MCQ Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 8

Below is given frequency distribution of marks (out of 100) obtained by the students.
Calculate mean marks scored by a student.

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of Students	3	5	7	10	12	15	12	6	2	8

0%
$48.75$ marks
0%
$51.75$ marks
0%
$54.75$ marks
0%
$57.75$ marks

Explanation

$ci$	$f_i$	$x_i$	$f_ix_i$
0-10	3	5	15
10-20	5	15	75
20-30	7	25	175
30-40	10	35	350
40-50	12	45	540
50-60	15	55	825
60-70	12	65	780
70-80	6	75	450
80-90	2	85	170
90-100	8	95	760

Consider the following table, to calculate mean:

$N=\Sigma f_i=80$

$\Sigma f_ix_i=4140$

Mean

$\overline x=\dfrac {\Sigma f_ix_i}{N}$

$\therefore \overline x=\dfrac{4140}{80}=51.75$

Mean marks scored by a student is

$51.75$

Hence, option

$B$ is correct.

Find the mode of the following:

$18,14,22,23,14,18,17,28,28,14,25,14$

0%
$12$
0%
$13$
0%
$14$
0%
$15$

Explanation

The numbers are:

$18,14,22,23,14,18,17,28,28,14,25,14$
Arranging in ascending order:

$14, 14, 14, 14, 17, 18, 18, 22, 23,25, 28, 28$

$14$ occurs

$4$ times and maximum number of times.

Hence, the mode of the above data is

$14$

$25$ is the arithmetic mean of

$X$ and

$46,$ then find

$X.$

0%
$2$
0%
$4$
0%
$8$
0%
$16$

Explanation

Since

$25$ is the arithmetic mean of

$X$ and

$46$ ,

$\therefore 25=\displaystyle \frac{(X+46)}{2}$

$\therefore X+46=50$

$\therefore X=4$

Deciles are the values dividing a given set of observations into

0%
10
0%
5
0%
6
0%
4

Following table gives age groupwise distribution of people suffering from 'Asthama' due to air pollution in certain city. Find mean age of person suffering from 'Asthama' .

Age (in years)	7 - 11	11 - 15	15 - 19	19 - 23	23 - 27	27 - 31	31 - 35	35 - 39
No. of People	5	9	13	21	16	15	12	9

0%
$23.88$ years
0%
$24.88$ years
0%
$25.88$ years
0%
$26.88$ years

Explanation

Consider the following table, to calculate mean:

$ci$	$f_i$	$x_i$	$f_ix_i$
7-11	5	9	45
11-15	9	13	117
15-19	13	17	221
19-23	21	21	441
23-27	16	25	400
27-31	15	29	435
31-35	12	33	396
35-39	9	37	333

$N=\Sigma f_i=100$

$\Sigma f_ix_i=2388$

Mean

$\overline x=\dfrac {\Sigma f_ix_i}{N}$

$\therefore \overline x=\dfrac{2388}{100}=23.88$

Mean age of people suffering from Asthama is

$23.88$ years

Hence, option

$A$ is correct.

Following table gives frequency distribution of milk (in litres) given per week by 50 cows.
Find average (mean) amount of milk given by a cow by 'shift of origin method'.

Milk (in litres)	24 - 30	30 - 36	36 - 42	42 - 48	48 - 54	54 - 60	60 - 66	66 - 72	72 - 78	78 - 84	84 - 90
No. of cows	1	3	8	5	5	5	8	4	6	2	3

0%
$51.12$ litres
0%
$54.12$ litres
0%
$57.12$ litres
0%
$60.12$ litres

Explanation

Consider the following table, to calculate mean by "shift of origin method":

$x_i$ =mid value of class interval

Assumed mean

$a=57$

$ci$	$f_i$	$x_i$	$d_i=x_i-a$	$f_id_i$
24-30	1	27	27-57= -30	-30
30-36	3	33	33-57= -24	-72
36-42	8	39	39-57= -18	-144
42-48	5	45	45-57= -12	-60
48-54	5	51	51-57= -6	-30
54-60	5	57	57-57=0	0
60-66	8	63	63-57=6	48
66-72	4	69	69-57=12	48
72-78	6	75	75-57=18	108
78-84	2	81	81-57=24	48
84-90	3	87	87-57=30	90

$N=\Sigma f_i=50$

$\Sigma f_id_i=6$

Mean

$\overline x=a +\dfrac {\Sigma f_id_i}{N}$

$\therefore \overline x=57 + \dfrac{6}{50}=57.12$

Average amount of milk given by cow is

$57.12$ litres

Hence, option

$C$ is correct.

