ hours

Explanation

Answer:-

Representing the data into frequency distribution table.

Daily Pocket allowance	No. of workers	$x_i$	$f_i x_i$
5-15	20	10	200
15-25	30	20	600
25-35	15	30	450
35-45	12	40	480
45-55	6	50	300
55-65	17	60	1020

	$\Sigma f_i=100$		$\Sigma f_i x_i = 3050$

$\therefore \; Mean=\cfrac{\Sigma f_i x_i}{\Sigma f_i}=\cfrac{3050}{100}=30.5 \simeq 31$

The average hours are 31.

Find the arithmetic mean of the children's height using direct method.

0%
$9.394$ cm
0%
$9.925$ cm
0%
$7.925$ cm
0%
$6.925$ cm

Explanation

Arithmetic mean using direct method formula is

$\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$

$\bar { x } = \dfrac { 1785 }{ 190 }$

$\bar { x } = 9.3947$
The average mean height of the children is

$9.39$ cm.

A group of

$50$ house owners contributes money towards children's education of their street. The amount of money collected is shown in the table below: (use direct method).

0%
Rs. $27$
0%
Rs. $17$
0%
Rs. $10$
0%
Rs. $23$

Explanation

Answer:- Given

$\Sigma f=50$

Using frequency distribution table

Amount	No. of house owners	$x_i$	$x_i f_i$
0-10	5	5	25
10-20	10	15	150
20-30	15	25	375
30-40	10	35	350
40-50	10	45	450
	$\Sigma f_i=50$		$\Sigma f_i x_i = 1350$

$\therefore$ Amount of money collected = Arithmetic mean

$\Rightarrow \; Mean = \cfrac{\Sigma f_i x_i}{\Sigma f_i} = \cfrac{1350}{50}=27$

$\therefore \;$ the amount of money collected = 27Rs.

A) Rs. 27

The fruit vendor has apples packed in boxes. Each box contains different numbers of apples. The following was the distribution of apples according to the number of boxes. Find the mean number of Apples kept in a packing box using the direct method.

0%
$39.5$
0%
$41.5$
0%
$38.5$
0%
$35.5$

Explanation

No. of apples	No. of boxes $f_i$	$x_i = \cfrac{\text{Upper limit+Lower limit}}{2}$	$f_i x_i$
30-33	60	31.5	1890
33-36	120	34.5	4140
36-39	150	37.5	5625
39-42	80	40.5	3160
42-45	50	43.5	2175
45-48	90	46.5	4185
	$\Sigma f_i = 550$		$\Sigma f_i x_i = 21175$

By direct method

$\text{Mean} = \cfrac{\Sigma f_i x_i}{\Sigma f_i} = \cfrac{21175}{550} = 38.5$

$\therefore$ Mean

$=$

$38.5$

$30$ people gave money for donation. Find the average amount of money collected. (use Assumed mean).

0%
$150$
0%
$180$
0%
$170$
0%
$175$

Explanation

Money	X	People (F)	d=X-A	Fd
75-125	100	7	-50	-350
125-175	150=A	8	0	0
175-225	200	8	100	800
225-275	250	7	150	1050
		$\Sigma F = 30$		$\Sigma Fd = 1500$

Mean (By shortcut method) =

$A+\cfrac{\Sigma Fd}{\Sigma F} = 100+\cfrac{1500}{30} = 100+50 = 150$

D) Mean = 150

Priya jogged for a total of

$30$ minutes. Her average speed for the first

$10$ minutes was

$5$ miles per hour. During the last 20 minutes, she jogged

$2.5$ miles. What was the average speed of her entire jog?

0%
$6\dfrac {1}{2} mph$
0%
$3\dfrac {1}{3} mph$
0%
$7\dfrac {1}{2} mph$
0%
$7\dfrac {1}{3} mph$
0%
$7\dfrac {2}{3} mph$

Explanation

Distance traveled in first

$10$ minutes is

$5 \times \dfrac 16 = \cfrac 56$

Distance traveled in next

$20$ minutes is

$2.5 \times \dfrac 13 = \dfrac 56$

Total distance traveled is

$\dfrac 56 + \dfrac 56 = \dfrac 53$

Average speed is

$\dfrac { \frac 53 }{ \frac 12 } = \dfrac {10}{3}$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 10 - 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 10 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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