Tr Jain & Vk Ohri (2017) for Class 11 Humanities Economics Chapter 11

Question 1:

5 students obtained following marks in statistics:
20, 35, 25, 30, 15
Find out range and coefficient of range.

Answer:

Here,

Highest value (H)= 35
Lowest value (L) = 15

Range= Highest value − Lowest value

i.e. R= H− L

Substituting the given values in the formula

R= 35 − 15= 20

Coefficient of Range is as follows:

$CR = \frac{H - L}{H + L} or, CR = \frac{35 - 15}{35 + 15} = \frac{20}{50} \Rightarrow CR = 0.4$

Hence, the range (R) of the above data is 20 and coefficient of Range (CR) is 0.4

Question 2:

Prices of shares of a company were note as under from Monday through Saturday. Find out range and the coefficient of range.

Day	Mon.	Tues.	Wed.	Thu.	Fri.	Sat.
Price (₹)	200	210	208	160	220	250

Answer:

Here,
Highest value among the prices of shares= 250
Lowest Value among the prices of shares= 160

Range (R) = Highest value (H)− Lowest Value (L)
or, R = 250 − 160
$\Rightarrow$ R = 90
$Coefficient of Range (CR) = \frac{H - L}{H + L} or, CR = \frac{250 - 160}{250 + 160} = \frac{90}{410} \Rightarrow CR = 0.219 or 0.22 (approx .)$

Hence, the Range (R) of the above data is 90 and Coefficient of Range (CR) is 0.22

Question 3:

You know share market is going bullish during the last several months. Collect weekly data on the share price of any two important industries during the past six months. Calculate the range of share prices. Comment on how volatile are the share prices.

Answer:

Month	Price of shares Tata Motors	Price of shares Reliance
Oct. Nov. Dec. Jan. Feb. Mar.	325 397 405 415 420 388	913.35 900.25 750.90 780.70 799.25 850.35

For Tata Motors
Highest Value=420
Lowest Value=325

Range (R) = Highest Value (H)− Lowest Value (L)
or, R_₁ = 420 − 325

\Rightarrow

R_₁ = 95

Coefficient of Range ({CR}_{1}) = \frac{H - L}{H + L} = \frac{420 - 325}{420 + 325} = \frac{95}{745} = 0.127

For Reliance
Highest Value= 913.35
Lowest Value= 750.90

Range (R)= Highest value (H)− Lowest value (L)
or, R_₂ = 913.35 − 750.90

\Rightarrow

R_₂ = 162.45

Coefficient of Range ({CR}_{2}) = \frac{H - L}{H + L} = \frac{913.35 - 750.90}{913.35 + 750.90} = \frac{162.45}{1664.25} = 0.097

From the above results we can observe that the prices of the Tata Motors shares are less volatile as compared to the prices of Reliance shares.

Question 4:

Calculate range and the coefficient of range of the following series:

Marks	10	20	30	40	50	60	70
Number of Students	15	18	25	30	16	10	9

Answer:

Marks	10	20	30	40	50	60	70
No. of Students	15	18	25	30	16	10	9

Here,
Highest value=70
Lowest value=10

Range (R) = Highest value (H) − Lowest Value (L)
= 70 − 10
= 60

Coefficient of Range (CR) = \frac{70 - 10}{70 + 10} = \frac{60}{80} = 0.75

Hence, the Range (R) of the above series is 60 and Coefficient of Range (CR) is 0.75

Question 5:

Find out the range and the coefficient of range from the following data:

Daily Wage (₹)	6	7	8	9	10	11	12	15
Number of Workers	10	15	12	18	25	20	10	4

Answer:

Daily Wage	6	7	8	9	10	11	12	15
No. of workers	10	15	12	18	25	20	10	4

Here,
Highest value =15
Lowest value = 6

Range (R) = Highest value (H) − Lowest value (L)
= 15 − 6
= 9

Coefficient of Range (CR) = \frac{H - L}{H + L} = \frac{15 - 6}{15 + 6} = \frac{9}{21} = 0.429

Hence, the Range (R) of the above series is 9 and Coefficient of Range (CR) is 0.429

Question 1:

Marks obtained by 100 students of a class are given below. Find out range and coefficient of range of the marks.

Marks	10−20	20−30	30−40	40−50	50−60	60−70	70−80	80−90	90−100
Number of Students	4	10	16	22	20	18	8	2	5

Answer:

Marks	No. of Student
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90 90 − 100	4 10 16 22 20 18 8 2 5

Range (R)= Upper limit of last class interval − Lower limit of the first class interval
(R)= 100 − 10
(R) = 90

Coefficient of Range (CR) = \frac{H - L}{H + L} or, CR = \frac{100 - 10}{100 + 10} or, CR = \frac{90}{110} \Rightarrow CR = 0.81

Hence, the Range (R) of the marks is 90 and Coefficient of Range (CR) is 0.8

Question 2:

In an examination, 25 students obtained the following marks. Find out coefficient of range of the marks.

Marks	5−9	10−14	15−19	20−24	25−29	30−34	35−39
Number of Students	1	3	8	5	4	2	2

Answer:

Converting the inclusive series into exclusive series,

Marks	No. of Students
4.5 − 9.5 9.5 − 14.5 14.5 −19.5 19.5 −24.5 24.5 −29.5 29.5 −34.5 34.5 −39.5	1 3 8 5 4 2 2

Here,
Highest Marks value (H)=39.5
Lowest Marks value (L)=4.5

Coefficient of Range (C R) = \frac{H - L}{H + L} or, CR = \frac{39.5 - 4.5}{39.5 + 4.5} or, CR = \frac{35}{44} \Rightarrow CR = 0.79

Hence, the Coefficient of range of marks in the above series is 0.79

Question 1:

Estimate quartile deviation and the coefficient of quartile deviation of the following data:
8, 9, 11, 12, 13, 17, 20, 21, 23, 25, 27
Show that QD is the average of the difference between two quartiles.

Answer:

Sr. No.	1	2	3	4	5	6	7	8	9	10	11
Data	8	9	11	12	13	17	20	21	23	25	27

In order to find the quartile deviation in case of individual series, find out the values of first quartile and third quartile using the following equations:

Q₁ = Size of ${(\frac{N + 1}{4})}^{th}$ item
or, Q₁     = Size of ${(\frac{11 + 1}{4})}^{th}$ item
or, Q₁     = Size of 3^rd item
$\Rightarrow$ Q₁     = 11

Q₃ = Size of $3 {(\frac{N + 1}{4})}^{th}$ item

or, Q₃ = Size of $3 {(\frac{11 + 1}{4})}^{th}$ item
or, Q₃ = Size of 9^th item
$\Rightarrow$    Q₃ = 23

Calculating Quartile Deviation and Coefficient of Quartile Deviation

$Quartile deviation (Q . D .) = \frac{Q_{3} - Q_{1}}{2} or, Q . D . = \frac{23 - 11}{2} or, Q . D . = \frac{12}{2} \Rightarrow Q . D . = 6 Coefficient of quartile deviation = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} o r, Coefficient of Q . D . = \frac{23 - 11}{23 + 11} = \frac{12}{34} = 0.353$

We know that the difference between the third quartile (Q₃₎and first Quartile (Q₁) of a series is called the inter Quartile range i.e.
                             Inter Quartile Range= Q_{3 $-$}Q₁
_{Where as Quartile deviation is half of the Inter Quartile Range. i.e.}_{Quartile Deviation (Q.D.)=} $\frac{Q_{3 -} Q_{1}}{2}$
Hence, from the above formula it is clear that quartile deviation is the average of the difference between the two quartiles.

Question 2:

Find out quartile deviation and coefficient of quartile deviation of the following series:
28, 18, 20, 24, 30, 15, 47, 27

Answer:

Arranging the data in ascending order.

