undefined | unit 2 measures of central tendency

Unit – 2

Measures of Central Tendency

Q1) Find the mean.

Xi	9	10	11	12	13	14	15
Freq (Fi)	2	5	12	17	14	6	3

A1)

Xi	Freq (Fi)	XiFi
9	2	18
10	5	50
11	12	132
12	17	204
13	14	182
14	6	84
15	3	45
	Fi = 59	XiFi= 715

Then, N = ∑ fi = 59, and ∑fi Xi=715

X = 715/59 = 12.11

Q2) The following data represent the income distribution of 100 families. Calculate mean income of 100 families?

Income	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of families	8	12	25	22	16	11	6

A2)

Income	No. of families	Xm (Mid-point)	fXm
30-40	8	35	280
40-50	12	34	408
50-60	25	55	1375
60-70	22	65	1430
70-80	16	75	1200
80-90	11	85	935
90-100	6	95	570
	n = 100		∑f Xm = 6198

X = ∑f Xm/n = 6330/100 = 63.30

Mean = 63.30

Q3) Calculate the mean number of hours per week spent by each student in texting message.

Time per week	0 - 5	5 - 10	10 - 15	15 - 20	20 - 25	25 – 30
No. of students	8	11	15	12	9	5

A3)

Time per week (X)	No. of students (F)	Mid point X	XF
0 - 5	8	2.5	20
5 – 10	11	7.5	82.5
10 - 15	15	12.5	187.5
15 - 20	12	17.5	210
20 - 25	9	22.5	202.5
25 – 30	5	27.5	137.5
	60		840

Mean = 840/60 = 14

Q4) The following table of grouped data represents the weights (in pounds) of all 100 babies born at a local hospital last year.

Weight (pounds)	Number of Babies
[3−5)	8
[5−7)	25
[7−9)	45
[9−11)	18
[11−13)	4

A4)

Weight (pounds)	Number of Babies	Mid point X	XF
[3−5)	8	4	32
[5−7)	25	6	150
[7−9)	45	8	360
[9−11)	18	10	180
[11−13)	4	12	48
	100		770

Mean = 770/100 = 7.7

Q5) Find the median score of 7 students in science class.

Score = 19, 17, 16, 15, 12, 11, 10

A5)

Median = (7+1)/2 = 4th value

Median = 15

Q6) Find the median of the table given below.

Marks obtained	No. of students
20	6
25	20
28	24
29	28
33	15
38	4
42	2
43	1

A6)

Marks obtained	No. of students	cf
20	6	6
25	20	26 (20+6)
28	24	50 (26+24)
29	28	78
33	15	93
38	4	97
42	2	99
43	1	100

Median = (n+1)/2 = 100+1/2 = 50.5

Median = (28+29)/2 = 28.5

Q7) Calculate the median.

Marks	No. of students
0-4	2
5-9	8
10-14	14
15-19	17
20-24	9

A7)

Marks	No. of students	CF
0-4	2	2
5-9	8	10
10-14	14	24
15-19	17	41
20-24	9	50
	50

n = 50

n = 50/2= 25

The category containing n/2 is 15 -19

Lb = 15

Cfp = 24

f = 17

ci = 4

Median = 15 + 25-24 *4 = 15.23

Q8) Given the below frequency table calculate median.

X	60 – 70	70 – 80	80- 90	90-100
F	4	5	6	7

A8)

X	F	CF
60 - 70	4	4
70 - 80	5	9
80 - 90	6	15
90 - 100	7	22

n = 22

n = 22/2= 11

The category containing n+1/2 is 80 - 90

Lb = 80

Cfp = 9

f = 6

ci = 10

Median = 80 + 11-9 *10 = 83.33

Q9) Calculate the median of grouped data.

Class interval	1-3	3-5	5-7	7-9	9-11	11-13
Frequency	4	12	13	19	7	5

CI	F	CF
1-3	4	4
3-5	12	16
5-7	13	29
7-9	19	48
9-11	7	55
11-13	5	60

n = 60

n = 60/2= 30

The category containing n+1/2 is 7-9

Lb = 7

Cfp = 29

f = 19

ci = 2

Median = 7 + 30-29 *2 = 7.105

Q10) Find the mode of scores of section A.

Scores = 25, 24, 24, 20, 17, 18, 10, 18, 9, 7

A10) – Mode is 24, 18 as both have occurred twice.

Q11) Find the mode.

Seconds	Frequency
51 - 55	2
56 - 60	7
61 - 65	8
66 - 70	4

A11) The group with the highest frequency is the modal group: - 61-65

D1 = 8-7 = 1

D2 = 8-4 = 4

Mode = L1 + (L2 – L1) d1

d1 +d2

mode = 61 + (65-61) 1 = 61+4 (1/5) = 61.8

1+4

Mode = 61.8

Q12) In a class of 30 students marks obtained by students in science out of 50 is tabulated below. Calculate the mode of the given data.

Marks obtained	No. of students
10 -20	5
20 – 30	12
30 – 40	8
40 - 50	5

A12)

The group with the highest frequency is the modal group: - 20 -30

D1 = 12 - 5 = 7

D2 = 12 - 8 = 4

Mode = L1 + (L2 – L1) d1

d1 +d2

Mode = 20 + (30-20) 7 = 20+10 (7/11) = 26.36

7+4

Mode = 61.8

Q13) Based on the group data below, find the mode.