The following observation are arranged in ascending order. If the median of the data is

$17$ , find

$x$ if the observations are

$6, 8, 9, 15, x,x + 2, 21, 22, 25, 29.$

0%
$x = 11$
0%
$x = 13$
0%
$x = 15$
0%
$x = 16$

Explanation

The observations are:

$6, 8, 9, 15, x,x+2,21,22,25,29$
The number of observations are

$10$ . Since, the observations are even, the median will be the mean of

$\cfrac{n}{2}$ and

$\cfrac{n}{2}+1$ observations
median

$=$ mean of

$\cfrac{10}{2}$ and

$\cfrac{10+2}{2}$ observations
median

$=$ mean of

$5^{th}$ and

$6^{th}$ observations
median =

$\cfrac{x + x+2}{2}$

$17 = x + 1$

$x = 16$

Find the median of:

$25,16,26,16,32,31,19,28$ and

$35$

0%
$16$
0%
$28$
0%
$26$
0%
$19$

Explanation

$\textbf{Step -1: Arrange the given data in ascending order.}$

$\text{Arranging data in ascending order:}$

$16,16,19,25,26,28,31,32,35$

$\textbf{Step -2: Find median of the given data.}$

$\text{Here, number of observations, }n=9(\text{odd number})$

$\mathbf{\because\text{Median}=\left(\dfrac{n+1}{2}\right)^\text{th}observation.}$

$\therefore \text{Median}=\left(\dfrac{9+1}{2}\right)^\text{th}\text{observation}$

$=\bigg(\dfrac{10}{2}\bigg)^\text{th}\text{observation}$

$=5^\text{th}\text{ observation}$

$=26$

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

Below is given frequency distribution of dividend in percentage declared by 120 companies.

Dividend (in %)	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79
No. of Companies	5	15	28	42	15	12	3

Obtain mean dividend declared by a company by step deviation method.

0%
$40.42$ %
0%
$42.42$ %
0%
$46.42$ %
0%
$48.42$ %

Explanation

Consider the following table, to calculate mean by "step deviation method":

$x_i$ =mid value of class interval

Assumed mean

$a=44.5$

class interval

$c=10$

$ci$	$f_i$	$x_i$	$u_i=\dfrac{x_i-a}{c}$	$f_iu_i$
$9.5-19.5$	$5$	$14.5$	$\dfrac{14.5-44.5}{10}=-3$	$-15$
$19.5-29.5$	$15$	$24.5$	$-2$	$-30$
$29.5-39.5$	$28$	$34.5$	$-1$	$-28$
$39.5-49.5$	$42$	$44.5$	$0$	$0$
$49.5-59.5$	$15$	$54.5$	$1$	$15$
$59.5-69.5$	$12$	$64.5$	$2$	$24$
$69.5-79.5$	$3$	$74.5$	$3$	$9$

$N=\Sigma f_i=120$

$\Sigma f_iu_i=-25$

Mean

$\overline x=a +\dfrac {\Sigma f_iu_i}{N}\times c$

$\therefore \overline x=44.5 + \dfrac{-25}{120} \times 10=42.4166667 \Rightarrow 42.42$

Hence, option

$B$ is correct.

Find the median of the following data:

$18,19, 20, 23, 22, 20, 17, 19, 25$ and

$21$

0%
$20$
0%
$21$
0%
$22$
0%
$23$

Explanation

The given data series is:

$18,19, 20, 23, 22, 20, 17, 19, 25. 21$
Arranging it in increasing order:

$17, 18, 19, 19, 20, 20, 21, 22, 23, 25$
The number of terms in the series are

$10$ , which is even. Hence, the median will be the mean of the two middle numbers.
Thus, median

$=$ mean of

$5^{th}$ term and

$6^{th}$ term
Median

$= \dfrac{20 + 20}{2}$
Median

$= 20$

What is the mean of first 8 whole numbers?