Sr. No.	1	2	3	4	5	6	7	8
Observation	15	18	20	24	27	28	30	47

In order to find the quartile deviation in case of individual series, we need to find out the values of third quartile and first quartile using the following equations:

Q₁ = Size of ${(\frac{N + 1}{4})}^{th}$ item

or, Q₁= Size of ${(\frac{8 + 1}{4})}^{th}$ item

or, Q₁= Size of 2.25^th item

or, Q₁= Size of 2^nd item + $\frac{1}{4} [Size of 3^{rd} item - Size of 2^{nd} item]$
$or, Q_{1} = 18 + \frac{1}{4} [20 - 18]$

$\Rightarrow$ Q₁ = 18.5

Q₃ = Size of $3 {(\frac{N + 1}{4})}^{th}$

or, Q₃= Size of 6.75^thitem
or, Q₃ = Size of 6^th item + $\frac{3}{4} [Size of 7^{th} item - Size of 6^{th} item]$
$or, Q_{3} = 28 + \frac{3}{4} [30 - 28] or, Q_{3} = 27 + \frac{3}{4} [2] or, Q_{3} = 28 + 1.5 \Rightarrow Q_{3} = 29.5 Quartile deviation (Q . D .) = \frac{Q_{3} - Q_{1}}{2} or, Q . D . = \frac{29.5 - 18.5}{2} or, Q . D . = \frac{11}{2} \Rightarrow Q . D . = 5.5 Coefficient of Quartile deviation = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} or, Coefficient of Q . D . = \frac{29.5 - 18.5}{29.5 + 18.5} or, Coefficient of Q . D . = \frac{11}{48} \Rightarrow Coefficient of Q . D . = 0.229 = 0.23$

Hence the Q.D. of the series is 5.5 and Coefficient of the Q.D. is 0.23

Question 1:

Find out quartile deviation and the coefficient of quartile deviation of the following data:

Age	20	30	40	50	60	70	80
Members	3	61	132	153	140	51	3

Answer:

Age	Numbers (f)	Cumulative Frequency (c.f.)
20 30 (Q₁)40 50 (Q₃)60 70 80	3 61 132 153 140 51 3	3 3 + 61 = 64 64 + 132 = 196 196 + 153 = 349 349 + 140 = 489 489 + 51 = 540 540 + 3 = 543
	$Σ f = N = 543$

Q₁ = Size of

{(\frac{N + 1}{4})}^{th}

item
= Size of

{(\frac{543 + 1}{4})}^{th}

item
= Size of 136^th item

136^th item lies in 196^th cumulative frequency of the series. Age corresponding to 196^th(c.f.) is 40.

Hence, Q₁ = 40

Q₃ = Size of

3 (\frac{N + 1}{4}) th

item
= Size of

3 (\frac{543 + 1}{4}) th

item
= Size of 408th item

408th item lies in the 489th c.f. of the series. Age corresponding to 489th c.f. is 60.

Hence, Q₃ = 60.

Calculating Quartile Deviation and Coefficient of Quartile Deviation:

Quartile deviation (Q . D .) = \frac{Q_{3} - Q_{1}}{2} or, Q . D . = \frac{60 - 40}{2} or, Q . D . = \frac{20}{2} \Rightarrow Q . D . = 10 Coefficient of quartile deviation = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} or, Coefficient of Q . D . = \frac{60 - 40}{60 + 40} or, Coefficient of Q . D . = \frac{20}{100} \Rightarrow Coefficient of Q . D . = 0.2

Hence, Q.D. of the series is 10 and Coefficient of Q.D. is 0.2

Question 2:

Estimate quartile deviation and the coefficient of quartile deviation of the following series:

Height (inches)	58	59	60	61	62	63	64	65	66
Number of Students	15	20	32	35	33	22	20	10	8

Answer:

Height (inches)	No. of student (f)	*Cumulative frequency (c.f.)*
58 59 60 61 62 63 64 65 66	15 20 32 35 33 22 20 10 8	15 15 + 20 = 35 35 + 32 = 67 67 + 35 = 102 102 + 33 = 135 135 + 22 = 157 157 + 20 = 177 177 + 10 = 187 187 + 8 = 195
	$Σ f = N = 195$

Q₁ = Size of

{(\frac{N + 1}{4})}^{th}

item
= Size of

{(\frac{195 + 1}{4})}^{th}

item
= Size of 49^th item

49^th item lies in 67^th c.f. of the series. Height corresponding to 67^th (c.f.) is 60.

Hence, Q₁ = 60 inches

Q₃ = Size of

3 {(\frac{N + 1}{4})}^{th}

item
= Size of

3 {(\frac{195 + 1}{4})}^{th}

item
= Size of 147^th item

147^thitem lies in 157^th c.f. of the series. Height corresponding to 157^th c.f. is 63.

Hence, Q₃ = 63 inches

Calculating Quartile Deviation and Coefficient of Quartile Deviation

Quartile deviation (Q . D .) = \frac{Q_{3} - Q_{1}}{2} or, Q . D . = \frac{63 - 60}{2} or, Q . D . = \frac{3}{2} \Rightarrow Q . D . = 1.5 Coefficient of quartile deviation = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} or, Coefficient of Q . D . = \frac{63 - 60}{63 + 60} or, Coefficient of Q . D . = \frac{3}{123} \Rightarrow Coefficient of Q . D . = 0.024

Hence, Q.D. of the series is 1.5 and Coefficient of Q.D. is 0.024

Question 1:

Given the following data, find out quartile deviation and the coefficient of quartile deviation:

Wages (₹)	0−5	5−10	10−15	15−20	20−25	25−30
Number of Workers	4	6	3	8	12	7

Answer:

Wage	No. of worker (f)	*Cumulative Frequency (c.f.)*
0 − 5 5 − 10 10 − 15 15 − 20 20 −25 25 − 30	4 6 3 8 12 7	4 4 + 6 = 10 10 + 3 = 13 13 + 8 = 21 21 + 12 = 33 33 + 7 = 40
	$Σ f = N = 40$

Q₁ = Size of

{(\frac{N}{4})}^{th}

item
= Size of

{(\frac{40}{4})}^{th}

item
= Size of 10^th item

10^thitem lies in group 5 − 10 and falls within 10^th c.f. of the series.

Q_{1} = l_{1} + \frac{\frac{N}{4} - c . f .}{f} \times i

Here,
l_₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval

Thus, Q_{1} = 5 + \frac{10 - 4}{6} \times 5 Q_{1} = 5 + \frac{6}{6} \times 5 \Rightarrow Q_{1} = 10

Like wise,
Q₃ = Size of

3 {(\frac{N}{4})}^{th}

item
= Size of 3

{(\frac{40}{4})}^{th} item

= Size of 30^thitem

30^th item lies in the group 20-25 within the 33^rdc.f. of the series.

Thus, Q_{3} = l_{1} + \frac{3 (\frac{N}{4}) - c . f}{f} \times i or, Q_{3} = 20 + \frac{30 - 21}{12} \times 5 or, Q_{3} = 20 + \frac{9 \times 5}{12} or, Q_{3} = 20 + 3.75 \Rightarrow Q_{3} = 23.75 Quartile deviation (Q . D .) = \frac{Q_{3} - Q_{1}}{2} = \frac{23.75 - 10}{2} = \frac{13.75}{2} = 6.87 Coefficient of quartile deviation = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} = \frac{23.75 - 10}{23.75 + 10} = \frac{13.75}{33.75} = 0.4

Question 2:

Find out quartile deviation and coefficient of quartile deviation from the following data:

Class Interval	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	4	8	5	4	9	10

Answer:

Class interval	Frequency (f)	Cumulative Frequency (c.f.)
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	4 8 5 4 9 10	4 4 + 8 = 12 12 + 5 = 17 17 + 4 = 21 21 + 9 = 30 30 + 10 = 40
	$Σ f$ =N = 40

Q₁= Size of

(\frac{N}{4}) th

item
= Size of

(\frac{40}{4}) th

item
= Size of 10th item

10th item lies in group 10 − 20 and fall within 12th c.f. of the series.

Q_{1} = l_{1} + \frac{\frac{N}{4} - c . f .}{f} \times i

Here,
l₁=Lower limit of the class interval
N= Sum total of the frequencies
c.f.=Cumulative frequency of the class preceding the first quartile class
f= Frequency of the quartile class
i= Class interval

T h u s, Q_{1} = 10 + \frac{10 - 4}{8} \times 10 o r, Q_{1} = 10 + \frac{60}{8} o r, Q_{1} = 10 + 7.5 \Rightarrow Q_{1} = 17.5

Likewise,
Q₃ = Size of

3 (\frac{N}{4}) th

item
= Size of

3 (\frac{40}{4}) th

item
= Size of 30th item

30th item lies in group 40-50 and fall within the 30th c.f. of the series

Thus, Q_{3} = l_{1} + \frac{3 (\frac{N}{4}) - c . f .}{f} \times i o r, Q_{3} = 40 + \frac{30 - 21}{9} \times 10 o r, Q_{3} = 40 + \frac{9}{9} \times 10 o r, Q_{3} = 40 + 10 \Rightarrow Q_{3} = 50 Quartile deviation (Q D) = \frac{Q_{3} - Q_{1}}{2} = \frac{50 - 17.5}{2} = \frac{32.5}{2} = 16.25 Coefficient of Q . D = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} = \frac{50 - 17.5}{50 + 17.5} = \frac{32.5}{67.5} = 0.48

Question 1:

Find out mean deviation of the monthly income of the five families given below, using arithmetic mean of the data:
852, 635, 792, 836, 750

Answer:

Sr. No.	Monthly Income (X)	Deviation from Arithmetic mean $\|d_{X}\| = \|X - X\|$ $\bar{X} =$ 773
1 2 3 4 5	852 635 792 836 750	852-773=79 138 19 63 23
N=5	ΣX = 3865	$Σ \|d_{\bar{X}}\| = 322$

Mean (X) = \frac{Σ X}{N} = \frac{3865}{5} = 773 Mean deviation M D_{\bar{x}} = \frac{Σ |d_{X}|}{N} = \frac{322}{5} = 64.4

Hence, mean deviation of the monthly income is 64.4.