Time to travel to work	frequency
1 – 10	8
11 -20	14
21 – 30	12
31 – 40	9
41 - 50	7

A13)

The group with the highest frequency is the modal group: - 11 - 20

D1 = 14 - 8 = 6

D2 = 14 - 12 = 2

Mode = L1 + (L2 – L1) d1

d1 +d2

Mode = 11 + (20-11) 6 = 11+9 (6/8) = 17.75

6+2

Q14) Compute the mode from the following frequency distribution.

CI	F
70-71	2
68-69	2
66-67	3
64-65	4
62-63	6
60-61	7
58-59	5

A14)

The group with the highest frequency is the modal group: - 60 - 61

D1 = 7 - 6 = 1

D2 = 7 - 5 = 2

Mode = L1 + (L2 – L1) d1

d1 +d2

Mode = 60 + (61-60) 1 = 60+1 (1/3) 60.85

1+2

Q15) Calculate the harmonic mean of the numbers 13.2, 14.2, 14.8, 15.2 and 16.1.

A15)

X	1/X
13.2	0.0758
14.2	0.0704
14.8	0.0676
15.2	0.0658
16.1	0.0621
Total	0.3147

H.M of X = 5/0.3147 = 15.88

Q16) Calculate harmonic mean.

Class	frequency
2-4	3
4-6	4
6-8	2
8-10	1

A16)

Class	frequency	x	f/x
2-4	3	3	1
4-6	4	5	0.8
6-8	2	7	0.28
8-10	1	9	0.11
	10		2.19

Harmonic mean = 10/2.19 = 4.55

Q17) Calculate quartile deviation from the following test scores.

Sl. N o	Test scores
1	17
2	17
3	26
4	27
5	30
6	30
7	31
8	37

A17)

First quartile (Q1)

Qi= [i * (n + 1) /4] th observation

Q1= [1 * (8 + 1) /4] th observation

Q1 = 2.25 th observation

Thus, 2.25 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 17 and 26

First quartile (Q1) is calculated as

Q1 = 2nd observation +0.75 * (3rd observation - 2nd observation)

Q1 = 17 + 0.75 * (26 – 17) = 23.75

Third quartile (Q3)

Qi= [i * (n + 1) /4] th observation

Q3= [3 * (8 + 1) /4] th observation

Q3 = 6.75 th observation

So, 6.75 th observation lies between the 6th and 7th value in the ordered group, between frequency 30 and 31

Third quartile (Q3) is calculated as

Q3 = 6th observation +0.25 * (7th observation – 6th observation)

Q3 = 30 + 0.25 * (31 – 30) = 30.25

Now using the quartiles values Q1 and Q3, we will calculate the quartile deviation.

QD = (Q3 - Q1) / 2

QD = (30.25 – 23.75) / 2 = 3.25

Q18) Computation of Mean deviation in grouped data.

Class interval	15 – 19	20 – 24	25 – 29	30 – 34	35 – 39	40 – 44	45 - 49
Frequency	1	4	6	9	5	3	2

A18)

Class Interval	F	X	FX	D	FD
15 – 19	1	17	17	15	15
20 – 24	4	22	88	10	40
25 – 29	6	27	162	5	30
30 - 34	9	32	288	0	0
35 - 39	5	37	185	5	25
40 - 44	3	42	126	10	30
45 - 49	2	47	94	15	30
	N = 30		∑fx = 960		= 170

Mean = 960/30 = 32

MD = 170 / 30 = 5.667

Coefficient of mean deviation

Coefficient of mean deviation = (5.67/32)*100 = 17.71

Q19) Calculate mean deviation from the median.

Class	5 -15	15 – 25	25 - 35	35 - 45	45 – 55
Frequency	5	9	7	3	8

A19)

x	f	cf	Mid point x	x –median	F(x-m)
5 -15	5	5	10	17.42	87.1
15 -25	9	14	20	7.42	66.78
25 -35	7	21	30	2.58	18.06
35 -45	3	24	40	12.58	37.74
45- 55	8	32	50	22.58	180.64
	32				390.32

Since n/2 = 32/2 = 16, therefore the class is 25 – 35 is the median.

Median =

Median = 25+16-14 *10 = 27.42

MD from median is 390.32/32 = 12.91

Q20) Calculate the standard deviation using the direct method.

Class interval	Frequency
30 – 39	3
40 – 49	1
50 – 59	8
60 – 69	10
70 – 79	7
80 – 89	7
90 – 99	4

A20)

Class interval	Frequency	Mid point x	Fx	X – x	(x – x ) 2	F (x – x ) 2
30 – 39	3	34.5	103.5	-33.5	1122.25	3366.75
40 – 49	1	44.5	44.5	-23.5	552.25	552.25
50 – 59	8	54.5	436.0	-13.5	182.25	1458
60 – 69	10	64.5	645.0	-3.5	12.25	122.5
70 – 79	7	74.5	521.5	6.5	42.25	295.75
80 – 89	7	84.5	591.5	16.5	272.25	1905.75
90 – 99	4	94.5	378.0	26.5	702.25	2809
	40		2720			10510

Mean = 2720/40 = 68

SD = √10510/40 = 16.20

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