0%
$3$
0%
$3.5$
0%
$4$
0%
$4.5$

Explanation

First

$8$ whole numbers are

$0,1,2,3,4,5,6,7$

We know that,

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$

$\therefore \ \text{Mean}=\dfrac {0+1+2+3+4+5+6+7}{8}$

$\Rightarrow \dfrac {28}{8}=3.5$

Hence, the mean of first

$8$ whole numbers is

$3.5$ .

Find the median of:

$233,173,189,208,194,204,194,185,200$ and

$220$

0%
$194$
0%
$185$
0%
$197$
0%
$220$

Explanation

Arrange the series in ascending order:

$173, 185, 189, 194, 194, 200, 204, 208, 220, 233$
The series has

$10$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers
Median

$=$ mean of

$5^{th}$ and

$6^{th}$ terms
Median

$= \dfrac{194 + 200}{2}$
Median

$= 197$

Find the median of:

$241,243,347,350,327,299,261,292,271,258$ and

$257$

0%
$271$
0%
$292$
0%
$257$
0%
$243$

Explanation

Arrange the series in ascending order:

$241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350$
The series has

$11$ terms. Hence, the median will be the middle term, i.e.

$6^{th}$ term.
Hence, the median is

$271$

The median of

$63,17,50,9,25,43,21,50,14$ and

$34$ is 29.5

0%
True
0%
False
0%
Can't determine
0%
None of these

Explanation

Arranging the numbers in ascending order:

$9, 14, 17, 21, 25, 34, 43, 50, 50, 63$
The series has

$10$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers
Median = mean of

$5^{th}$ and

$6^{th}$ terms
Median =

$\frac{25 + 34}{2}$
Median =

$29.5$

A study related to the time(in months) taken to settle a dispute in a lower court resulted the following data.

Time (in months)	$0 - 2$	$2 - 4$	$4 - 6$	$6 - 8$	$8 - 10$	$10 - 12$
No. of persons	$15$	$90$	$120$	$75$	$70$	$50$

Find the meantime taken to settle a dispute in a lower court.

0%
$6.9$ months
0%
$6.16$ months
0%
$6.66$ months
0%
None of these

Explanation

Consider the following table to calculate mean:

Class Interval	$f_i$	$x_i$	$f_ix_i$
$0-2$	$15$	$1$	$15$
$2-4$	$90$	$3$	$270$
$4-8$	$120$	$5$	$600$
$8-6$	$75$	$7$	$525$
$6-10$	$70$	$9$	$630$
$10-12$	$50$	$11$	$550$

$N=\sum f_i$

$=420$

$\sum f_ix_i=2590$

$\overline x=\dfrac {\sum f_ix_i}{N}$

$\Rightarrow \overline x=\dfrac{2590}{420}$

$\Rightarrow \overline x=6.16$

The mean time to settle a dispute in a lower court is

$6.16$ months.

Hence, option

$C$ is correct.

Find the arithmetic mean for the following grouped frequency distribution:

Class-intervals	$6-10$	$10-14$	$14-18$	$18-22$	$22-26$	$26-30$
Frequency	$4$	$6$	$9$	$12$	$7$	$2$

0%
$15.8$
0%
$16.8$
0%
$17.8$
0%
None of these

Explanation

Consider the following table, to calculate mean:

$ci$	$f_i$	$x_i$	$f_ix_i$
$6-10$	$4$	$8$	$32$
$10-14$	$6$	$12$	$72$
$14-18$	$9$	$16$	$144$
$18-22$	$12$	$20$	$240$
$22-26$	$7$	$24$	$168$
$26-30$	$2$	$28$	$56$

From the table,

$N=\Sigma f_i=4+6+9+12+7+2=40$

$\Sigma f_ix_i=32+72+144+240+168+56=712$

Mean

$\overline x=\dfrac {\Sigma f_ix_i}{N}$

$\therefore \overline x=\dfrac{712}{40}=17.8$

Hence, option

$C$ is correct.

Frequency distribution of duration of

$500$ telephone calls received at a telephone exchange on a certain day is given below :

Duration of call (in sec.)	15 - 29	30 - 44	45 - 59	60 - 74	75 - 89	90 - 104	105 - 119	120 - 134
No. of calls	8	12	61	153	190	57	10	9

Find mean duration of telephone call received at the telephone exchange.

