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

Class interval	$10-15$	$15-20$	$20-25$	$25-30$
Frequency	$7$	$12$	$8$	$10$

Find the mean of the following data using Assumed mean method.

0%
$19.781$
0%
$20.836$
0%
$20.337$
0%
$21.005$

Explanation

Answer:- Using Shortcut method

Class Interval	$x$	Frequency $f$	$d=x-A$	$f_d$
$10-15$	$13$	$7$	$-4$	$-28$
$15-20$	$17=A$	$12$	$0$	$0$
$20-25$	$23$	$8$	$6$	$48$
$25-30$	$27$	$10$	$10$	$100$
		$\Sigma f=37$		$\Sigma f_d = 120$

Mean =

$A + \cfrac{\Sigma f_d}{\Sigma f} = 17+\cfrac{120}{37} = 17 + 3.337 = 20.337$

Hence option

$(C)$ is correct

What is the average weight of

$40$ people whom are diagnosed for thyroid? (use Assumed mean method).

0%
$90.75$
0%
$101.45$
0%
$87.25$
0%
$95$

Explanation

Weight	X = $\cfrac{U.L.+L.L.}{2}$	Patients (F)	d=X-A	Fd
65-75	70	3	-30	-30
75-85	80	13	-20	-260
85-95	90	12	-10	-120
95-105	100 = 100	8	0	0
105-115	110	4	10	40
		$\Sigma F = 40$		$\Sigma Fd = -430$

Mean by Shortcut Method =

$A+\cfrac{\Sigma Fd}{\Sigma F} = 100+\cfrac{-370}{40} = 100-9.25 = 90.75$

A) Mean = 90.75

Estimate the mean of the given data using Assumed meant method.

0%
$50$
0%
$49.14$
0%
$51.83$
0%
$61.37$

Explanation

Answer:- Using Shortcut method

Interval	$x = \cfrac{U.L. + L.L.}{2}$	Frequency f	d=x-A	fd
22-32	27	12	-30	-360
32-42	37	7	-20	-140
42-52	47	11	-10	-110
52-62	57=A	3	0	0
62-72	67	19	10	190
72-82	77	6	20	120
		$\Sigma f=58$		$\Sigma fd = -300$

Mean =

$A + \cfrac{\Sigma fd}{\Sigma f} = 57+\cfrac{-300}{58} = 57 - 5.17 = 51.827 \simeq 51.83$

C)51.83

The marks obtained by

$20$ students of Class

$X$ of a certain school in a English paper consisting of

$100$ marks are presented in table below. Find the mean of the marks obtained by the students using step deviation method.

0%
$61$
0%
$62$
0%
$63$
0%
$64$

Explanation

Answer:- Using Shortcut Method

Class interval width (i) =

$50-40 = 10$

Marks	X	F	$d=\cfrac{X-A}{i}$	Fd
40-50	45	5	-2	-10
50-60	55	4	-1	-4
60-70	65 = A	3	0	0
70-80	75	6	1	6
80-90	85	2	2	4
		$\Sigma F = 20$		$\Sigma Fd = -4$

Mean ( By step deviation formula) =

$A+\cfrac{\Sigma Fd}{\Sigma F} \times i = 65+\left(\cfrac{-4}{20} \times 10 \right) = 65=2 = 63$

C) Mean =

$63$

The marks obtained by

$40$ students of Class XII of a certain school in a Science paper consisting of

$200$ marks are presented in table below. Find the mean of the marks obtained by the students using step deviation method.

0%
$153.5$
0%
$154.5$
0%
$150.2$
0%
$156.5$

Explanation

Answer:- Using Shortcut Method

Class interval width (i) =

$120-100 = 20$

Marks	X	Students (F)	$d=\cfrac{X-A}{i}$	Fd
100-120	110	4	-2	-8
120-140	130	5	-1	-5
140-160	150 = A	11	0	0
160-180	170	20	1	20
		$\Sigma F = 40$		$\Sigma Fd = 7$

Mean ( By step deviation formula) =

$A+\cfrac{\Sigma Fd}{\Sigma F} \times i = 150+\left(\cfrac{7}{40} \times 20 \right) = 150+3.5+ = 153.5$

A) Mean =

$153.5$

The marks scored by

$25$ students in the exam are as follows. Find the average using Assumed mean method.

