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#### Question 33:

The following are the scores made by two batsmen A and B in a series of innings:

 A : 12 115 6 73 7 19 119 36 84 29 B : 47 12 76 42 4 51 37 48 13 0

Who is better as a run-getter ? Who is more consistent?

For Batsman A

 S. No. Score (XA) Deviation from mean $\left({x}_{\mathrm{A}}={X}_{\mathit{A}}\mathit{-}{\overline{)\mathit{X}}}_{\mathit{A}}\right)$ ${x}_{\mathrm{A}}^{2}$ 1 2 3 4 5 6 7 8 9 10 12 115 6 73 7 19 119 36 84 29 −38 65 −44 23 −43 −31 69 −14 34 −21 144 4225 1936 529 1849 961 4761 196 1156 441 N = 10 ΣxA = 500 $\mathrm{\Sigma }{x}_{\mathrm{A}}^{2}=17498$

For Batsman B
 S. No. Score XB Deviation from Mean $\left({x}_{\mathrm{B}}={X}_{\mathrm{B}}-{\overline{)\mathit{X}}}_{\mathrm{B}}\right)$ ${x}_{\mathrm{B}}^{2}$ 1 2 3 4 5 6 7 8 9 10 47 12 76 42 4 51 37 48 13 0 14 −21 43 9 −29 18 4 15 −20 −33 196 441 1849 81 841 324 16 225 400 1089 N = 10 ΣxB = 330 $\mathrm{\Sigma }{x}_{\mathrm{B}}^{2}=5462$

Since average score of A is more than B, he is better as run-getter but since, coefficient of variance is less for B, therefore B is more consistent.

#### Question 1:

The daily wages of ten workers are given below. Find out range and its coefficient.

 No. of Workers : A B C D E F G H I J Wages in (₹) : 175 50 50 55 100 90 125 145 70 60

Range = Largest item (L) − Smallest item (S)
=175 − 50 = 125

#### Question 2:

Following are the marks obtained by students in Sec . A and Sec . B . Compare the range of marks of students in two sections:

 Marks ( Section A) : 20 25 28 45 15 30 Marks ( Section B) : 45 52 36 42 28 25

For Section A
Range = Largest item (L) − Smallest item (S)
= 45 − 15 = 30

For Section B
Range = Largest item (L) − Smallest item (S)
= 52 − 25 = 27

Thus, range of marks of students in section A is greater than range of marks of students in section B.

#### Question 3:

Find range and coefficient of range of the following:
(a) Per day earning of seven agricultural labourers in ₹:

 60, 72, 36, 85, 35, 52, 72

(b)  Mean temperature deviation from normal (2002).
 Jan. Feb. Mar. Apr. May June +1.5 +2.4 +3.1 −1.5 −0.4 +3.3 July Aug. Sep. Oct. Nov. Dec. −0.1 −0.6 −1.5 −0.6 −1.9 −6.1

(a) Range = Largest item (L) − Smallest item (S)

= 85 − 35 = 50

(b) Range = Largest item (L) − Smallest item (S)
= +3.3 − (−6.1)
= +3.3 + 6.1 = +9.4

#### Question 4:

Calculate range and coefficient of range for the following data:

 Income (₹) : 50 70 80 90 100 120 130 150 No of workers : 2 8 12 7 4 3 8 6

Range = Largest item (L) − Smallest item (S)
= 150 − 50 = 100

#### Question 5:

Calculate range and coefficient of range of the following data:

 X : 10 15 20 30 40 50 f : 4 12 7 3 5 2

Range = Largest Value (L) − Smallest Value (S)
= 50 − 10 = 40

#### Question 6:

Find the range and coefficient of range of the following:

 Age in Years : 5-10 10-15 15-20 20-25 Frequency : 10 15 20 5

Range = Largest item (L) − Smallest item (S)
= 25 − 5 = 20

#### Question 7:

Calculate range and coefficient of range for the following distribution:

 Class : 1-5 6-10 11-15 16-20 21-25 26-30 Frequency : 2 8 15 35 20 10

The given series is an inclusive series. For the calculation of range the given series must first be converted into exclusive series as follows:

 Inclusive class Exclusive class Frequency 1−5 0.5−5.5 2 6−10 5.5−10.5 8 11−15 10.5−15.5 15 16−20 15.5−20.5 35 21−25 20.5−25.5 20 26−30 25.5−30.5 10

Range = Largest item (L) − Smallest item (S)
= 30.5 − 0.5 = 30

#### Question 8:

The following table gives the height of 100 persons . Calculate dispersion by Range Method.