Question 2:

Weight of nine students of a class is given below. Calculate mean deviation, using median and arithmetic mean of the series. Also calculate coefficient of mean deviation:
Weight (kg) : 47, 50, 58, 45, 53, 59, 47, 60, 49

Answer:

Sr. No	Weight (X)	Deviation from Median $\|d m\| = \|X - M\|$ (M = 50)
1 2 3 4 5 6 7 8 9	45 47 47 49 50 53 58 59 60	5 3 3 1 0 3 8 9 10
N=9		$Σ \|d m\| = 42$

Sr. No	Weight (X)	Deviation from Mean $\|d_{X}\| = \|X - \bar{X}\| (X = 52)$
1 2 3 4 5 6 7 8 9	45 47 47 49 50 53 58 59 60	7 5 5 3 2 1 6 7 8
N=9	$\sum_{} X$ =468	$Σ \|d_{\bar{x}}\| = 44$

From Median
(a) M = Size of

(\frac{N + 1}{2}) th

item
= Size of

(\frac{9 + 1}{2}) th

item
= Size of 5th item
= 50 kgs

(b)

{MD}_{m} = \frac{Σ |d m|}{N} = \frac{42}{9} = 4.666 = 4.67 k g s a p p r o x .

(c)

Coefficient of {MD}_{m} = \frac{{MD}_{m}}{M} = \frac{4.67}{50} = 0.0934

From Mean

(a)

X = \frac{Σ X}{N} = \frac{468}{9} = 52 k g s

(b)

{MD}_{X} = \frac{Σ |d_{X}|}{N} = \frac{44}{9} = 4.89 k g s

(c)

Coefficient of {MD}_{X} = \frac{{MD}_{X}}{X} = \frac{4.89}{52} = 0.094

Question 1:

Find out mean deviation and its coefficient of the following data:

Items	5	10	15	20	25	30	35	40
Frequency	8	16	18	22	14	9	6	7

Answer:

Items (X)	Frequency (f)	fX	Deviation from Mean $\|X - X\|$ = $\|d_{X}\|$	$f \|d_{X}\|$
5 10 15 20 25 30 35 40	8 16 18 22 14 9 6 7	40 160 270 440 350 270 210 280	15.2 10.2 5.2 0.2 4.8 9.8 14.8 19.8	121.6 163.2 93.6 4.4 67.2 88.2 88.8 138.6
	Σf = 100	Σfx = 2020		$Σ f \|d X\| = 765.6$

(a)

Mean (\bar{X}) = \frac{Σ f x}{Σ f} = \frac{2020}{100} = 20.2

(b) Mean deviation from Arithmetic Mean

M D_{X} = \frac{Σ f |d_{X}|}{Σ f} = \frac{765.6}{100} = 7.656

(c) Coefficient of

M D_{\bar{X}}

= \frac{M D_{X}}{X} = \frac{7.656}{20.2} = 0.379 = 0.38 a p p r o x .

Question 2:

Calculate mean deviation from the following data, using mean and median, respectively.

Size	4	6	8	10	12	14	16
Frequency	2	4	5	3	2	1	4

Answer:

Calculation of mean deviation from median

Size (X)	Frequency (f)	Cumulative frequency (c.f.)	Deviation from median \|d_M\| = \|X − M\| M = 8	*f\|d_M\|*
4 6 8 10 12 14 16	2 4 5 3 2 1 4	2 2 + 4 = 6 6 + 5 = 11 11 + 3 = 14 14 + 2 = 16 16 + 1 = 17 17 = 4 = 21	4 2 0 2 4 6 8	8 8 0 6 8 6 32
	Σf=N= 21			Σf\|d_M\| = 68

(a) Median or (M) = Size of

(\frac{N + 1}{2}) th

item
= Size of

(\frac{21 + 1}{2}) th

item
= Size of 11th item
= 8

(b) Mean Deviation from median

{MD}_{M} = \frac{Σ f |d_{M}|}{N} = \frac{68}{21} = 3.24

Calculation of mean deviation from mean

Size (X)	Frequency (f)	fX	From mean deviation from mean $\|d_{X}\| = \|X - X\| (X = 9.71)$	$f \|d_{\bar{X}}\|$
4 6 8 10 12 14 16	2 4 5 3 2 1 4	8 24 40 30 24 14 64	5.71 3.71 1.71 0.29 2.29 4.29 6.29	11.42 14.84 8.55 0.87 4.58 4.29 25.16
	Σf = 21	ΣfX = 204		$Σ f \|d \bar{_{X}}\|$ = 69.71

(a)

Mean (\bar{X}) = \frac{Σ f X}{Σ f} = \frac{204}{21} = 9.71

(b) Mean deviation from A.M.

{MD}_{X} = \frac{Σ f |d_{X}|}{Σ f} = \frac{69.71}{21} = 3.32

Question 1:

The following table gives distribution of marks for 50 students of a class. Calculate mean deviation from the mean and median respectively from the data:

Marks Obtained	140−150	150−160	160−170	170−180	180−190	190−200
Frequency	4	6	10	18	9	3

Answer:

Calculation of M.D from Median

Marks	Mid value m	Frequency (f)	Cumulative frequency	\|d_M\| = \|m − M\| M = 172.78	*f\|d_M\|*
140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200	145 155 165 175 185 195	4 6 10 18(f) 9 3	4 4 + 6 = 10 10 + 10 = 20(c.f) 20 + 18 = 38 38 + 9 = 47 47 + 3 = 50	27.78 17.78 7.78 2.22 12.22 22.22	111.12 106.68 77.8 39.96 109.98 66.66
		Σf = N = 50			Σf\|d_M\| = 512.2

Median = Size of

(\frac{N}{2}) th

item
= Size of

(\frac{50}{2}) th

item
= Size of 25th item

25th item lies under the 38th cumulative frequency therefore 170 − 180 is the median class interval.

Thus, median value will be as follows:

M = l_{1} + \frac{\frac{N}{2} - c . f}{f} \times i or, M = 170 + \frac{25 - 20}{18} \times 10 or, M = 170 + \frac{50}{18} or, M = 170 + 2.78 \Rightarrow M = 172.78 H e n c e, Mean deviation {MD}_{M} = \frac{Σ f |d_{M}|}{Σ f} = \frac{512.2}{50} = 10.24

Calculation of M.D from Mean

Marks	Mid value m	Frequency (f)	fm	$\|d_{X}\| = \|m - X\| X = 171.2$	$f \|d_{X}\|$
140 − 150 150 − 160 160 − 170 170 − 180 180 − 190 190 − 200	145 155 165 175 185 195	4 6 10 18(f) 9 3	580 930 1650 3150 1665 585	26.2 16.2 6.2 3.8 13.8 23.8	104.8 97.2 62 68.4 124.2 71.4
		Σf = 50	Σfm = 8560		$Σ f \|d_{X}\| = 528$

Mean (X) = \frac{Σ f m}{Σ f} = \frac{8560}{50} = 171.2

Hence, Mean deviation from the arithmetic mean is:

{MD}_{X} = \frac{Σ f |d_{X}|}{Σ f} = \frac{528}{50} = 10.56

Question 2:

Estimate the coefficient of mean deviation from the median from the following data:

Age Group	20−30	30−40	40−50	50−60	60−70
Number of Workers	8	12	20	16	4

Answer:

Calculation of MD. from Median

Age	Mid value (m)	No. of workers (f)	Cumulative Frequency	\|dm\| = \|m − M\| M = 45	f\|dm\|
20 − 30 30 − 40 40 − 50 50 − 60 60 −70	25 35 45 55 65	8 12 20 16 4	8 8 + 12 = 20 20 + 20 =40 40 + 16 =56 56 + 4 = 60	20 10 0 10 20	160 120 0 160 80
		N=Σf = 60			Σf\|dm\| = 520

Median = Size of

{(\frac{N}{2})}^{th}

item
= Size of

{(\frac{60}{2})}^{th}

item
= Size of 39^thitem

39^thitem lies under the 40^th cumulative frequency of the series. So, 40 − 50 will be the median class interval.