0%
$74.83$ seconds
0%
$79.83$ seconds
0%
$83.83$ seconds
0%
$86.83$ seconds

Explanation

$ci$	$f_i$	$x_i$	$f_ix_i$
14.5-29.5	8	22	176
29.5-44.5	12	37	444
44.5-59.5	61	52	3172
59.5-74.5	153	67	10251
74.5-89.5	190	82	15580
89.5-104.5	57	97	5529
104.5-119.5	10	112	1120
119.5-134.5	9	127	1143

$N=\Sigma f_i=500$

$\Sigma f_ix_i=37415$

Mean

$\overline x=\dfrac {\Sigma f_ix_i}{N}$

$\therefore \overline x=\dfrac{37415}{500}=74.83$

Hence, option

$A$ is correct.

Median of a data set is a number which has an equal number of observations below and above it. The median of the data

$1, 9, 4, 3, 7, 6, 8, 8, 12, 15$ is

0%
$7.5$
0%
$7$
0%
$8$
0%
Any number between $7$ and $8$

Explanation

The data given here is:

$1, 9, 4, 3, 7, 6, 8, 8, 12, 15$
After arranging in ascending order

$1,3,4,6,7,8,8,9,12,15$
Since the number of terms are even, median

$= \dfrac{4^{th} + 5^{th}}{2}$

$= \dfrac{7 + 8}{2}$

$= \dfrac{15}{2}$

$= 7.5$
Hence, option

$A$ is correct.

Find the median of 63, 17, 50, 9, 25, 43, 21, 50, 14

$and\ 34$

0%
$29.5$
0%
$29$
0%
$20$
0%
$30$

Explanation

Arranging the numbers in ascending order:

$9, 14, 17, 21, 25, 34, 43, 50, 50, 63$
The series has

$10$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers
Median = mean of

$5^{th}$ and

$6^{th}$ terms
Median =

$\dfrac{25 + 34}{2}$
Median =

$29.5$

The monthly salaries (in Rs.) of

$10$ employees of a factory are:

$12000,8500,9200,7400,11300,12700,7800,11500,10320,8100$ . The median salary is:

0%
Rs. $9670$
0%
Rs. $9760$
0%
Rs. $6970$
0%
Rs. $7960$

Explanation

Arranging the observation in ascending order :

$7400, 7800, 8100, 8500, 9200, 10320, 11300, 11500, 12000, 12700$
Total number of observations

$(n) = 10$

$\displaystyle \therefore median = \frac{1}{2} \left[\left(\frac{n}{2}\right)^{th} observation + \left(\frac{n}{2} + 1 \right)^{th} observation \right]$

$\displaystyle = \frac{1}{2} \left[\left(\frac{10}{2}\right)^{th} observation + \left(\frac{10}{2} + 1 \right)^{th} observation \right]$

$\displaystyle = \frac{1}{2} \left[5^{th} observation + 6^{th} observation \right]$
Median

$\displaystyle = \frac{1}{2} \left[9200 + 10320 \right] = \frac{19520}{2}$

$\implies$ Median

$= 9760$

$\therefore$ Median salary

$= Rs. 9760$

Frequency distribution of distance travelled in kms. per liter of petrol by different mopeds is given below.

Distance travelled (in km/lit)	62 - 65	65 - 68	68 - 71	71 - 74	74 - 77	77 - 80	80 - 83
No.of mopeds	5	8	12	28	35	10	2

Find mean distance travelled per liter of petrol by a moped.

0%
$71.04$ kms/lit
0%
$73.04$ kms/lit
0%
$75.04$ kms/lit
0%
$79.04$ kms/lit

Explanation

Consider the following table, to calculate mean:

$ci$	$f_i$	$x_i$	$f_ix_i$
$62-65$	$5$	$63.5$	$317.5$
$65-68$	$8$	$66.5$	$532$
$68-71$	$12$	$69.5$	$834$
$71-1-74$	$28$	$72.5$	$2030$
$74-77$	$35$	$75.5$	$2642.5$
$77-80$	$10$	$78.5$	$785$
$80-83$	$2$	$81.5$	$163$

$N=\Sigma f_i=100$

$\Sigma f_ix_i=7304$

Mean

$\overline x=\dfrac {\Sigma f_ix_i}{N}$

$\therefore \overline x=\dfrac{7304}{100}=73.04$

Mean distance travelled per liter of petrol by a moped is

$73.04$ kms/lit

Hence, option

$B$ is correct.