 Height (in centimetres) No. of Persons Below 162 2 Below 163 8 Below 164 19 Below 165 32 Below 166 45 Below 167 58 Below 168 85 Below 169 93 Below 170 100

 Height Cumulative Frequency Frequency 161−162 2 2 162−163 8 8−2 = 6 163−164 19 19−8 = 11 164−165 32 32−19 = 13 165−166 45 45−32 = 13 166−167 58 58−45 = 13 167−168 85 85−58 = 27 168−169 93 93−85 = 8 169−170 100 100−93 = 7

For the calculation of range, the given data must be converted in continuous series as done above.

Range = Largest item (L) − Smallest item (S)
= 170 − 161 = 9

#### Question 9:

Calculate Quartile Deviation and its Coefficient of Rajesh's daily income.

 Months : 1 2 3 4 5 6 7 8 9 10 11 Income (₹) : 239 250 251 251 257 258 260 261 262 262 273

 Month Income (Rs) Month Income (Rs) 1 239 7 260 2 250 8 261 3 251 9 262 4 251 10 262 5 257 11 273 6 258

N = 11

Thus, the quartile deviation and coefficient of quartile deviation are 5.5 and 0.021 respectively.

#### Question 10:

Find the Quartile Deviation and its Coefficient from the following data relating to the daily wages of seven workers:

 Daily Wages ( in ₹) : 50 90 70 40 80 65 60

 S. No. Daily Wages 1 40 2 50 3 60 4 65 5 70 6 80 7 90

For the calculation of quartile deviation, we first arrange the given data in ascending order as above. The quartile deviation is then calculated in the following manner.

Thus, the quartile deviation and coefficient of quartile deviation are 15 and 0.23 respectively.

#### Question 11:

Find out Quartile Deviation and Coefficient of Quartile Deviation of the following items:

 145 130 200 210 198 234 159 160 178 257 260 300 345 360 390

 S. No. Item 1 130 2 145 3 159 4 160 5 178 6 198 7 8 9 10 11 12 13 14 15 200 210 234 257 260 300 345 360 390

We first arrange the given data in ascending order as above.

Thus, the quartile deviation and coefficient of quartile deviation are 70 and 0.304 respectively.

#### Question 12:

Find out Quartile Deviation , Coefficient of Quartile Deviation of the following data:

 X : 10 15 20 25 30 35 40 45 f : 6 17 29 38 25 14 9 1

 X f Cumulative Frequency 10 6 6 15 17 23 20 29 52 25 38 90 30 25 115 35 14 129 40 9 138 45 1 139 N = 139

The 35th item corresponds to 52 in the cumulative frequency.
$\therefore {Q}_{1}=20$

The 105th item corresponds to 115 in the cumulative frequency.
$\therefore {Q}_{3}=30$

Thus, coefficient of quartile deviation is 0.2.

#### Question 13:

Calculate the Quartile Deviation and Coefficient of Quartile Deviation of the following data:

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

 Marks Students (f) Cumulative Frequency 5−9 1 1 10−14 3 1 + 3 = 4 15−19 8 4 + 8 = 12 20−24 5 12 + 5 = 17 25−29 4 17 + 4 = 21 30−34 2 21 + 2 = 23 35−39 2 23 + 2 = 25 N = Ʃf = 25

This corresponds to the class interval (15−19).

This corresponds to the class interval (20−24).

This corresponds to the class interval (25−29).

#### Question 14:

Calculate the Semi-interquartile Range and its Coefficient of the following data:

 Marks : 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of Students : 4 8 10 14 10 9 5

 Marks No. of Students (f) c.f 0−10 4 4 10−20 8 12 20−30 10 22 30−40 14 36 40−50 10 46 50−60 9 55 60−70 5 60 N = 60
The first quartile class is given by the size of the ${\left(\frac{N}{4}\right)}^{\mathrm{th}}$item, i.e. the ${\left(\frac{60}{4}\right)}^{\mathrm{th}}$item, which is the 15th item.
This corresponds to the class interval 20−30, so this is the first quartile class.

