Step deviation method of finding mean - class-VIII

Question 1 Multiple Choice (Single Answer)

The mean weight of 9 students is 24 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

25 kg
24 kg
26 kg
23 kg

Question 2 Multiple Choice (Single Answer)

The arithmetic mean in a measure of central tedency and is popularlyknown as mean. Arithmetic mean is obtained by dividing the sum of the values of all items of a series by the number of items of that series. Normally, arithmetic mean is denoted by $\bar X$ which is red as '$X$ bar'. It can be computed for unclassified or ungrouped data or individual series as well as classified or grouped data or discrete or continuous series.

From the following data calculate arithmentic mean.

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

$25$
$40$
$15$
$35$

Marks		No. of	(X-45)/10
X	m	students f	d	fd
0-10	5	10	-4	-40
10-20	15	20	-3	-60
20-30	25	30	-2	-60
30-40	35	50	-2	-50
40-50	45	40	0	0
50-60	55	30	+1	30

Question 3 Multiple Choice (Single Answer)

Following table gives frequency distribution of trees planted by different housing societies in a particular locality.
Find mean number of trees planted by housing society by using 'step deviation method'.

No. of tress	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35	35 - 40
No. of Societies	2	7	9	8	6	4

$25.42$ trees
$27.42$ trees
$29.42$ trees
$31.42$ trees

$ci$	$f _i$	$x _i$	$d _i=\dfrac{x _i-a}{c}$	$f _id _i$
10-15	2	12.5	$\dfrac{12.5-27.5}{5}=$-3	-6
15-20	7	17.5	-2	-14
20-25	9	22.5	-1	-9
25-30	8	27.5	0	0
30-35	6	32.5	1	6
35-40	4	37.5	2	8

Question 4 Multiple Choice (Single Answer)

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

$23.88$years
$24.88$ years
$25.88 $years
$26.88$ years

$ci$	$ 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

Question 5 Multiple Choice (Single Answer)

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 $	$2$

$15.8$
$16.8$
$17.8$
None of these

$ci$	$f _i$	$x _i$	$f _ix _i$
$6-10$	$4$	$ 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$	$28$	$56$

Question 6 Multiple Choice (Single Answer)

Following table shows frequency distribution of advertisement on T.V. by shift of origin ans scale method.

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 advertisement on T.V. by shift of origin and scale method.

$31.97$ seconds
$32.97$ seconds
$34.97$ seconds
$35.97$ seconds

$ci$	$f _i$	$x _i$	$d _i=x _i-a$	$f _id _i$
$25-30$	$10$	$27.5$	$-15$	$-150$
$30-35$	$32$	$32.5$	$-10$	$-320$
$35-40$	$15$	$37.5$	$-5$	$-75$
$40-45$	$9$	$42.5$	$0$	$0$
$45-50$	$7$	$47.5$	$5$	$35$
$50-55$	$2$	$52.5$	$10$	$20$

Question 7 Multiple Choice (Single Answer)

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.

$2222$ calories
$2548$ calories
$2563$ calories
$2761$ calories

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

Question 8 Multiple Choice (Single Answer)

Consider the table given below

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of Students	12	18	27	20	17	6

The arithmetic mean of the marks given above is

$18$
$28$
$27$
$6$

Marks	No. of students $ \left( f _i \right) $	$ x _i $	$ x _i f _i $
0-10	12	5	60
10-20	18	15	270
20-30	27	25	675
30-40	20	35	700
40-50	17	45	765
50-60	6	55	330
	$ \Sigma{f _i} = 100 $		$ \Sigma{x _i f _i} = 2800 $

Question 9 Multiple Choice (Single Answer)

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.

$61$
$62$
$63$
$64$

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
		$\Sigma F = 20$		$\Sigma Fd = -4$

Question 10 Multiple Choice (Single Answer)

Find the arithmetic mean using step deviation method for the following 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

$15$
$16$
$17$
$19$

Distance (km)	Mid point (X)	Number of passengers(F)	i=class interval width	A=11 Assumed mean	d'=$\dfrac{x-A}{i}$	fd'
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
		$\Sigma f=40$				$\Sigma f d'=55$

Question 11 Multiple Choice (Single Answer)

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.

$2002$
$2003$
$2004$
$2005$

X	Midpoint (X)	Frequency (F)	i=class interval width	A=2003.5 Assumed mean	d'=$\dfrac{x-A}{i}$	fd'
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
		$\Sigma f=280$				$\Sigma fd'=280$

Question 12 Multiple Choice (Single Answer)

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.

$58$
$48$
$69$
$71$

X	Mid point(x)	Frequency (F)	i=class interval width	A=65 Assumed mean	d'=$\dfrac{x-A}{i} $	fd'
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
		$\Sigma f=60$				$Sigma fd'=22$

Question 13 Multiple Choice (Single Answer)

Using step deviation method find the mean.

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

$40$
$50$
$60$
$70$

X	Mid point (X)	Frequency (F)	i=class interval width	A=50 Assumed Means	d'=$\dfrac{x-A}{i}$	fd'
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
		$\Sigma f=40$				$\Sigma f d'=40$

Question 14 Multiple Choice (Single Answer)

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:

$54$
$55$
$56$
$57$

X	Mid Point (X)	Frequency (F)	i=class interval width	A=45 Assumed mean	d'=$\dfrac{x-A}{i}$	fd'
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
		$\Sigma f=70$				$\Sigma fd'=63$

Question 15 Multiple Choice (Single Answer)

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)

$29$
$31$
$35$
$37$

Step deviation method of finding mean - class-VIII

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