Hence, the median value of the series is:

M = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i or, M = 40 + \frac{30 - 20}{20} \times 10 or, M = 40 + \frac{10 \times 10}{20} \Rightarrow M = 45 Thus, Mean Deviation from median {MD}_{M} = \frac{Σ f |d m|}{Σ f} = \frac{520}{60} = 8.67 Coefficient of {MD}_{M} = \frac{{MD}_{M}}{Median} = \frac{8.67}{45} = 0.19

Question 1:

The following table gives marks obtained by 7 students of a class. Find out standard deviation of the marks.
40, 42, 38, 44, 46, 48, 50

Answer:

S. No.	Marks (X)	Deviation $(x = X - X) X = 44$	Square of the deviation $x^{2} = {(X - X)}^{2}$
1 2 3 4 5 6 7	40 42 38 44 46 48 50	−4 −2 −6 0 2 4 6	16 4 36 0 4 16 36
N = 7	ΣX = 308		Σx² = 112

Mean (X) = \frac{Σ X}{N} = \frac{308}{7} = 44 Standard deviation (σ) = \sqrt{\frac{x^{2}}{N}} = \sqrt{\frac{112}{7}} = \sqrt{16} = 4

Hence, standard deviation of the marks is 4.

Question 2:

Weight of some students is given below in kilograms. Find out standard deviation.
41, 44, 45, 49, 50, 53, 55, 55, 58, 60

Answer:

S. No.	Weight (X)	Deviation $(x = X - X) (X = 51)$	Square of deviation $x^{2} = {(X - X)}^{2}$
1 2 3 4 5 6 7 8 9 10	41 44 45 49 50 53 55 55 58 60	−10 −7 −6 −2 −1 2 4 4 7 9	100 49 36 4 1 4 16 16 49 81
N = 10	ΣX = 510		Σx² = 356

Mean (X) = \frac{Σ X}{N} = \frac{510}{10} = 51 Standard deviation (σ) = \sqrt{\frac{x^{2}}{N}} = \sqrt{\frac{356}{10}} = \sqrt{35.6} = 5.97

Hence, the Standard deviation is 5.97.

Question 3:

Using step-deviation method, calculate standard deviation of the following series:

Marks	0−10	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Number of Students	5	10	20	40	30	20	10	4

Answer:

Marks (X)	Mid value (m)	Frequency (f)	fm	Deviation from mean value $(x = X - X)$ $\bar{X}$ =39.39	x²	fx²
0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 15 25 35 45 55 65 75	5 10 20 40 30 20 10 4	25 150 500 1400 1350 1100 650 300	−34.39 −24.39 −14.39 −4.39 5.61 15.61 25.61 35.61	1182.67 594.87 207.07 19.27 31.47 243.67 655.87 1268.07	5913.35 5948.7 4141.4 770.8 944.1 4873.4 6558.7 5072.28
		Σf = 139	Σfm = 5475			Σfx² = 34222.73

Mean (X) = \frac{Σ f m}{N} = \frac{5475}{139} = 39.39 Standard deviation (σ) = \sqrt{\frac{Σ f x^{2}}{N}} = \sqrt{\frac{34222.73}{139}} = \sqrt{246.21} = 15.7

Hence, the standard deviation of the above series is 15.7

Question 1:

Find out standard deviation of the savings of the following 10 persons:

Persons	1	2	3	4	5	6	7	8	9	10
Savings (₹)	114	108	100	98	101	109	117	119	121	126

Answer:

Persons	Savings (X)	Deviation $(x = X - X)$ $\bar{X}$ =111.3	Square of the deviation x²
1 2 3 4 5 6 7 8 9 10	114 108 100 98 101 109 117 119 121 126	2.7 −3.3 −11.3 −13.3 −10.3 −2.3 5.7 7.7 9.7 14.7	7.29 10.89 127.69 176.89 106.09 5.29 32.49 59.29 94.09 216.09
N = 10	ΣX = 1113		Σx² = 836.1

Mean (X) = \frac{Σ X}{N} = \frac{1113}{10} = 111.3 Standard deviation (σ) = \sqrt{\frac{Σ x^{2}}{N}} = \sqrt{\frac{836.1}{10}} = \sqrt{83.61} = 9.14

Hence, standard deviation of the savings is 9.14

Question 2:

Find out standard deviation and its coefficient of the following series:

Size	10	20	30	40	50	60	70
Frequency	6	8	16	15	33	11	12

Answer:

Size (X)	Frequency (f)	fX	Deviation from mean $(x = X - X)$	Square of deviation x²	fx²
10 20 30 40 50 60 70	6 8 16 15 33 11 12	60 160 480 600 1650 660 840	−34.06 −24.06 −14.06 −4.06 5.94 15.94 25.94	1160.08 578.88 197.68 16.48 35.28 254.08 672.88	6960.48 4631.04 3162.88 247.2 1164.24 2794.88 8074.60
	Σf = 101	Σfx = 4450			Σfx² = 27035.32

Mean (X) = \frac{Σ f X}{Σ f} = \frac{4450}{101} = 44.06 Standard deviation (σ) = \sqrt{\frac{Σ f x^{2}}{Σ f}} = \sqrt{\frac{27035.32}{101}} = \sqrt{267.68} = 16.36 Coefficient of S . D . = \frac{σ}{X} = \frac{16.36}{44.06} = 0.37

Hence, the S.D. is 16.36 and Coefficient of S.D. is 0.37

Question 3:

Find out standard deviation of the distribution of population in 104 villages of a Tehsil, as given below by step-deviation method.

Distribution of Population

Population	No. of Villages
0−200 200−400 400−1,000 1,000−2,000 2,000−5,000	10 28 42 18 6

Answer:

Population	Mid Value (m)	No. of villages (f)	Deviation assumed from average $d_{x} = m - A$	$d_{x}' = \frac{d_{x}}{i} = \frac{d_{x}}{40}$	$f d_{x}' = f \times d_{x}'$	$f d_{x}'^{2} = f \times d_{x}'^{2}$
0-200 200-400 400-1000 1000-2000 2000-5000	100 300 $700 = A$ 1500 3500	10 28 42 18 6	-600 -400 0 800 2800	-15 -10 0 20 70	-150 -280 0 360 420	2250 2800 0 7200 29400
		N=104			$\sum_{} f d_{x}'$ = $-$ 350	$\sum_{} f d_{x}'^{2}$ =41650

Calculating Standard deviation through Step deviation method:

$Standard Deviation (σ) = \sqrt{\frac{Σ f d_{x}'^{2}}{N} - {(\frac{Σ f d_{x}'}{N})}^{2}} \times i or, σ = \sqrt{\frac{41650}{104} - {(\frac{- 350}{104})}^{2}} \times 40 or, σ = \sqrt{400.480 - 11.296} \times 40 or, σ = \sqrt{389.184} \times 40 or, σ = 19.72775 \times 40 \Rightarrow σ = 789.11$

Hence, standard deviation of the above series is 789.11

Question 1:

Calculate standard deviation of the following series:

Size	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	2	5	7	13	21	16	8	3

Answer:

Size	Mid Value m	Frequency	Deviation from assumed Average (dx = m − A)	$d x' = \frac{d x}{C} C = 5$	fdx'	fdx'²
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40	2.5 7.5 12.5 17.5(A) 22.5 27.5 32.5 37.5	2 5 7 13 21 16 8 3	−15 −10 −5 0 5 10 15 20	−3 −2 −1 0 1 2 3 4	−6 −10 −7 0 21 32 24 12	18 20 7 0 21 64 72 48
		N = 75			Σfdx' = 66	Σfdx'² = 250

Calculating Standard deviation through Step deviation method:

Standard deviation (σ) = \sqrt{\frac{Σ f d x'^{2}}{N} - {(\frac{Σ f d x'}{N})}^{2}} \times C or, σ = \sqrt{\frac{Σ f d x'^{2}}{N} - {(\frac{Σ f d x'}{N})}^{2}} \times 5 or, σ = \sqrt{\frac{250}{75} - {(\frac{66}{75})}^{2}} \times 5 or, σ = \sqrt{3.33 - 0.7744} \times 5 or, σ = \sqrt{2.5556} \times 5 or, σ = 1.598 \times 5 \Rightarrow σ = 7.99

Hence, standard deviation of the above series is 7.99.