0%
$73.1$
0%
$75.8$
0%
$77.4$
0%
$76.9$

Explanation

Marks	$X =$ $\cfrac{U.L.+L.L.}{2}$	Students $(F)$	$d=X-A$	$Fd$
50-60	55	3	-20	-60
60-70	65	4	-10	-40
70-80	75 $= A$	8	0	0
80-90	85	8	10	80
90-100	95	2	20	40
		$\Sigma F = 25$		$\Sigma Fd = 20$

Mean by Shortcut Method

$=A+\cfrac{\Sigma Fd}{\Sigma F}$

$= 75+\cfrac{20}{25}$

$=75+\cfrac{4}{5}$

$= 75+0.8$

$= 75.8$

Class interval	$4-8$	$8-12$	$12-16$	$16-20$
No. of employees	$5$	$10$	$15$	$20$

Find the mean of the following frequency distribution using step deviation method.

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

Explanation

Answer:- By shortcut Method(Step deviation mean method)

Class interval width

$(w) =8-4 = 4$

Class interval $(i)$	$X$	No. of employees $F$	$d=\cfrac{X-A}{i}$	$Fd$
$4-8$	$6$	$5$	- $1$	$-5$
$8-12$	$10=A$	$10$	$0$	$0$
$12-16$	$14$	$15$	$1$	$15$
$16-20$	$18$	$20$	$2$	$40$
		$\Sigma F = 50$		$\Sigma Fd = 50$

Mean =

$A+\cfrac{\Sigma Fd}{\Sigma f} \times i=10+\left(\cfrac{50}{50} \times 4\right) = 10+4 = 14$

Hence option (D) is correct

Find the mean of the above frequency distribution using the step deviation method.

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

Class interval width

$i = 4-2 = 2$

Class interval	$X_i$	Frequency $F_i$	$u=\cfrac{X-A}{i}$	$F_iu_i$
$2-4$	$3$	$11$	$-1$	$-11$
$4-6$	$5=A$	$22$	$0$	$0$
$6-8$	$7$	$33$	$1$	$33$
$8-10$	$9$	$44$	$2$	$88$
		$\sum F_i = 110$		$\sum F_iu_i = 110$

$\overline x = A+\cfrac{\sum fu}{\sum f} \times i$

$=5+\left(\cfrac{110}{110} \times 2\right)$

$= 5+2$

$= 7$

Hence, option

$C$ is correct.

An Egg Seller distributes eggs to the shop in a city. The number of eggs he distributes for each shop has been recorded and the data obtained was grouped into a class shown in the table below. Find the mean using sdirect method.

0%
$91.5$
0%
$83.07$
0%
$87.34$
0%
$89.76$

Explanation

Number of eggs	Mid-Value	${ f }_{ i }$	${ f }_{ i }{ x }_{ i }$
$0-30$	$15$	$40$	$600$
$30-60$	$45$	$26$	$1170$
$60-90$	$75$	$38$	$2850$
$90-120$	$105$	$12$	$1260$
$$120-150$	$135$	$25$	$3375$
$150-180$	$165$	$30$	$4950$
		$\sum { { f }_{ i }=171 }$	$\sum { { f }_{ i }{ x }_{ i }=14205 }$

$Mean=\cfrac { \sum { { f }_{ i }x_{ i } } }{ \sum { { f }_{ i } } } \\ =\cfrac { 14205 }{ 171 } \\ =83.07$

Find the mean number of leaves per tree using step deviation method.

0%
$10.4$
0%
$11$
0%
$12$
0%
$13$

Explanation

Answer:-

Class interval width (i) = 4

No. of leaves	X	No. of trees F	$u=\cfrac{X-A}{i}$	Fu
0-4	2	10	-1	-10
4-8	6 = A	20	0	-0
8-12	10	30	1	30
12-16	14	40	2	80
		$\Sigma f = 100$		$\Sigma fu = 110$

Mean =

$A+\cfrac{\Sigma fu}{\Sigma f}=6+\cfrac{110}{100} \times 4 ={6+4}=10.4$

10.4 leaves

The percentage of marks obtained by the students in a class of

$50$ are given below. Find the mean for the following data.

