test: NCERT Solutions for Class 11th: Ch 5 Measures of Central Tendency

NCERT Solutions for Class 11th: Ch 5 Measures of Central Tendency Statistics for Economics

Page No: 71

Exercises

1. Which average would be suitable in the following cases?

(i) Average size of readymade garments.

► Mode

(ii) Average intelligence of students in a class.

► Median

(iii) Average production in a factory per shift.

► Arithmetic average

(iv) Average wages in an industrial concern.

► Arithmetic average

(v) When the sum of absolute deviations from average is least.

► Median

(vi) When quantities of the variable are in ratios.
► Arithmetic average

(vii) In case of open-ended frequency distribution.
► Median

2. Indicate the most appropriate alternative from the multiple choices provided against each question.

(i) The most suitable average for qualitative measurement is

(a) arithmetic mean
(b) median
(c) mode
(d) geometric mean
(e) none of the above

► (b) median

Page No: 72

(ii) Which average is affected most by the presence of extreme items?

(a) median
(b) mode
(c) arithmetic mean
(d) none of the above

► (c) arithmetic mean

(iii) The algebraic sum of deviation of a set of n values from A.M. is
(a) n
(b) 0
(c) 1
(d) none of the above

► (b) 0

3. Comment whether the following statements are true or false.

(i) The sum of deviation of items from median is zero.

► False

(ii) An average alone is not enough to compare series.

► True

(iii) Arithmetic mean is a positional value.

► False

(iv) Upper quartile is the lowest value of top 25% of items.

► True

(v) Median is unduly affected by extreme observations.

► False

4. If the arithmetic mean of the data given below is 28, find (a) the missing frequency, and (b) the median of the series:

Profit per retail shop (in Rs)	0-10	10-20	20-30	30-40	40-50	50-60
Number of retail shops	12	18	27	-	17	6

Answer

(a) Let the missing frequency be x
Arithmetic mean = 28 (given)

Profit per retail shop (in Rs) Class Interval	No. of retail shops Frequency (f)	Mid Value (m)	fm
0-10	12	5	60
10-20	18	15	270
20-30	27	25	675
30-40	x	35	35x
40-50	17	45	765
50-60	6	55	330
	Σf = 80 + x		Σfx = 2100 + 35x

Mean = Σfx/Σf
⇒ 28 = 2100 + 35x/80 + x

⇒ 2240 + 28x = 2100 + 35

⇒ 2240 - 2100 = 35x - 25x

⇒ 140 = 7x

⇒ x = 140/7 = 20

Missing frequency = 20

(b)

Class Interval	Frequency (f)	Cumulative frequency (CF)
0-10	12	12
10-20	18	30
20-30	27	57
30-40	x	77
40-50	17	94
50-60	6	100
Total	Σf = 100

Median = Size of (N/2)th item
= 100/2 = 50th item
It lies in class 20-30.

5. The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers	A	B	C	D	E	F	G	H	I	J
Daily Income (in Rs)	120	150	180	200	250	300	220	350	370	260

Answer

Workers	Daily Income (in Rs) X
A	120
B	150
C	180
D	200
E	250
F	300
G	220
H	350
I	370
J	260
Total	ΣX = 2400

N = 10
Arithmetic Mean = ΣX/N
= 2400/10
= 240
Arithmetic Mean = 240

6. Following information pertains to the daily income of 150 families. Calculate the arithmetic mean.

Income (in Rs)	Number of families
More than 75	150
More than 85	140
More than 95	115
More than 105	95
More than 115	70
More than 125	60
More than 135	40
More than 145	25

Answer

Income	No. of families Frequency (f)	Mid Class (x)	fx
75-85	10	80	800
85-95	25	90	2250
95-105	20	100	2000
105-115	25	110	2750
115-125	10	120	1200
125-135	20	130	2600
135-145	15	140	2100
145-155	25	150	3750
	150		17450

Arithmetic Mean = Σfx/Σf
= 17450/150
= 116.33

7. The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Size of Land Holdings (in acres)	Less than 100	100-200	200-300	300-400	400 and above
Number of families	40	89	148	64	39

Answer

Size of Land Holdings Class Interval	Number of families (f)	Cumulative frequency (CF)
0-100	40	40
100-200	89	129
200-300	148	277
300-400	64	341
400-500	39	380
Total	Σf = 380

Σf = N = 380
Median = Size of (N/2)th item
= 380/2 = 190th item
It lies in class 200-300.

Median size of land holdings = 241.22 acres

8. The following series relates to the daily income of workers employed in a firm. Compute (a) highest income of lowest 50% workers (b) minimum income earned by the top 25% workers and (c) maximum income earned by lowest 25% workers.

Daily Income (in Rs)	10-14	15-19	20-24	25-29	30-34	35-39
Number of workers	5	10	15	20	10	5

(Hint: Compute median, lower quartile and upper quartile.)

Answer

Daily Income (in Rs) Class Interval	No of workers (f)	Cumulative frequency (CF)
9.5-14.5	5	5
14.5-19.5	10	15
19.5-24.5	15	30
24.5-29.5	20	50
29.5-34.5	10	60
34.5-39.5	5	65
Total	Σf = 65

(a) Σf = N = 65
Median = Size of (N/2)th item
= 65/2 = 32.5th item
It lies in class 24.5-29.5.

Highest income of lowest 50% workers = Rs 25.12

(b) First, we need to find Q₁
Class interval of Q₁ = (N/4)th items
= (65/4)th items = 16.25th item
It lies in class 19.5-24.5.

Minimum income earned by the top 25% workers = Rs 19.92

(c) First, we need to find Q₃
Class interval of Q₃= 3 (N/4)th items
= 3 (65/4)th items = 3 × 16.25th item
= 48.75th item
It lies in class 24.5-29.5.

Maximum income earned by lowest 25% workers = Rs 29.19

9. The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode values.

Production yield (kg. per hectare)	50-53	53-56	56-59	59-62	62-65	65-68	68-71	71-74	74-77
Number of workers	3	8	14	30	36	28	16	10	5

Answer

Production Yield (kg. per hectare)	No. of farms Frequency (f)	Mid Class (x)	fx	Cumulative frequency (CF)
50-53	3	51.5	154.5	3
53-56	8	54.5	436	11
56-59	14	57.5	805	25
59-62	30	60.5	1815	55
62-65	36	63.5	2286	91
65-68	28	66.5	1862	119
68-71	16	69.5	1112	135
71-74	10	72.5	725	145
74-77	5	75.5	377.5	150
	Σf = 150		Σfx = 9573