Tr Jain & Vk Ohri (2017) Solutions for Class 11 Humanities Economics Chapter 11 Measures Of Dispersion are provided here with simple step-by-step explanations. These solutions for Measures Of Dispersion are extremely popular among Class 11 Humanities students for Economics Measures Of Dispersion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Tr Jain & Vk Ohri (2017) Book of Class 11 Humanities Economics Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Tr Jain & Vk Ohri (2017) Solutions. All Tr Jain & Vk Ohri (2017) Solutions for class Class 11 Humanities Economics are prepared by experts and are 100% accurate.

#### Question 1:

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

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:

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

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
$⇒$   R  = 90

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.

 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   = 420 − 325
$⇒$  R   = 95

For Reliance
Highest Value= 913.35
Lowest Value= 750.90

Range (R)= Highest value (H)− Lowest value (L)
or, R   = 913.35 − 750.90
$⇒$ R   = 162.45

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

 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

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

 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

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

 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

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

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

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.

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

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

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

Calculating Quartile Deviation and Coefficient of Quartile Deviation

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= Q3$-$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

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 ${\left(\frac{N+1}{4}\right)}^{\mathrm{th}}$ item

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

or, Q1= Size of 2.25th item

or, Q1= Size of 2nd item +

$⇒$ Q1 = 18.5

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

or, Q3 = Size of 6.75th item
or, Q3 = Size of 6th item +

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

 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 $\mathrm{\Sigma }f=N=543$

Q1 = Size of item
= Size of ${\left(\frac{543+1}{4}\right)}^{\mathrm{th}}$ item
= Size of 136th item

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

Hence, Q1 = 40

Q3 = Size of $3\left(\frac{N+1}{4}\right)\mathrm{th}$ item
= Size of $3\left(\frac{543+1}{4}\right)\mathrm{th}$ 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:

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

 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 $\mathrm{\Sigma }f=N=195$

Q1 = Size of ${\left(\frac{N+1}{4}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{195+1}{4}\right)}^{\mathrm{th}}$ item
= Size of 49th item

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

Hence, Q1 = 60 inches

Q3 = Size of $3{\left(\frac{\mathrm{N}+1}{4}\right)}^{\mathrm{th}}$ item
= Size of $3{\left(\frac{195+1}{4}\right)}^{\mathrm{th}}$ item
= Size of 147th item

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

Hence, Q3 = 63 inches

Calculating Quartile Deviation and Coefficient of Quartile Deviation

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

 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 $\mathrm{\Sigma }f=N=40$

Q1 = Size of ${\left(\frac{N}{4}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{40}{4}\right)}^{\mathrm{th}}$ item
= Size of 10th item

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

Here,
l=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

Like wise,
Q3 = Size of $3{\left(\frac{\mathrm{N}}{4}\right)}^{\mathrm{th}}$ item
= Size of 3
= Size of 30th item

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

#### 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

 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 $\Sigma f$=N = 40

Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{40}{4}\right)\mathrm{th}$ item
= Size of 10th item

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

${\mathrm{Q}}_{1}={l}_{1}+\frac{\frac{\mathrm{N}}{4}-c.f.}{f}×i\phantom{\rule{0ex}{0ex}}$

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

Likewise,
Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item
= Size of $3\left(\frac{40}{4}\right)\mathrm{th}$ item
= Size of 30th item

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

#### 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

 Sr. No. Monthly Income (X) Deviation from Arithmetic mean $\left|{\mathbit{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}\right|$  $\overline{\mathbit{X}}=$773 1 2 3 4 5 852 635 792 836 750 852-773=79 138 19 63 23 N=5 ΣX = 3865 $\mathrm{\Sigma }\left|{d}_{\overline{X}}\right|=322$

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

 Sr. No Weight (X) Deviation from Median $\left|\mathbit{d}\mathbit{m}\right|\mathbf{=}\left|\mathbit{X}\mathbf{-}\mathbf{M}\right|$ (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 $\mathrm{\Sigma }\left|dm\right|=42$

 Sr. No Weight (X) Deviation from Mean $\left|{\mathbit{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbit{X}\mathbf{-}\overline{\mathbf{X}}\right|\phantom{\rule{0ex}{0ex}}\left(\overline{)\mathbit{X}}\mathbf{=}\mathbf{52}\right)$ 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 $\mathrm{\Sigma }\left|{d}_{\overline{x}}\right|=44$

From Median
(a) M = Size of $\left(\frac{N+1}{2}\right)\mathrm{th}$ item
= Size of $\left(\frac{9+1}{2}\right)\mathrm{th}$ item
= Size of 5th item
= 50 kgs