The following table shows the frequency distribution of advertisements on T.V.

Duration (in sec.)	$25 - 30$	$30 - 35$	$35 - 40$	$40 - 45$	$45 - 50$	$50 - 55$
No. of advertisements	$10$	$32$	$15$	$9$	$7$	$2$

Obtain mean duration of an advertisement on T.V. by assumed mean method

0%
$31.97$ seconds
0%
$32.97$ seconds
0%
$34.97$ seconds
0%
$35.97$ seconds

Explanation

Consider the following table, to calculate mean by "assumed mean method":

$x_i=$ mid value of class interval

Let the assumed mean be

$a=42.5$

Class interval	$f_i$	$x_i$	$d_i=x_i-a$	$f_id_i$
$25-30$	$10$	$27.5$	$-15$	$-150$
$30-35$	$32$	$32.5$	$-10$	$-320$
$35-40$	$15$	$37.5$	$-5$	$-75$
$40-45$	$9$	$42.5$	$0$	$0$
$45-50$	$7$	$47.5$	$5$	$35$
$50-55$	$2$	$52.5$	$10$	$20$

$N=\Sigma f_i=75$

$\Sigma f_id_i=-490$

Mean,

$\overline x=a +\cfrac {\Sigma f_id_i}{N}$

$\therefore \overline x=42.5 + \cfrac{-490}{75}=35.966667$

$\implies x= 35.97$

The mean duration of an advertisement on T.V is

$35.97$ seconds

Hence, option

$D$ is correct.

The following data have been arranged in ascending order of magnitude.

$63, 66, 69, x, x + 2, 76, 89$ and

$103$ .
If the median of the given data is

$71$ , find the value of

$x$ .

0%
$60$
0%
$69$
0%
$70$
0%
None of these

Explanation

median of the data:

$63,66,69,x,x+2,76,89,103$ is

$71$
The number of observations in the data are 8. Hence, the median will be the mean of

$3^{rd}$ term and

$4^{th}$ term,
Thus,

$\cfrac{x + x +2}{2} = 71$

$\Rightarrow 2x + 2 = 142$

$\Rightarrow x = 70$

The data is

$17, 26, 60, 45, 33, 32,29, 34$ and

$56$ . If

$26$ is replaced by

$62$ , then the new median is

0%
$17$
0%
$29$
0%
$33$
0%
$34$

Explanation

Arranging data in ascending order-

$17,26,29,32,33,34,45,56,60$
Count of observations

$=9$
Median for odd number of data is the middle most observation
So median $= 5^{th}$ observation $= 33$
If

$26$ is replaced by

$62$ , then the observations in ascending order will be-

$17,29,32,33,34,45,56,60,62$ .
Again Median for odd number of data is the middle most observation
So new median $= 5^{th}$ observation $= 34$

The mean, the median, the mode and the range of the following data:

$84, 56, 39, 45, 54, 39, 56, 54, 84, 21, 77, 56$

0%
Median $=55$
0%
Mode $= 56$
0%
Mean $=58$
0%
Range $=63$

Explanation

Arranging the given series in ascending order.

$21, 39, 39, 45, 54, 54, 56, 56, 56, 77, 84, 84$
Number of terms

$= 12$

$\therefore$ Mode

$= 56$
Range

$=$ Highest value

$-$ Lowest value

$= 84 - 21$

$=$

$63$
Median

$= \displaystyle \frac{6^{th} \text {term} + 7^{th} \text { term}}{2} = \frac{54 + 56}{2} = 55$
Therefore, mean

$= \dfrac{665}{12} = 55.42$

Number of calories (in' 00) consumed daily by a sample of 15 years old boys are given below.

Calories	1000 - 1500	1500 - 2000	2000 - 2500	2500 - 3000	3000 - 3500	3500 - 4000	4000 - 4500
No. of boys	5	13	16	18	27	10	4

Obtain mean calories consumed daily by a boy by step deviation method.