Question 2:

Calculate standard deviation of the following data, using step-deviation method.

Age	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Frequency	3	61	132	153	140	51	2

Answer:

Age	Mid value m	Frequency (f)	Deviation assumed from average (d_x = m − A)	$d_{x}' = \frac{d_{x}}{c} = \frac{d_{x}}{10}$	$f d_{x}' = f \times d_{x}'$	$f d_{x}'^{2} = f \times d_{x}'^{2}$
20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90	25 35 45 $55 = A$ 65 75 85	3 61 132 153 140 51 2	−30 −20 −10 0 10 20 30	−3 −2 −1 0 1 2 3	−9 −122 −132 0 140 102 6	27 244 132 0 140 204 18
		N = 542			Σfd_x_' = -15	Σfd_x_'² = 765

Standard deviation (σ) = \sqrt{\frac{Σ f d_{x}'^{2}}{N} - {(\frac{Σ f d_{x}'}{N})}^{2}} \times c or, σ = \sqrt{\frac{765}{542} - {(\frac{- 15}{542})}^{2}} \times 10 or, σ = \sqrt{1.41 - {(- 0.03)}^{2}} \times 10 or, σ = \sqrt{1.41 - 0.0009} \times 10 or, σ = 1.19 \times 10 \Rightarrow σ = 11.9

Hence, standard deviation of the above data is 11.9

Question 4:

Calculate mean and standard deviation of the following data by short-cut method:

Class Interval	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Frequency	5	10	15	20	25	18	7

Answer:

C.I.	Mid Value (m)	Frequency (f)	Deviation from Assumed mean d_x= m − A (A = 45)	*fd_x*	$f \times {d_{x}}^{2} = f {d_{x}}^{2}$
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	15 25 35 $45 = A$ 55 65 75	5 10 15 20 25 18 7	−30 −20 −10 0 10 20 30	−150 −200 −150 0 250 360 210	4500 4000 1500 0 2500 7200 6300
		Σf = N = 100		Σfd_x = 320	Σfd_x² = 26000

Calculating Mean and standard deviation through Short - cut method : Mean (X) = A + \frac{Σ f d_{x}}{Σ f} or, X = 45 + \frac{320}{100} or, X = 45 + 3.2 \Rightarrow X = 48.2 S \tan d a r d d e v i a t i o n (σ) = \sqrt{\frac{Σ f {d_{x}}^{2}}{N} - {(\frac{Σ f d_{x}}{N})}^{2}} or, σ = \sqrt{\frac{26000}{100} - {(\frac{320}{100})}^{2}} or, σ = \sqrt{260 - {(3.2)}^{2}} or, σ = \sqrt{260 - 10.24} or, σ = \sqrt{249.76} \Rightarrow σ = 15.80

Question 1:

Following are the marks obtained by 20 students in statistics. Find out coefficient of variation of the marks.

62	85	73	81	74	58	66	72	54	84
65	50	83	62	85	52	80	86	71	75

Answer:

S. No.	Marks (X)	Deviation from mean $(x = X - X)$	Square deviation x²
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20	62 85 73 81 74 58 66 72 54 84 65 50 83 62 85 52 80 86 71 75	−8.9 +14.1 +2.1 10.1 3.1 −12.9 −4.9 1.1 −16.9 13.1 −5.9 −20.9 12.1 −8.9 14.1 −18.9 9.1 15.1 .1 4.1	79.21 198.81 4.41 102.01 9.61 166.41 24.01 1.21 285.61 171.61 34.81 436.81 146.81 79.21 198.81 357.21 82.81 228.01 .01 16.81
	ΣX = 1418		Σx² = 2623.8

Mean (X) = \frac{Σ X}{N} = \frac{1418}{20} = 70.9 Standard deviation (σ) = \sqrt{\frac{x^{2}}{N}} or, σ = \sqrt{\frac{2623.8}{20}} or, σ = \sqrt{131.19} \Rightarrow σ = 11.45 Coefficient of variation (CV) = \frac{σ}{X} \times 100 or, CV = \frac{11.45}{70.9} \times 100 \Rightarrow CV = 16.14

Hence, coefficient of variation (CV) of the marks is 16.14

Question 2:

Calculate coefficient of variation of the following data:

S. No.	1	2	3	4	5	6	7	8
Value	25	42	33	48	45	29	43	39

Answer:

S. No.	Value (X)	Deviation from mean $(x = X - X)$	x²
1 2 3 4 5 6 7 8	25 42 33 48 45 29 43 39	−13 4 −5 10 7 −9 5 1	169 16 25 100 49 81 25 1
N = 8	ΣX = 304		Σx² = 466

Mean (X) = \frac{ΣX}{N} = \frac{304}{8} = 38 Standard deviation (σ) = \sqrt{\frac{Σ x^{2}}{N}} = \sqrt{\frac{466}{8}} = \sqrt{58.25} = 7.63 Coefficient of variation (CV) = \frac{σ}{X} \times 100 = \frac{7.63}{38} \times 100 = \frac{763}{38} = 20.08

Hence, coefficient of variation is 20.08

Question 3:

Given the following data, calculate coefficient of variation:

Age	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Number of Students	3	61	132	153	140	51	2

Answer:

Age	Mid Value (m)	Frequency (f)	Deviation from Assumed mean (d_x = X − A) (A = 55)	$d_{x}' = \frac{d_{x}}{i} = \frac{d_{x}}{10}$	$f \times d_{x}' = f d_{x}'$	$f \times d_{x}'^{2} = f d_{x}'^{2}$
20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90	25 35 45 $55 = A$ 65 75 85	3 61 132 153 140 51 2	−30 −20 −10 0 10 20 30	−3 −2 −1 0 1 2 3	−9 −122 −132 0 140 102 6	27 244 132 0 140 204 18
		N = 542			Σfd_x'= -15	Σfd_x'² =765

Mean (X) = A + \frac{Σ f d_{x}'}{N} \times i o r, X = 55 + \frac{(- 15)}{542} \times 10 o r, X = 55 - . 28 \Rightarrow X = 54.72 Standard Deviation (σ) = \sqrt{\frac{Σ f d_{x}'^{2}}{N} - {(\frac{Σ f d_{x}'}{N})}^{2}} \times i o r, σ = \sqrt{\frac{765}{542} - {(\frac{- 15}{542})}^{2}} \times 10 o r, σ = \sqrt{1.41 - {(- 0.03)}^{2}} \times 10 o r, σ = \sqrt{1.41 - 0.0009} \times 10 o r, σ = 1.187 \times 10 \Rightarrow σ = 11.87 Coefficient of Variation (C V) = \frac{σ}{X} \times 100 o r, C V = \frac{11.87}{54.72} \times 100 o r, C V = \frac{1190}{54.72} \Rightarrow C V = 21.69

Hence, Coefficient of variation is 21.69

Question 2:

Make a Lorenz curve of the following data:

Income	500	1,000	2,000	3,000	3,500
Number in Class 'A' (000)	4	6	8	12	10
Number in Class 'B' (000)	8	7	5	3	2

Answer:

			CLASS A			CLASS B
Income (x)	Cumulative sum	Cumulative %	No. of class	c.f.	Cumulative %	No. of class	c.f.	cumulative
500 1000 2000 3000 3500	500 1500 3500 6500 10,000	5 15 35 65 100	4 6 8 12 10	4 10 18 30 40	10 25 45 75 100	8 7 5 3 2	8 15 20 23 25	32 60 80 92 100

Question 1:

The following table shows number of firms in two different areas 'A' and 'B' according to their annual profits. Present the data by way of Lorenz curve.