(b)

(c)

From Mean

(a)

(b)

(c)

#### 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

 Items (X) Frequency (f) fX Deviation from Mean $\left|\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right|$= $\left|{\mathbit{d}}_{\overline{)\mathbf{X}}}\right|$ $\mathbit{f}\left|{\mathbit{d}}_{\overline{)\mathbf{X}}}\right|$ 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 $\mathrm{\Sigma }f\left|d\overline{)\mathrm{X}}\right|=765.6$

(a)

(b) Mean deviation from Arithmetic Mean

$M{D}_{\overline{)X}}=\frac{\mathrm{\Sigma }f\left|{d}_{\overline{)X}}\right|}{\mathrm{\Sigma }f}=\frac{765.6}{100}=7.656$

(c) Coefficient of $M{D}_{\overline{\mathit{X}}}$

#### 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

Calculation of mean deviation from median

 Size (X) Frequency (f) Cumulative frequency (c.f.) Deviation from median |dM| = |X − M| M = 8 f|dM| 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|dM| = 68

(a) Median or (M) = Size of $\left(\frac{N+1}{2}\right)\mathrm{th}$ item
= Size of $\left(\frac{21+1}{2}\right)\mathrm{th}$ item
= Size of 11th item
= 8

(b) Mean Deviation from median

${\mathrm{MD}}_{\mathrm{M}}=\frac{\mathrm{\Sigma }f\left|{d}_{M}\right|}{\mathrm{N}}=\frac{68}{21}=3.24$

Calculation of mean deviation from mean
 Size (X) Frequency (f) fX From mean deviation from mean $\left|{\mathbit{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right|\phantom{\rule{0ex}{0ex}}\left(\overline{)\mathbf{X}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{71}\right)$ $f\left|{d}_{\overline{X}}\right|$ 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 $\Sigma f\left|d\overline{{}_{X}}\right|$= 69.71

(a)

(b) Mean deviation from A.M.

${\mathrm{MD}}_{\overline{)\mathrm{X}}}=\frac{\mathrm{\Sigma }f\left|{d}_{\overline{)X}}\right|}{\mathrm{\Sigma }f}=\frac{69.71}{21}=3.32$

#### 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

Calculation of M.D from Median

 Marks Mid value m Frequency (f) Cumulative frequency |dM| = |m − M| M = 172.78 f|dM| 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|dM| = 512.2

Median = Size of $\left(\frac{\mathrm{N}}{2}\right)\mathrm{th}$ item
= Size of $\left(\frac{50}{2}\right)\mathrm{th}$ 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:

Calculation of M.D from Mean

 Marks Mid value m Frequency (f) fm $\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\overline{)\mathbf{X}}\right|\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{X}}\mathbf{=}\mathbf{171}\mathbf{.}\mathbf{2}$ $\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$ 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 $\mathrm{\Sigma }f\left|{d}_{\overline{)\mathrm{X}}}\right|=528$

Hence, Mean deviation from the arithmetic mean is:

${\mathrm{MD}}_{\overline{)\mathrm{X}}}=\frac{\mathrm{\Sigma }f\left|{d}_{\overline{)X}}\right|}{\mathrm{\Sigma }f}=\frac{528}{50}=10.56$

#### 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

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 ${\left(\frac{\mathrm{N}}{2}\right)}^{\mathrm{th}}$ item
= Size of ${\left(\frac{60}{2}\right)}^{\mathrm{th}}$ item
= Size of 39th item

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

Hence, the median value of the series is:

#### 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

 S. No. Marks (X) Deviation $\left(\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right)\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{X}}\mathbf{=}\mathbf{44}$ Square of the deviation ${\mathbit{x}}^{\mathbf{2}}\mathbf{=}{\left(\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right)}^{\mathbf{2}}$ 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

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

 S. No. Weight (X) Deviation $\left(\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right)\phantom{\rule{0ex}{0ex}}\left(\overline{)\mathbf{X}}\mathbf{=}\mathbf{51}\right)$ Square of deviation ${\mathbit{x}}^{\mathbf{2}}\mathbf{=}{\left(\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right)}^{2}$ 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

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

 Marks (X) Mid value (m) Frequency (f) fm Deviation from mean value $\left(\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right)$ $\overline{\mathbit{X}}$=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

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

 Persons Savings (X) Deviation $\left(\mathbit{x}\mathbf{=}\mathbf{X}\mathbf{-}\overline{)\mathbf{X}}\right)$ $\overline{\mathbf{X}}$=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