0%
$2222$ calories
0%
$2548$ calories
0%
$2563$ calories
0%
$2761$ calories

Explanation

Consider the following table, to calculate mean by "step deviation method":

$x_i$ =mid value of class interval

Assumed mean

$a=2750$

class interval

$c=500$

$ci$	$f_i$	$x_i$	$u_i=\dfrac{x_i-a}{c}$	$f_iu_i$
$1000-1500$	$5$	$1250$	$-3$	$-15$
$1500-2000$	$13$	$1750$	$-2$	$-26$
$2000-2500$	$16$	$2250$	$-1$	$-16$
$2500-3000$	$18$	$2750$	$0$	$0$
$3000-3500$	$27$	$3250$	$1$	$27$
$3500-4000$	$10$	$3750$	$2$	$20$
$4000-4500$	$4$	$4250$	$3$	$12$

$N=\Sigma f_i=93$

$\Sigma f_iu_i=2$

Mean

$\overline x=a +\dfrac {\Sigma f_iu_i}{N}\times c$

$\therefore \overline x=2750 + \dfrac{2}{93} \times 500=2760.75 \approx 2761$

mean calories consumed daily by a boy is

$2761$ calories.

Hence, option

$D$ is correct.

The A.M. of the first ten odd numbers is

0%
$10$
0%
$100$
0%
$1000$
0%
$1$

Explanation

First ten odd numbers are

$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$ respectively.

$A.M. (\overline {x}) = \dfrac{\text {sum of all the terms}}{\text {total number of terms}}$

$A.M. \left(\overline { x } \right) = \displaystyle\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10}$

$= \displaystyle\frac{100}{10}$

$=10$

$\therefore$ the

$A.M$ of first ten odd numbers is

$10$

A sample of

$25$ scores has a mean

$75$ , median

$79$ and standard deviation of

$8$ . If you increase every score by

$10$ , which of the following is true?

$I$ . The new mean is

$85$

$II$ . The new median is

$89$

$III$ . The new standard deviation is

$18$

0%
$I$ only
0%
$II$ only
0%
$I$ and $II$ only
0%
none
0%
$I, II$ and $III$

Explanation

New Mean = Old Mean +

$10=75+10=85$

New standard deviation= old standard deviation

$=8$

New median = Old median

$+10=79+10=89$

Number of plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

mean is

$8.4$

Is it true or false?

0%
True
0%
False
0%
Ambiguous
0%
Data insuffcient

Explanation

For grouped data when Class Interval is given, we find the Class Mark.

Class Mark

$=$ mid point of an interval

$= \dfrac { \text{Lower limit + Upper limit} }{ 2 }$

Class Mark is taken as

${ x }_{ i }$ for each class interval.

Find

${ x }_{ i }{ f }_{ i }$

From the table, we have

$\Sigma { x }_{ i }{ f }_{ i } = 162$ &

$\Sigma { f }_{ i } = 20$

Mean

$=\dfrac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } }$

$= \dfrac { 162 }{ 20 } = 8.1$

Class Interval	Frequency
100-12	12
120-140	14
140-160	8
160-180	6
180-200	10

mean is 145.20
If true then enter

$1$ and if false then enter

$0$

0%
non of thease
0%
0
0%
1
0%
can not det

Explanation

$For\quad grouped\quad data\quad when\quad Class\quad Interval\quad \quad is\quad given,\quad we\quad find\quad the\quad Class\quad Mark.\\ Class\quad Mark\quad =\quad mid\quad point\quad of\quad an\quad interval\quad =\quad \frac { \left( Lower\quad limit\quad +\quad Upper\quad limit \right) }{ 2 } \\ Class\quad Mark\quad is\quad taken\quad as\quad { x }_{ i }for\quad each\quad class\quad interval.\quad Find\quad { x }_{ i }{ f }_{ i }.\quad \\ From\quad the\quad table,\quad we\quad have\quad \Sigma { x }_{ i }{ f }_{ i }\quad =\quad 7260\quad \& \quad \quad \Sigma { f }_{ i }\quad =\quad 50\\ Mean\quad =\quad \frac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } } \quad =\quad \frac { 7260 }{ 50 } \quad =\quad 145.20$

CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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