Profit ('000' ₹)	6	25	60	84	105	150	170	400
Firms in Area A	6	11	13	14	15	17	10	14
Firms in Area B	2	38	52	28	38	26	12	4

Answer:

			Firm in Area A			Firms in Area B
Profit (x)	Cumulative sum	Cumulative %	No. of firm	c.f	Cumulative %	No. of firms	c.f.	Cumulative %
6 25 60 84 105 150 170 400	6 31 91 175 280 430 600 1000	.6 3.1 9.1 17.5 28 43 60 100	6 11 13 14 15 17 10 14	6 17 30 44 59 76 86 100	6 17 30 44 59 76 86 100	2 38 52 28 38 26 12 4	20 40 92 120 158 184 196 200	10 20 46 60 79 82 98 100
Σx = 1000

Question 1:

Calculate range and coefficient of range from the following data:
4, 7, 8, 46, 53, 77, 8, 1, 5, 13

Answer:

Given:
Highest value (H) = 77
Lowest value (L) = 1
Range = Highest value − Lowest value
i.e. R = H − L
Substituting the given values in the formula.
R = 77 − 1 = 76
$Coefficient of Range = \frac{H - L}{H + L} = \frac{77 - 1}{77 + 1} = \frac{76}{78} = 0.97$
Thus, range is 76 and coefficient of range is 0.97

Question 2:

Given the following data-set, calculate range and the coefficient of range:

Size	4.5	5.5	6.5	7.5	8.5	9.5	10.5	11.5
Frequency	4	5	6	3	2	1	3	5

Answer:

Highest value (H) = 11.5
Lowest value (L) = 4.5
Range = Highest value − Lowest value
i.e. R = H − L
or, R = 11.5 − 4.5 = 7
$Coefficient of Range = \frac{H - L}{H + L} = \frac{11.5 - 4.5}{11.5 + 4.5} = \frac{7}{16} = 0.437$
Note- As the given data series is a discrete series, where in order to calculate Range, frequencies are not taken into account. Hence, in this case, the formula for Range remains the same as it is for the individual series.

Question 3:

Find out quartile deviation and the coefficient of range, given the following data-set:

Class Interval	1−5	6−10	11−15	16−20	21−25	26−30	31−35
Frequency	2	8	15	35	20	10	14

Answer:

In order to calculate range and its coefficient, we must first convert the given inclusive class intervals in exclusive class intervals.

Class Interval	Exclusive Class Interval	$Mid Value = \frac{L_{1} + L_{2}}{2}$	Frequency
1 − 5 6 − 10 11 − 15 16 − 20 21 − 25 26 − 30 31 − 35	0.5 − 5.5 5.5 − 10.5 10.5 − 15.5 15.5 − 20.5 20.5 − 25.5. 25.5. − 30.5 30.5 − 35.5	3 8 13 18 23 28 33	2 8 15 35 20 10 14

Range = Upper Limit of the Highest Class Interval - Lower Limit of the Lowest Class Interval or, Range = 35.5 - 0.5 = 35

Coefficient of Range = \frac{H - L}{H + L} = \frac{35.5 - 0.5}{35.5 + 0.5} = \frac{35}{36} = 0.972

Question 4:

Find out quartile deviation and the coefficient of quartile deviation of the following series.
Wages of 9 workers in Rupees:
170, 82, 110, 100, 150, 150, 200, 116, 250

Answer:

Arranging the data in the ascending order as presented below.
82, 100, 110, 116, 150, 150, 170, 200, 250
Here, N = 9
$Q_{1} = size of {(\frac{9 + 1}{4})}^{t h} = size of 2 . 5^{th} item = size of 2^{n d} item + 0.5 (size of 3^{r d} item - size of 2^{n d} item) = 100 + 0.5 (110 - 100) = 105 Q_{3} = size of 3 {(\frac{9 + 1}{4})}^{t h} item = size of 7 . 5^{th} item = size of 7^{t h} item + 0.5 (size of 8^{t h} item - size of 7^{th} item) = 170 + 0.5 (200 - 170) = 185 Q . D . = \frac{Q_{3} - Q_{1}}{2} = \frac{185 - 105}{2} = 40$

$Coefficient of Q . D . = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} = \frac{185 - 105}{185 + 105} = 0.276$

Question 5:

Given the following data, estimate the coefficient of QD:
15, 20, 23, 23, 25, 25, 27, 40

Answer:

Given data series:
15, 20, 23, 23, 25, 25, 27, 40
Here, N = 8
$Q_{1} = size of {(\frac{8 + 1}{4})}^{t h} = size of 2 . 25^{th} item = size of 2^{n d} item + 0.25 (size of 3^{r d} item - size of 2^{n d} item) = 20 + 0.25 (23 - 20) = 20.75 Q_{3} = size of 3 {(\frac{8 + 1}{4})}^{t h} item = size of 6 . 75^{th} item = size of 6^{t h} item + 0.75 (size of 7^{t h} item - size of 6^{th} item) = 25 + 0.75 (27 - 25) = 26.5$
$Coefficient of Q . D . = \frac{Q_{3} - Q_{1}}{Q_{3} + Q_{1}} = \frac{26.5 - 20.75}{26.5 + 20.75} = 0.121$

Question 6:

Find out mean deviation of the following series from mean and median:

Size	4	6	8	10	12	14	16
Frequency	2	4	5	31	2	1	4

Answer:

x	f	fx	$d_{X} = \|X - X\|$	$f d_{X}$
4 6 8 10 12 14 16	2 4 5 31 2 1 4	8 24 40 310 24 14 64	5.87 3.87 1.87 0.13 2.13 4.13 6.13	11.74 15.48 9.35 4.03 4.26 4.13 24.52
	∑f = 49	∑fx = 484		$\sum_{} f d_{X}$ = 73.51

Mean (X) = \frac{\sum_{} f x}{\sum_{} f} = \frac{484}{49} = 9.87 Thus calculate the deviation of the values from 9.87 . Mean deviation from mean (M D_{X}) = \frac{\sum_{} f d_{X}}{\sum_{} f} = \frac{73.51}{49} = 1.50

x	f	c.f.	d_M = \|x − M\|	*f\|d_M\|*
4 6 8 10 12 14 16	2 4 5 31 2 1 4	2 6 11 42 44 45 49	6 4 2 0 2 4 6	12 16 10 0 4 4 24
	∑f = 49			∑f\|d_M\| = 70

N = 49

Median = size of {(\frac{N + 1}{2})}^{t h} or, Median = size of {(\frac{49 + 1}{2})}^{t h} item = size of 25^{t h} item

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25^th is 42 (in the c.f. column), which is corresponding to 10.
Hence, median is 10.
Thus, we calculate the deviation of the values from 10.

Mean Deviation from median = (M D_{M}) = \frac{\sum_{} f |d_{M}|}{\sum_{} f} = \frac{70}{49} = 1.428 (or 1.43 approximately)

Question 7:

Calculate mean deviation and coefficient of mean deviation with the help of median:

Class Interval	0−10	10−20	20−30	30−40	40−50	50−60
Frequency	15	19	14	20	18	14

Answer:

x	Mid Value (m)	f	c.f.	\|m−M\| = \|d_M\|	*f\|d_M\|*
0 − 10 10 − 20 20 − 30	5 15 25	15 19 14	15 34 $48$	26 16 6	390 304 80
$30 - 40$	35	$20$	68	4	252
40 − 50 50 − 60	45 55	18 14	86 100	4 14	252 336
		∑fx = 100			∑f\|d_M\| = 1446

Here, N or ∑fx = 100
Median class is given by the size of

{(\frac{N}{2})}^{t h}

item, i.e.

{(\frac{100}{2})}^{t h}

item, which is 50^th item.
This corresponds to the class interval of 30 − 40, so this is the median class.

Median = l_{1} + \frac{\frac{N}{2} - c . f .}{f} \times i so, Median = 30 + \frac{\frac{100}{2} - 48}{20} \times 10 or, Median = 30 + \frac{2}{20} \times 10 or, Median = 30 + 1 Therefore, Median = 31

Thus, we calculate the deviation of values from 31.

Mean Deviation from median = (M D_{M}) = \frac{\sum_{} f |d_{M}|}{\sum_{} f} = \frac{1446}{100} = 14.46 Coefficient of Mean Deviation = \frac{Mean Deviation}{Median} = \frac{14.46}{31} = 0.466

Question 8:

Calculate mean deviation from mean of the following series:

Size of Items	3−4	4−5	5−6	6−7	7−8	8−9	9−10
Frequency	3	7	22	60	85	32	9

Answer:

Class Interval	Mid Value (m)	f	fm	$\|m - X\| = \|d_{\bar{X}}\|$	$f \|d_{X}\|$
3 − 4 4 − 5 5 − 6 6 − 7 7 − 8 8 − 9 9 − 10	3.5 4.5 5.5 6.5 7.5 8.5 9.5	3 7 22 60 85 32 9	10.5 31.5 121 390 637.5 272 85.5	3.6 2.6 1.6 0.6 0.4 1.4 2.4	10.8 18.2 35.2 36 34 44.8 21.6
		∑f = 218	∑fm = 1548		$\sum_{} f \|d_{X}\|$ = 200.6

Mean (X) = \frac{\sum_{} f m}{\sum_{} f} = \frac{1548}{218} = 7.10 Thus, we calcualte the value of deviations from 7.10 Mean Deviation from Mean (M D_{X}) = \frac{\sum_{} f |d_{X}|}{\sum_{} f} = \frac{200.6}{218} = 0.920

Question 9:

Given below are the marks obtained by the students of a class. Calculate mean deviation, and its coefficient, using median of data.