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

 Size (X) Frequency (f) fX Deviation from mean $\left(x=\mathrm{X}-\overline{)\mathrm{X}}\right)$ 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

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

 Population Mid Value (m) No. of villages (f) Deviation assumed from average ${\mathbit{d}}_{\mathbit{x}}\mathbf{=}\mathbit{m}\mathbf{-}\mathbit{A}$ ${d}_{x}\text{'}=\frac{{d}_{x}}{i}=\frac{{d}_{x}}{40}$ $\mathbit{f}{\mathbit{d}}_{\mathbf{x}}\mathbf{\text{'}}\mathbf{=}\mathbit{f}\mathbf{×}{\mathbit{d}}_{\mathbf{x}}\mathbf{\text{'}}$ $\mathbit{f}{\mathbit{d}}_{\mathbf{x}}{\mathbf{\text{'}}}^{\mathbf{2}}\mathbf{=}\mathbit{f}\mathbf{×}{\mathbit{d}}_{\mathbf{x}}{\mathbf{\text{'}}}^{\mathbf{2}}$ 0-200 200-400 400-1000 1000-2000 2000-5000 100 300 $\overline{)\mathbf{700}\mathbf{=}\mathbf{A}}$ 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 $\sum _{}f{d}_{x}\text{'}$=$-$350 $\sum _{}f{d}_{x}{\text{'}}^{2}$=41650

Calculating Standard deviation through Step deviation method:

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

 Size Mid Value m Frequency Deviation from assumed Average (dx = m − A) 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:

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

 Age Mid value m Frequency (f) Deviation assumed from average (dx = m − A) ${\mathbit{d}}_{\mathbf{x}}\mathbf{\text{'}}\mathbf{=}\frac{{\mathbit{d}}_{\mathbf{x}}}{\mathbit{c}}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{x}}}{\mathbf{10}}$ $\mathbit{f}{\mathbit{d}}_{\mathbit{x}}\mathbf{\text{'}}\mathbf{=}\mathbit{f}\mathbf{×}{\mathbit{d}}_{\mathbit{x}}\mathbf{\text{'}}$ $\mathbit{f}{\mathbit{d}}_{\mathbit{x}}{\mathbf{\text{'}}}^{\mathbf{2}}\mathbf{=}\mathbit{f}\mathbf{×}{\mathbit{d}}_{\mathbit{x}}{\mathbf{\text{'}}}^{\mathbf{2}}$ 20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90 25 35 45   $\overline{)55=A}$ 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 Σfdx' = -15 Σfdx'2 = 765

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

 C.I. Mid Value (m) Frequency (f) Deviation from Assumed mean dx = m − A (A = 45) fdx $f×{{d}_{x}}^{2}=f{{d}_{x}}^{2}$ 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 15 25 35 $\overline{)45=\mathrm{A}}$ 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 Σfdx = 320 Σfdx2 = 26000

#### 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

 S. No. Marks (X) Deviation from mean $\left(\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{)\mathbf{X}}\right)$ 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

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

 S. No. Value (X) Deviation from mean $\left(x=X-\overline{)X}\right)$ 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

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

 Age Mid Value (m) Frequency (f) Deviation from Assumed mean (dx = X − A) (A = 55) ${d}_{x}\text{'}=\frac{{d}_{x}}{i}=\frac{{d}_{x}}{10}\phantom{\rule{0ex}{0ex}}$ $f×{d}_{x}\text{'}=f{d}_{x}\text{'}\phantom{\rule{0ex}{0ex}}$ $f×{d}_{x}{\text{'}}^{2}=f{d}_{x}{\text{'}}^{2}\phantom{\rule{0ex}{0ex}}$ 20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90 25 35 45 $\overline{)55=\mathrm{A}}$ 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 Σfdx'= -15 Σfdx'2 =765

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

 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

 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

Given:
Highest value (H) = 77
Lowest value (L) = 1
Range = Highest value − Lowest value
i.e. R = HL
Substituting the given values in the formula.
R = 77 − 1 = 76

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

Highest value (H) = 11.5
Lowest value (L) = 4.5
Range = Highest value − Lowest value
i.e. R = HL
or, R = 11.5 − 4.5 = 7

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

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 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

#### 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

Arranging the data in the ascending order as presented below.
82, 100, 110, 116, 150, 150, 170, 200, 250
Here, N = 9

#### Question 5:

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

Given data series:
15, 20, 23, 23, 25, 25, 27, 40
Here, N = 8

#### 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

 x f fx ${\mathbit{d}}_{\overline{)\mathit{X}}}\mathbit{=}\left|\mathit{X}\mathit{-}\overline{)\mathit{X}}\right|$ $\mathbit{f}{\mathbit{d}}_{\overline{)\mathit{X}}}$ 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 $\mathbf{\sum }_{}\mathbit{f}{\mathbit{d}}_{\overline{)\mathit{X}}}$ = 73.51

 x f c.f. dM = |x − M| f|dM| 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|dM| = 70

N = 49

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25th 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.