Marks $(\%)$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Number of students	$6$	$12$	$14$	$8$	$6$	$4$

0%
$64.6\%$
0%
$65.6\%$
0%
$66.6\%$
0%
$67.6\%$

Explanation

Marks

$f_i$

$x_i$

$f_ix_i$

$40-50$

$6$

$45$

$270$

$50-60$

$12$

$55$

$660$

$60-70$

$14$

$65$

$910$

$70-80$

$8$

$75$

$600$

$80-90$

$6$

$85$

$510$

$90-100$

$4$

$95$

$380$

$\Sigma f_i=50$

$\Sigma f_ix_i=3330$

We know that, the arithmetic mean using direct method is

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

$\overline{x}=\dfrac{3330}{50}$

$\Rightarrow 66.6$

Therefore, the mean for the given data is

$66.6\%$ .

Calculate the mean for the following table using step deviation method.

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

Explanation

Answer:- By shortcut Method

Class interval width (i) =

$5-2 = 3$

Age	No. of children F	Mid Value (X)	$u=\cfrac{X-A}{i}$	Fu
2-5	1	3.5	-1	-1
5-8	2	6.5 = A	0	0
8-11	3	9.5	1	3
11-14	4	12.5	2	8
14-17	5	15.5	3	15
	$\Sigma F = 15$			$\Sigma Fu = 25$

By step deviation method

Mean =

$A+\cfrac{\Sigma fu}{\Sigma f} \times i=6.5+\cfrac{25}{15} \times 3 ={6.5+5}=11.5 \simeq 12$

B) 12

The average (arithmetic mean) of all the grades on a certain algebra test was

$90$ . If the average of the

$8$ males grades was

$87$ , and the average of the females grades was

$92$ , calculate the number of female students.

0%
8
0%
9
0%
10
0%
11
0%
12

Explanation

Let the female students be

$x$ and male students is

$8$ , so the total number of students is

$x+8$

$\Rightarrow 90(x+8)=87 \times 8 + 92x$

$\Rightarrow 2x = 24$

$\Rightarrow x = 12$

Find the Arithmetic mean of the following data by Assumed mean method.

0%
$21$
0%
$22$
0%
$30$
0%
$33$

Explanation

The above table represents the given data.
Let us take the assumed mean,

$a = 25$

We know that, by assumed mean method, the mean of the distribution is given by:

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

$\therefore \ \bar { x } =25+\left[ \dfrac { 520 }{ 65 } \right] =25+8$

$\bar { x } =33$

Hence, the mean of the given distribution, by assumed mean method, is

$33$ .

Find the mean of following frequency distribution using step deviation method.

0%
$44.66$
0%
$40.66$
0%
$42.66$
0%
$45.66$

Explanation

Answer:-

Class interval width

$(W)=21-11 = 10$

Converting the given data into frequency distribution table:

1. Write the mid-point of each interval

$(X_i)$ .

2. Consider any value of the

$X_i$ as

$A$ or Assumed mean.

3. Find deviation

$(d_i)$ of each value of

$X$ from assumed mean as defined in the table.

4. Multiply each valye of

$d_i$ with the corresponding frequency

$f_i$ .

Class interval	$X_i$	No. of employees $$f_i$	$d_i=\cfrac{X-A}{W}$	$f_id_i$
11-21	16	2	-2	-4
21-31	26	4	-1	-4
31-41	36=A	6	0	0
41-51	46	8	1	8
51-61	56	10	2	20
		$\Sigma f = 30$		$\Sigma fd = 20$

5. Calculate mean using the formula below:

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f}=36+\cfrac{20}{30} \times 10 ={36+6.66}=42.66$

Hence, option C is correct.

Calculate the average income of the workers per week using step deviation method.