Marks

17

35

38

16

42

27

19

11

40

25

Answer:

The given data is an individual series, therefore, to calculate median, we first need to arrange the data in ascending order. This is done as below.

$11, 16, 17, 19, 25, 27, 35, 38, 40, 42 Here, N = 10 (e v e n) Median = \frac{size of {(\frac{N}{2})}^{t h} item + size of {(\frac{N}{2} + 1)}^{t h} item}{2} or, Median = \frac{size of 5^{t h} item + size of 6^{t h} item}{2} = \frac{25 + 27}{2} = 26 so, Median is 26 .$
Thus, we calculate the deviation of values from 26

X	\|x - M*\| = \|d_M\|*
11 16 17 19 25 27 35 38 40 42	15 10 9 7 1 1 9 12 14 16
	∑*\|d_M\|* = 94

Mean Deviation from Median (M D_{M}) = \frac{\sum_{} |d_{M}|}{n} = \frac{94}{10} = 9.4 C oefficient of M D_{M} = \frac{M D_{M}}{M e d i a n} = \frac{9.4}{26} = 0.36

Question 10:

Nine students of a class obtained following marks. Calculate mean deviation from median.

S. No.	1	2	3	4	5	6	7	8	9
Marks	68	49	32	21	54	38	59	66	41

Answer:

The given data series is discrete in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.
$21, 32, 38, 41, 49, 54, 59, 66, 68 Here, N = 9 (o d d) Median = size of (\frac{N + 1}{2}) = (\frac{9 + 1}{2}) = size of 5$ ^th item
Thus, we calculate the deviation of values from 49.

Marks	Ascending Order	*\|X- M\| = \|d_M\|*
68 49 32 21 54 38 59 66 41	21 32 38 41 49 54 59 66 68	28 17 11 8 0 5 10 17 19
		∑*\|d_M\|* = 115

Mean Deviation from Median (M D_{M}) = \frac{\sum_{} |d_{M}|}{N} = \frac{115}{9} = 12.77

Question 11:

Following data relate to the age-difference of husbands and wives of a particular community. Find out mean deviation from mean.

Age-difference	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	449	705	507	281	109	52	16	4

Answer:

X	Mid Value m	f	fm	$\|m - X\| = \|d_{X}\|$	$f \|d_{X}\|$
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40	2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5	449 705 507 281 109 52 16 4	1122.5 5287.5 6337.5 4917.5 2452.5 1430 520 150	7.96 2.96 2.04 7.04 12.02 17.04 22.04 27.04	3574.04 2086.8 1034.28 1978.24 1312.36 886.08 352.64 108.16
		∑f = 2123	∑fm = 22217.5		∑ $f \|d_{X}\|$ = 11332.6

Mean (X) = \frac{\sum_{} f m}{\sum_{} f} = \frac{22217.5}{2123} = 10.46 Mean Deviation from Mean (M D_{X}) = \frac{\sum_{} f |d_{X}|}{X} = \frac{11332.6}{2123} = 5.34

Question 12:

Find out the mean deviation and its coefficient using median of the following data:

S. No.	1	2	3	4	5	6	7	8	9	10	11	12
Number of Victims of Accidents	16	21	10	17	8	4	2	1	2	2	2	2

Answer:

The given data series is individual in nature, therefore, to calculate median, we first have to arrange the data in ascending order. This is done as below.

$1, 2, 2, 2, 2, 2, 4, 8, 10, 16, 17, 21 Here, N = 12 (e v e n) Median = \frac{size of {(\frac{N}{2})}^{t h} item + size of {(\frac{N}{2} + 1)}^{t h} item}{2} or, Median = \frac{size of 6^{t h} item + size of 7^{t h} item}{2} = \frac{2 + 4}{2} = 3 so, Median is 3 .$

No. of Victims	Ascending Order	\|x - M*\| = \|d_M\|*
16 21 10 17 8 4 2 1 2 2 2 2	1 2 2 2 2 2 4 8 10 16 17 21	2 1 1 1 1 1 1 5 7 13 14 18
		∑*\|d_M\|* = 65

M e a n D e v i a t i o n f r o m M e d i a n (M D_{M}) = \frac{\sum_{} |d_{M}|}{n} = \frac{65}{12} = 5.42 C oefficient of M D_{M} = \frac{M D_{M}}{M e d i a n} = \frac{5.42}{3} = 1.81

Question 13:

Calculate standard deviation, given the following data:
10, 12, 14, 16, 18, 22, 24, 26, 28

Answer:

X	$x = (X - X)$	x²
10 12 14 16 18 22 24 26 28	−8.88 −6.88 −4.88 −2.88 −.88 3.12 5.12 7.12 9.12	78.85 47.33 23.81 8.29 0.77 9.73 26.21 50.69 83.17
∑X = 170		∑x² = 328.85

Mean (X) = \frac{\sum_{} X}{N} = \frac{170}{9} = 18.88 Standard Deviation (σ) = \sqrt{\frac{\sum_{} x^{2}}{N}} = \sqrt{\frac{328.85}{9}} = 6.045

Question 14:

Calculate standard deviation and the coefficient of standard deviation, given the following data:

Income (₹)	5	10	15	20	25	30	35	40
Number of Workers	26	29	40	35	26	18	14	12

Answer:

Income (X)	No. of workers (f)	fX	$x = X - X$	x²	fx²
5 10 15 20 25 30 35 40	26 29 40 35 26 18 14 12	130 290 600 700 650 540 490 280	−14.4 −9.4 −4.4 0.6 5.6 10.6 15.6 20.6	207.36 88.36 19.36 0.36 31.36 112.36 243.36 424.36	5391.36 2562.44 774.4 12.6 815.36 2022.48 3407.04 5092.32
	∑f = 200	∑fX = 3880			∑fx² = 20078

X = \frac{\sum f X}{\sum f_{}} = \frac{3880}{200} = 19.4 Standard Deviation (σ) = \sqrt{\frac{\sum f x^{2}}{\sum f}} = \sqrt{\frac{20078}{200}} = 10.02 Coefficient of Standard Deviation = \frac{σ}{X} = \frac{10.02}{19.4} = 0.516

Question 15:

Of the two sets of income distribution of five and seven persons respectively, as given below calculate standard deviation:

(i) Income (₹)	4,000	4,200	4,400	4,600	4,800
(ii) Income (₹)	3,000	4,000	4,200	4,400	4,600	4,800	5,800

Answer:

For Income Group I

X₁	$x_{1} = X_{1} - X_{1}$	$x_{I}^{2}$
4000 4200 4400 4600 4800	−400 −200 0 200 400	160000 40000 0 40000 160000
∑X₁ = 22,000		∑ $x_{I}^{2}$ = 4,00,000

X_{1} = \frac{\sum_{} X_{1}}{n_{1}} = \frac{22000}{5} = 4400 Standard Deviation (σ_{1}) = \sqrt{\frac{\sum_{} x_{1}^{2}}{n_{1}}} = \sqrt{\frac{4, 00, 000}{5}} = 282.84 Thus, Standard Deviation for the first group of income is Rs 282.84

For Income Group II

X₂	$x_{2} = X_{2} - X_{2}$	$x_{2}^{2}$
3000 4000 4200 4400 4600 4800 5800	−1400 −400 −200 0 200 400 1400	1960000 160000 40000 0 40000 160000 1960000
30,800		∑ $x_{2}^{2}$ = 4320000

X_{2} = \frac{\sum_{} X_{2}}{n_{2}} = \frac{30, 800}{7} = 4, 400 Standard Deviation (σ_{2}) = \sqrt{\frac{\sum_{} x_{2}^{2}}{n_{2}}} = \sqrt{\frac{4, 32, 000}{7}} = 785.58 Thus, Standard Deviation for the second group of income is Rs 785.58

Question 16:

Find out the standard deviation of the marks secured by 10 students:

S. No.	1	2	3	4	5	6	7	8	9	10
Marks	43	48	65	57	31	60	37	48	78	59

Answer:

S.No.	Marks (X)	$x = (X - X)$	x²
1 2 3 4 5 6 7 8 9 10	43 48 65 57 31 60 37 48 78 59	−9.6 −4.6 12.4 4.4 −21.6 7.4 −15.6 −4.6 25.4 6.4	92.16 21.16 153.76 19.36 466.56 54.76 243.36 21.16 645.16 40.96
	∑X = 526		∑x² = 1758.4

Mean (X) = \frac{\sum_{} X}{N} = \frac{526}{10} = 52.6 Standard Deviation (σ) = \sqrt{\frac{\sum_{} x^{2}}{N}} = \sqrt{\frac{1758.4}{10}} = 13.26

Question 17:

Data of daily sale proceeds of a shop are as below. Calculate mean deviation and standard deviation.