#### 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

 x Mid Value (m) f c.f. |m−M| = |dM| f|dM| 0 − 10 10 − 20 20 − 30 5 15 25 15 19 14 15 34 $\overline{)48}$ 26 16 6 390 304 80 $\overline{)30}-40$ 35 $\overline{)20}$ 68 4 252 40 − 50 50 − 60 45 55 18 14 86 100 4 14 252 336 ∑fx = 100 ∑f|dM| = 1446
Here, N or ∑fx = 100
Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$ item, i.e.${\left(\frac{100}{2}\right)}^{th}$ item, which is 50th item.
This corresponds to the class interval of 30 40, so this is the median class.

Thus, we calculate the deviation of values from 31.

#### 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

 Class Interval Mid Value (m) f fm $\left|m-\overline{)X}\right|=\left|{\mathbit{d}}_{\overline{\mathbit{X}}}\right|$ $\mathbit{f}\left|{\mathbit{d}}_{\overline{)\mathbit{X}}}\right|$ 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 ${\mathbf{\sum }}_{}\mathbit{f}\left|{\mathbit{d}}_{\overline{)\mathbit{X}}}\right|$ = 200.6

#### 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

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.

Thus, we calculate the deviation of values from 26

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

#### 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

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.
th item
Thus, we calculate the deviation of values from 49.

 Marks Ascending Order |X- M| = |dM| 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 ∑|dM| = 115

#### 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

 X Mid Value m f fm $\left|m-\overline{)X}\right|=\left|{\mathbit{d}}_{\overline{)\mathbit{X}}}\right|$ $\mathbit{f}\left|{\mathbit{d}}_{\overline{)\mathbit{X}}}\right|$ 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 ∑$\mathbit{f}\left|{\mathbit{d}}_{\overline{)\mathbit{X}}}\right|$ = 11332.6

#### 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

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.

 No. of Victims Ascending Order |x - M| = |dM| 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 ∑|dM| = 65

#### Question 13:

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

 X $x\mathit{=}\left(\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}\right)$ 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

#### 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

 Income (X) No. of workers (​f) fX $x\mathit{=}X\mathit{-}\overline{)X}$ 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

#### 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

For Income Group I

 X1 ${x}_{1}\mathit{=}{X}_{1}\mathit{-}\overline{){X}_{1}}$ ${\mathbit{x}}_{\mathbf{I}}^{\mathbf{2}}$ 4000 4200 4400 4600 4800 −400 −200 0 200 400 160000 40000 0 40000 160000 ∑X1 = 22,000 ∑${\mathbit{x}}_{\mathbf{I}}^{\mathbf{2}}$ = 4,00,000

For Income Group II
 X2 ${x}_{2}\mathit{=}{X}_{2}\mathit{-}\overline{){X}_{2}}$ ${\mathbit{x}}_{\mathbf{2}}^{\mathbf{2}}$ 3000 4000 4200 4400 4600 4800 5800 −1400 −400 −200 0 200 400 1400 1960000 160000 40000 0 40000 160000 1960000 30,800 ∑${\mathbit{x}}_{\mathbf{2}}^{\mathbf{2}}$ = 4320000

#### 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

 S.No. Marks (X) $x\mathit{=}\left(\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}\right)$ 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

#### 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

 Daily Sales (X) Days (​f) fX $\left|X\mathit{-}\overline{)X}\right|\mathbit{=}\left|{d}_{\overline{)X}}\right|$ $\mathbit{f}\left|{d}_{\overline{)X}}\right|$ or $f\left|X\mathit{-}\overline{)X}\right|$ 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 ∑$\mathbit{f}\left|{d}_{\overline{)X}}\right|$ = 1119.02 ∑ = 3547.92

#### 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

 X $x\mathit{=}\left(\mathit{X}\mathit{-}\overline{)\mathit{X}}\right)$ 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.

#### 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

 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

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

 X Mid Values m f fX 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

#### 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

 X f fX 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

#### 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