0%
$500$
0%
$640$
0%
$480$
0%
$320$

Explanation

Weekly Income h=200	Mid-point	Number of workers	${ d }_{ i }={ x }_{ i }-a$	${ u }_{ i }=\cfrac { { x }_{ i }-a }{ h }$	${ f }_{ i }{ u }_{ i }$
200-400	300	4	-200	-1	-4
400-600	500(a)	6	0	0	0
600-800	700	2	200	1	2
800-1000	900	8	400	2	16
		$\sum { { f }_{ i } } =20$			14

Mean

$=a+\cfrac { \sum { { f }_{ i }u_{ i } } }{ \sum { { f }_{ i } } } \times h\\ =500+\cfrac { 14 }{ 20 } \times 200\\ =500+140\quad \\ =640Ans$

The distribution of income of

$60$ teachers in a school are given below. Find the Arithmetic mean of the following data by Asumed mean method.

0%
$256$
0%
$230$
0%
$265$
0%
$254$

Explanation

Let us first construct the table.
Let us take the assumed mean,

$a = 275$

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

$\bar { x } =275+\left[ \dfrac { -600 }{ 60 } \right] =275-10$

$\bar { x } =265$

Find the mean using step deviation method.

0%
$14$
0%
$13$
0%
$12$
0%
$10$

Explanation

x	Mid-point	(frequency)	${ d }_{ i }={ x }_{ i }-6$	${ u }_{ i }=\cfrac { { x }_{ i }-6 }{ 4 }$	${ f }_{ i }{ u }_{ i }$
0-4	2	1	-4	-1	-1
4-8	6 (a)	2	0	0	0
8-12	10	3	4	1	3
12-16	14	4	8	2	8
		10			$\sum { { f }_{ i }{ u }_{ i } } =10$ =10

Mean

$=a+\cfrac { \sum { { f }_{ i }u_{ i } } }{ \sum { { f }_{ i } } } \times h\\ =6+\cfrac { 10 }{ 10 } \times 4\\ =10\quad Ans$

Calculate the mean of the following data with the help of Assumed method.

0%
$7.58$
0%
$5.78$
0%
$6.87$
0%
$8.67$

Explanation

Let us first construct the table.

Let us take the assumed mean,

$a = 5$

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

$\bar { x } =5+\left[ \dfrac { 85.12 }{ 33 } \right] =5+2.58$

$\bar { x } =7.58$

1006764_453123_ans_f87089ffeb724315980797473e287318.png

In a study on a certain population, the following data was given.

Population (X)	2000-2001	2001-2002	2002-2003	2003-2004	2004-2005	2005-2006	2006-2007
Number of people	10	20	30	40	50	60	70

Find the average number of population using step deviation method.

0%
$2002$
0%
$2003$
0%
$2004$
0%
$2005$

Explanation

X	Midpoint (X)	Frequency (F)	i=class interval width	A=2003.5 Assumed mean	u= $\dfrac{x-A}{i}$	fu
2000-2001	2000.5	10	1	2003.5	-3	-30
2001-2002	2001.5	20	1	2003.5	-2	-40
2002-2003	2002.5	30	1	2003.5	-1	-30
2003-2004	2003.5=A	40	1	2003.5	0	0
2004-2005	2004.5	50	1	2003.5	1	50
2005-2006	2005.5	60	1	2003.5	2	120
2006-2007	2006.5	70	1	2003.5	3	210
		$\Sigma f=280$				$\Sigma fu=280$

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

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

$i =$ Class interval width

$\overline {X}$ = arithmetic mean

$\overline {X} =2003.5 +\dfrac{280}{280}\times 1$

$= 2003.5 + 1$

$= 2004.5$

$\approx$

$2005$

Find the arithmetic mean of

$\dfrac {1}{2}, \dfrac {1}{3}, 2n$ and

$m$ .

0%
$\dfrac {5 + 12n + 6m}{24}$
0%
$\dfrac {5 + 8n + 4m}{24}$
0%
$\dfrac {5 + 2n + m}{4}$
0%
$\dfrac {5 + 12n + 6m}{6}$
0%
$\dfrac {5 + 2n + m}{12}$

Explanation

The arithmetic mean of

$\dfrac {1}{2}, \dfrac {1}{3}, 2n, m$ is

$\dfrac { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 3 } +2n+m }{ 4 }$

$=\dfrac { \dfrac { 5 }{ 6 } +2n+m }{ 4 }$

$=\dfrac { 5+12n+6m }{ 24 }$

$M$ is a set consisting of a finite number of consecutive integers. If the median of the numbers in set

$M$ is equal to one of the numbers in set

$M$ , which of the following must be true?