Daily Sales	102	100	110	114	118	122	126
Days	3	9	25	35	17	10	1

Answer:

Daily Sales (X)	Days (f)	fX	$\|X - X\| = \|d_{X}\|$	$f \|d_{X}\|$ or $f \|X - X\|$	$f {\|X - X\|}^{2} or f {\|d_{X}\|}^{2}$
102 100 110 114 118 122 126	3 9 25 35 17 10 1	306 900 2750 3990 2006 1220 126	10.98 12.98 2.98 1.02 5.02 9.02 13.02	32.94 116.82 745 35.7 85.34 90.2 13.02	361.68 1516.32 222 36. 428.4 813.6 169.52
	∑f = 100	∑fX = 11298		∑ $f \|d_{X}\|$ = 1119.02	∑ = 3547.92

Mean (X) = \frac{\sum_{} f X}{\sum_{} f} = \frac{11298}{100} = 112.98 Mean Deviation from mean (M D_{X}) = \frac{\sum_{} f d_{X}}{\sum_{} f} = \frac{1119.02}{100} = 11.19 Standard Deviation (σ) = \sqrt{\frac{\sum_{} f {|X - X|}^{2}}{\sum_{} f}} = \sqrt{\frac{3547.92}{100}} = 5.96

Question 18:

Calculate range, standard deviation and coefficient of variation of marks secured by students.

50

55

57

49

54

61

64

59

58

56

Answer:

X	$x = (X - X)$	x²
50 55 57 49 54 61 64 59 58 56	−6.3 −1.3 .7 −7.3 −2.3 4.7 7.7 2.7 1.7 −.3	39.69 1.69 .49 53.29 5.29 22.09 59.29 7.29 2.89 .09
$\sum_{}^{} X = 563$		$\sum_{}^{} x^{2} = 192.1$

Here,
N = 10
Highest value (H) = 64
Lowest value (Z) = 49
Range (R) = H − L = 64 − 49 = 15
So, range is 15.

Mean (X) = \frac{\sum_{} X}{N} = \frac{563}{10} = 56.3 Standard Deviation (σ) = \sqrt{\frac{\sum_{} x^{2}}{N}} = \sqrt{\frac{192.1}{10}} = 4.38 Coefficient of Variation = \frac{σ}{\bar{X}} \times 100 = \frac{4.38}{56.3} \times 100 = 7.78

Question 19:

Following data show the number of runs made by Sachin and Sourabh in different innings. Find out who is a good scorer and who is a consistent player?

Sachin	92	17	83	56	72	76	64	45	40	32
Sourabh	28	70	31	00	59	108	82	14	3	95

Answer:

Sachin			Saurav
X₁	x₁	x₁²	X₂	x₂	x₂²
92 17 83 56 72 76 64 45 40 30	34.3 −40.7 25.3 −1.7 14.3 18.3 6.3 −12.7 −17.7 −25.7	1176.49 1656.49 640.09 2.89 204.49 334.89 39.69 161.29 313.29 660.49	28 70 31 00 59 108 82 14 3 95	−27 25 −24 −55 4 53 27 −41 −52 40	729 625 576 3025 16 2809 729 1681 2704 1600
∑X₁= 577		∑x₁² = 5190.1	∑X₂ = 550		∑x₂² = 14494

X_{S a c h i n} = \frac{\sum_{}^{} X_{1}}{n_{1}} = \frac{577}{10} = 57.7 σ_{S a c h i n} = \sqrt{\frac{\sum_{}^{} x_{1}^{2}}{n_{1}}} = \sqrt{\frac{5190.1}{10}} = 22.78 Coefficient of Variation (Sachin) = \frac{σ_{S a c h i n}}{X_{S a c h i n}} \times 100 = \frac{22.78}{57.7} \times 100 = 39.48 X_{S a u r a v} = \frac{\sum_{}^{} X_{2}}{n_{2}} = = \frac{550}{10} = 55 σ_{S a u r a v} = \sqrt{\frac{\sum_{}^{} x_{2}^{2}}{n_{2}}} = \sqrt{\frac{14494}{10}} = 38.07 Coefficient of Variation (Saurav) = \frac{σ_{S a u r a v}}{X_{S a u r a v}} \times 100 = \frac{38.07}{55} \times 100 = 69.21

Observation and Conclusion

In order to regard anyone as a better batsman, we have to analyse the mean runs of both the players. Analysing the mean runs, we can say that Sachin is a better scorer, since his mean runs (average of 57.7) is higher than that of Saurav (average of 55).
In order to regard one as more consistent, we have to take the help of coefficient of variation. The one who has lower coefficient of variation will be regarded as more consistent than the other. In this way, we can say that Sachin is more consistent than Saurav, as the coefficient of variation of Sachin (39.48) is lower than that of Saurav's (69.21).

Thus, on the basis of the above two observations, we can conclude that Sachin is not only a better scorer, but also a more consistent player than Saurav.

Question 20:

Calculate standard deviation of marks secured by 100 examinees in the examination:

Marks	10−20	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Number of Examinees	19	3	2	49	24	2	0	1

Answer:

X	Mid Values m	f	fX	$x = X - X$	x²	fx²
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 − 90	15 25 35 45 55 65 75 85	19 3 2 49 24 2 0 1	285 75 70 2205 1320 130 0 85	−26.70 −16.70 −6.7 3.3 13.3 23.3 33.3 43.3	712.89 278.89 44.89 10.89 176.89 542.89 1108.89 1874.89	13544.91 836.67 89.78 533.61 4245.36 1085.78 0 1874.89
		∑f = 100	∑fX = 4170			∑fx² = 22211

X = \frac{\sum_{}^{} f X}{\sum_{}^{} f} = \frac{4170}{100} = 41.70 Standard Deviation (σ) = \sqrt{\frac{\sum_{}^{} f x^{2}}{\sum_{}^{} f}} = \sqrt{\frac{22211}{100}} = 14.90

Question 21:

Calculate coefficient of variation from the following data:

Variables	10	20	30	40	50	60	70
Frequencies	6	8	16	15	32	11	12

Answer:

X	f	fX	$x = X - X$	x²	fx²
10 20 30 40 50 60 70	6 8 16 15 32 11 12	60 160 480 600 1600 660 840	−34 −24 −14 −4 6 16 26	1156 576 196 16 36 256 676	6936 4608 3136 240 1152 2816 8112
	∑f = 100	∑fX = 4400			∑f x² = 27000

X = \frac{\sum_{}^{} f X}{\sum_{}^{} f} = \frac{4400}{100} = 44 Standard Deviation (σ) = \sqrt{\frac{\sum_{}^{} f x^{2}}{\sum_{}^{} f}} = \sqrt{\frac{27000}{100}} = 16.43 Coefficient of Variation = \frac{σ}{X} \times 100 = \frac{27000}{100} \times 100 = 37.34

Question 22:

Estimate coefficient of variation of the following data:

Weight (kg)	0−20	20−40	40−60	60−80	80−100
Number of Persons	81	40	66	49	14

Answer:

Class Interval	Mid Value (m)	f	fm	$x = X - X$	x²	fx²
0 − 20 20 − 40 40 − 60 60 − 80 80 − 100	10 30 50 70 90	81 40 66 49 14	810 1200 3300 3430 1260	−30 −10 10 30 50	900 100 100 900 2500	72900 4000 6600 44100 35000
		∑f = 250	∑fm = 10000			∑fx² = 162600

X = \frac{\sum_{}^{} f m}{\sum_{}^{} f} = \frac{10000}{250} = 40 Standard Deviation (σ) = \sqrt{\frac{\sum_{}^{} f x^{2}}{\sum_{}^{} f}} = \sqrt{\frac{162600}{250}} = 25.50 Coefficient of Variation = \frac{σ}{X} \times 100 = \frac{25.50}{40} \times 100 = 63.75

Tr Jain & Vk Ohri (2017) for Class 11 Humanities Economics Chapter 11 - Measures Of Dispersion

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