The third quartile class is given by the size of the 3${\left(\frac{N}{4}\right)}^{\mathrm{th}}$item, i.e. the 3${\left(\frac{60}{4}\right)}^{\mathrm{th}}$item, which is the 45th item.
This corresponds to the class interval 40−50, so this is the third quartile class.

#### Question 15:

Calculate the Interquartile Range for the data given below:

 Class : 35-40 30-35 25-30 20-25 15-20 10-15 5-10 0-5 Frequency : 1 4 9 11 10 6 5 4

 Class Frequency (f) c.f. 0−5 4 4 5−10 5 9 10−15 6 15 15−20 10 25 20−25 11 36 25−30 9 45 30−35 4 49 35−40 1 50 N = 50

The first quartile class is given by the size of the ${\left(\frac{N}{4}\right)}^{\mathrm{th}}$item, i.e. the ${\left(\frac{50}{4}\right)}^{\mathrm{th}}$item, which is the 12.5th item.
This corresponds to the class interval 10−15, so this is the first quartile class.

The third quartile class is given by the size of the 3${\left(\frac{N}{4}\right)}^{\mathrm{th}}$item, i.e. the 3${\left(\frac{50}{4}\right)}^{\mathrm{th}}$item, which is the 37.5th item.
This corresponds to the class interval 25−30, so this is the third quartile class.

#### Question 16:

Calculate the Mean deviation from median of the following data:

 Marks : 41 66 59 38 54 21 32 49 68

 Marks (X) Deviation from Median D= |X−M| (M = 49) 21 28 32 17 38 11 41 8 49 0 54 5 59 10 66 17 68 19 ΣD = 115

For the calculation of mean deviation, the given data is first arranged in the ascending order as done above. Then median is calculated in the following manner.

Thus, mean deviation from median is 12.77.

#### Question 17:

Find  Coefficient of Mean deviation from median :

 Price (Rs) : 25 28 32 32 36 48 44 45 50 50

 Price Deviation from Median D = |X−M| M = 40 25 15 28 12 32 8 32 8 36 4 44 4 45 5 48 8 50 10 50 10 ΣD = 84

Thus, coefficient of mean deviation from median is 0.2.

#### Question 18:

Calculate the Mean Deviation from mean and median of the following data:

 X : 54 71 57 52 49 45 72 57 47

 X Deviation from Median DM =|X−M| M = 54 Deviation from Mean 45 9 11 47 7 9 49 5 7 52 2 4 54 0 2 57 3 1 57 3 1 71 17 15 72 18 16 ΣX = 504 Σ|X−Md| = 64 $\mathrm{\Sigma }\left|X\mathit{-}\overline{)\mathit{X}}\right|=66$

#### Question 19:

Calculate  Mean deviation from (i) arithmetic mean, (ii) Median:

 Marks : 7 4 10 9 15 12 7 9 7

 Marks (X) Deviation from Mean Deviation from Median DM =|X−M| 4 4.89 5 7 1.89 2 7 1.89 2 7 1.89 2 9 0.11 0 9 0.11 0 10 1.11 1 12 3.11 3 15 6.11 6 ΣX = 80 $\mathrm{\Sigma }\left|X\mathit{-}\overline{)\mathit{X}}\right|=21.11$ Σ|X−Md| = 21

#### Question 20:

Find out the average deviation from median for the following distribution:

 Values : 6 12 18 24 30 36 42 Frequency : 4 7 9 18 15 10 5

 Value (X) Frequency (f) c.f Deviation from Median |DM|=|X−M| (M = 24) f|DM| 6 4 4 18 72 12 7 11 12 84 18 9 20 6 54 24 18 38 0 0 30 15 53 6 90 36 10 63 12 120 42 5 68 18 90 Σf = N = 68 Σf|dm| = 510

This corresponds to value 24. Thus, median is 24.

#### Question 21:

Calculate Mean Deviation from median:

 No. of tomatoes per plant : 0 1 2 3 4 5 6 7 8 9 10 No. of Plants : 2 5 7 11 18 24 12 8 6 4 3

 Tomatoes Frequency (f) c.f Deviation from Median |DM|=|X−M| (M = 5) f|DM| 0 2 2 5 10 1 5 7 4 20 2 7 14 3 21 3 11 25 2 22 4 18 43 1 18 5 24 67 0 0 6 12 79 1 12 7 8 87 2 16 8 6 93 3 18 9 4 97 4 16 10 3 100 5 15 Σf = N = 100 Σf|dm|=168

This corresponds to 5. Thus, median is 5.