$I.$ The average (arithmetic mean) of the numbers in set

$M$ equals the median.

$II.$ The number of numbers in set

$M$ is odd.

$III.$ The sum of the smallest number and the largest number in set

$M$ is even

0%
None of these
0%
$II$ only
0%
$I$ and $II$ only
0%
$I$ and $III$ only
0%
$I, II$ and $III$

The frequency distribution of marks in English are given in the table:

Marks	50-60	60-70	70-80	80-90
Number of students	12	24	14	10

Find the mean by step deviation method.

0%
$58$
0%
$48$
0%
$69$
0%
$71$

Explanation

$X$	Mid point(x)	Frequency (F)	i=class interval width	A=65 Assumed mean	u= $\dfrac{x-A}{i}$	fu
$50-60$	$55$	$12$	$10$	$65$	$-1$	$-12$
$60-70$	$65=A$	$24$	$10$	$65$	$0$	$0$
$70 -80$	$75$	$14$	$10$	$65$	$1$	$14$
$80-90$	$85$	$10$	$10$	$65$	$2$	$20$
		$\Sigma f=60$				$\Sigma fu=22$

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

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

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

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

$i =$ Class interval width

$\overline {X}=$ arithmetic mean

$\overline {X} =65 +\dfrac{22}{60}\times 10$

$= 65 + 3.666$

$= 68.666$

$\approx$

$69$ marks

Calculate the arithmetic mean for the following birthday gifts using direct method.

Children	0-12	12-24	24-36	36-48	48-60
Birthday gifts	3	5	7	9	11

0%
$36.85$
0%
$23.56$
0%
$24.87$
0%
$12.24$

Explanation

Children	Birthday gifts $({ f }_{ i }$ )	$x_{ i }$	${ f }_{ i }x_{ i }$
0-12	3	6	18
12-24	5	18	90
24-36	7	30	210
36-48	9	42	378
48-60	11	54	594
		$\sum { { f }_{ i } } =35$	$\sum { { f }_{ i }x_{ i } } =1290$

$Mean=\cfrac { \sum { { f }_{ i }x_{ i } } }{ \sum { { f }_{ i } } } \\ =\cfrac { 1290 }{ 35 } =36.85$

Find the arithmetic mean using step deviation method for the following. The data shows distance covered by

$40$ passengers to perform their work. (Round off your answer to the nearest whole number).

Distance (km)	1-5	5-9	9-13	13-17	17-21	21-25	25-29
Number of passengers	2	4	6	8	10	5	5

0%
$15$
0%
$16$
0%
$17$
0%
$19$

Explanation

Distance (km)	Mid point $x$	Number of passengers $f$	$i$ =class interval width	$A$ = Assumed mean	$u=\dfrac{x-A}{i}$	$fu$
1-5	3	2	4	11	-2	-4
5-9	7	4	4	11	-1	-4
9-13	11	6	4	11	0	0
13-17	15	8	4	11	1	8
17-21	19	10	4	11	2	20
21-25	23	5	4	11	3	15
25-29	27	5	4	11	4	20
		$\Sigma f=40$				$\Sigma f u=55$

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

$\overline {x} =A + \frac{\sum fu}{\sum f} \times i$
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

$\overline {x} =11 +\frac{55}{40}\times 4$

$= 11 + 5.5$

$= 16.5$

$\approx$

$17$

Calculate the mean for the following data using step deviation method:

Class interval (X)	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	15	10	15	20	25	30	35

Round off your answer to the nearest whole number.

0%
$41$
0%
$42$
0%
$43$
0%
$44$

Explanation

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

$x_i$ =mid value of class interval

Assumed mean

$a=35$

class interval

$c=10$

$ci$	$f_i$	$x_i$	$u_i=\dfrac{x_i-a}{c}$	$f_iu_i$
$0-10$	$15$	$5$	$-3$	$-45$
$10-20$	$10$	$15$	$-2$	$-20$
$20-30$	$15$	$25$	$-1$	$-15$
$30-40$	$20$	$35$	$0$	$0$
$40-50$	$25$	$45$	$1$	$25$
$50-60$	$30$	$55$	$2$	$60$
$60-70$	$35$	$65$	$3$	$105$

$N=\Sigma f_i=150$

$\Sigma f_iu_i=110$

Mean

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

$\therefore \overline x=35 + \dfrac{110}{150} \times 10=42.3334 \approx 42$

Hence, option

$B$ is correct.