#### Question 22:

Calculate (a) Median Coefficient of dispersion from the following data:

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

 Size of Item (X) Frequency (f) c.f Deviation from Median |DM|=|X−M| (M = 8) f|DM| 4 2 2 4 8 6 4 6 2 8 8 5 11 0 0 10 3 14 2 6 12 2 16 4 8 14 1 17 6 6 16 4 21 8 32 Σf = N = 21 Σf|dm| = 68

This corresponds to size 8. Thus, median is 8.

 Size (X) Frequency (f) fX Deviation from Mean $\left|{D}_{\overline{)X}}\right|$ $\overline{)\mathit{X}}=9.71$ $f\left|{D}_{\overline{)\mathit{X}}}\right|$ 4 2 8 5.71 11.42 6 4 24 3.71 14.84 8 5 40 1.71 8.55 10 3 30 0 .29 0.87 12 2 24 2.29 4.58 14 1 14 4.29 4.29 16 4 64 6.29 25.16 Σf = 21 ΣfX = 204 $\mathrm{\Sigma }f\left|d\overline{)\mathrm{X}}\right|=69.71$

#### Question 23:

Calculate Mean and Mean Deviation and coefficient  of M.D. for the following distribution:

 Weekly wages : 20-40 40-60 60-80 80-100 Workers : 20 40 30 10

 Weekly wages (X) Mid value m f fm $\left|{\mathbit{D}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbit{m}\mathbit{-}\overline{)X}\right|\phantom{\rule{0ex}{0ex}}\overline{)X}\mathbf{=}\mathbf{56}$ $\mathbit{f}\left|{\mathbit{D}}_{\overline{)\mathbf{X}}}\right|$ 20−40 30 20 600 26 520 40−60 50 40 2000 6 240 60−80 70 30 2100 14 420 80−100 90 10 900 34 340 Σf = 100 Σfm = 5600 $\mathrm{\Sigma }\mathbit{f}\left|{\mathbit{D}}_{\overline{)\mathbf{X}}}\right|=1520$

#### Question 24:

Find Mean Deviation from median of the marks secured by 100 students in a class-test as given below:

 Marks : 60-63 63-66 66-69 69-72 72-75 No. of Students : 5 18 42 27 8

 Marks Mid Value f c.f Deviation from Median DM = |m−M| (M= 67.93) f|DM| 60−63 61.5 5 5 6.43 32.15 63−66 64.5 18 23 3.43 61.74 66−69 67.5 42 65 0 .43 18.06 69−72 70.5 27 92 2.57 69.39 72−75 73.5 8 100 5.57 44.56 Σf = 100 Σf|DM| = 225.9
The median class is given by the size of theitem, i.e. the ${\left(\frac{100}{2}\right)}^{\mathrm{th}}$item, which is the 50th item.
This corresponds to the class interval 66−69, so this is the median class.

#### Question 25:

Using Mean Deviation from median of the income group of 5 and 7 members given below, compare which of the group has more variability?

 Group A : 4000 4200 4400 4600 4800 Group B : 3000 4000 4200 4400 4600 4800 5800

For Group A

 Income (X) Deviation from median DM=|X−M| (M=4400) 4000 400 4200 200 4400 0 4600 200 4800 400 Σ|DM|=1200

For Group B
 Income Deviation from median DM=|X−M| (M=4400) 3000 1400 4000 400 4200 200 4400 0 4600 200 4800 400 5800 1400 Σ|DM|=4000

In case of group B, as the coefficient of MD is more, so group B has greater variation.

#### Question 26:

Calculate the standard Deviation of wage earner's daily earnings:

 Week : 1 2 3 4 5 6 7 8 9 10 Earnings (in ₹) : 54 62 63 65 68 71 73 78 82 84

 S.No. (Week) Farming (X) Deviation $\left(X\mathit{-}\overline{)\mathit{X}}\right)$ $\left(\overline{)\mathit{X}}=70\right)$ Square of deviation ${\left(X\mathit{-}\overline{)\mathit{X}}\right)}^{2}$ 1 2 3 4 5 6 7 8 9 10 54 62 63 65 68 71 73 78 82 84 −16 −8 −7 −5 −2 1 3 8 12 14 256 64 49 25 4 1 9 64 144 196 ΣX = 700 $\mathrm{\Sigma }{\left(\mathit{X}\mathit{-}\overline{)\mathit{X}}\right)}^{2}=812$

The mean of the given series can be calculated in the following manner.
$\overline{)X}=\frac{\Sigma X}{N}=\frac{700}{10}=70\phantom{\rule{0ex}{0ex}}$
To calculate standard deviation, we use the following formula.