Marks	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40	40-45	45-50
Frequency	3	5	7	8	10	11	14	19	15	13

For the following distribution, find the mean using step deviation method. (Round off your answer to the nearest whole number)

0%
$29$
0%
$31$
0%
$35$
0%
$37$

Explanation

X	Mid point (x)	Frequency (F)	i=class interval width	A=22.5 Assumed mean	u= $\dfrac{x-A}{i}$	fu
0-5	2.5	3	5	22.5	-4	-12
5-10	7.5	5	5	22.5	-3	-15
10 -15	12.5	7	5	22.5	-2	-14
15-20	17.5	8	5	22.5	-1	-8
20-25	22.5=A	10	5	22.5	0	0
25-30	27.5	11	5	22.5	1	11
30-35	32.5	14	5	22.5	2	28
35-40	37.5	19	5	22. 5	3	57
40-45	42.5	15	5	22.5	4	60
45-50	47.5	13	5	22.5	5	65
		$\Sigma f=105$				$\Sigma fu=172$

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

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

$i =$ Class interval width

$\overline {X}$ = arithmetic mean

$\overline {X} =22.5 +\dfrac{172}{105}\times 5$

$= 22.5 + 8.19$

$= 30.69$

$\approx$

$31$

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70- 80	80-90	90-100
Frequency	3	5	6	7	8	9	10	12	6	4

Find the mean mark using step deviation method:

0%
$54$
0%
$55$
0%
$56$
0%
$57$

Explanation

X	Mid Point (X)	Frequency (F)	i=class interval width	A=45 Assumed mean	u= $\dfrac{x-A}{i}$	fu
0-10	5	3	10	45	-4	-12
10-20	15	5	10	45	-3	-15
20-30	25	6	10	45	-2	-12
30-40	35	7	10	45	-1	-7
40-50	45=A	8	10	45	0	0
50-60	55	9	10	45	1	9
60-70	65	10	10	45	2	20
70-80	75	12	10	45	3	36
80-90	85	6	10	45	4	24
90-100	95	4	10	45	5	20
		$\Sigma f=70$				$\Sigma fu=63$

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

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

$i =$ Class interval width

$\overline {X}$ = arithmetic mean

$\overline {X} =45 +\dfrac{63}{70}\times 10$

$= 45 + 9$

$= 54$

Brian got grades of

$92,89$ and

$86$ on his first three math tests. What grade must he get on his final test to have an overall average of

$90$ ?

0%
90
0%
93
0%
87
0%
85

Explanation

Let the grade he got on final test be

$x$
The average of all four grades is

$90$ .
So, we have

$\dfrac {(92+89+86+x)}{4}=90$

$\Rightarrow x = 360-267 = 93$

You grew sixty plants using special soil. You measure their lengths and the values are tabulated below.

Length(cm)	$120-125$	$125-130$	$130-135$	$135-140$	$140-145$	$145-150$
Number of plants	$8$	$10$	Y	$16$	$8$	$12$

The mean length of the plants is $136$ cm. Find y in the table.

0%
$10$
0%
$8$
0%
$6$
0%
$4$

Explanation

Length(cm)

$(f_i)$

$x_i$

$f_ix_i$

$120-125$

$8$

$122.5$

$122.5\times 8=980$

$125-130$

$10$

$127.5$

$127.5\times 10=1275$

$130-135$ Y

$132.5$

$y\times 132.5=132.5y$

$135-140$

$16$

$137.5$

$137.5\times 16=2200$

$140-145$

$8$

$142.5$

$142.5\times 8=1140$

$145-150$

$12$

$147.5$

$147.5\times 12=1770$

$\Sigma f_i=60$

$\Sigma f_ix_i=7365+795y$

Arithmetic mean using direct method formula is

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

$136 = (7365 + 795y)/60$

$136 \times 60 = 7365 + 132.5y$

$8160 - 7365 = 132.5y$

$y = \frac{795}{132.5}$

$y = 6$

Therefore, the value of y is

$6$ .

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

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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