Thus, standard deviation of wage earner's daily earnings is 9.01.

#### Question 27:

Calculate Standard Deviation of the following two series . Which series has more variability:

 A: 58 59 60 65 66 52 75 31 46 48 B: 56 87 89 46 93 65 44 54 78 68

For A

 S. No. (X) Deviation $\left(x=\mathrm{X}-\overline{)\mathrm{X}}\right)\phantom{\rule{0ex}{0ex}}\overline{)\mathrm{X}}=56$ x2 1 2 3 4 5 6 7 8 9 10 58 59 60 65 66 52 75 31 46 48 2 3 4 9 10 −4 19 −25 −10 −8 4 9 16 81 100 16 361 625 100 64 ΣX = 560 Σx2 = 1376

For B
 S. No. Y Deviation $\left(y=Y-\overline{)y}\right)\phantom{\rule{0ex}{0ex}}\left(\overline{)y}=68\right)$ y2 1 2 3 4 5 6 7 8 9 10 56 87 89 46 93 65 44 54 78 68 −12 19 21 −22 25 −3 −24 −14 10 0 144 361 441 484 625 9 576 196 100 0 ΣY = 680 Σy2 = 2936

Since the coefficient of variation of series B is more, it is said to have more variability.

#### Question 28:

Calculate Mean and Standard Deviation of income of 8 employees of  a firm:

 Income (₹): 100 120 140 120 180 140 120 150

 S. No. (X) Deviation $x=X\mathit{-}\overline{)X}\phantom{\rule{0ex}{0ex}}\overline{)X}=133.75$ x2 100 120 140 120 180 140 120 150 −33.75 −13.75 6.25 −13.75 46.25 6.25 −13.75 16.25 1139.06 189.06 39.06 189.06 2139.06 39.06 189.06 264.06 ΣX= 1070 Σx2 = 4187.48

$\overline{)X}=\frac{\mathrm{\Sigma }X}{N}=\frac{1070}{8}=133.75\phantom{\rule{0ex}{0ex}}SD=\sqrt{\frac{\mathrm{\Sigma }{x}^{2}}{N}}=\sqrt{\frac{4187.48}{8}}=\sqrt{523.43}=22.87$
Thus, mean standard deviation is 22.87.

#### Question 29:

Calculate Mean and Standard Deviation from the following data:

 Marks ( Above) : 0 10 20 30 40 50 60 70 No. of Students : 100 90 75 50 25 15 5 0

As we are given marks (above) series, we can convert it into classes and frame the following table.

 Marks Mid Value (m) cf Frequency (f) fm Deviation from mean $x=\left(m\mathit{-}\overline{)X}\right)\phantom{\rule{0ex}{0ex}}\overline{)X}=31$ 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 75 100 90 75 50 25 15 5 0 10 15 25 25 10 10 5 0 50 225 625 875 450 550 325 0 -26 -16 -6 4 14 24 34 44 676 256 36 16 196 576 1156 1936 6760 3840 900 400 1960 5760 5780 0 Σf = 100 Σfm = 3100 Σfx2=25400

#### Question 30:

Calculate Mean and Variance for the following frequency distribution:

 X : 1 2 3 4 5 6 7 8 9 10 f : 2 5 8 16 21 18 13 10 4 3

 X f fX Deviation $x=X\mathit{-}\overline{)\mathit{X}}\phantom{\rule{0ex}{0ex}}\overline{)\mathit{X}}=5.5$ x2 fx2 1 2 3 4 5 6 7 8 9 10 2 5 8 16 21 18 13 10 4 3 2 10 24 64 105 108 91 80 36 30 −4.5 −3.5 −2.5 −1.5 −.5 .5 1.5 2.5 3.5 4.5 20.25 12.25 6.25 2.25 .25 .25 2.25 6.25 12.25 20.25 40.5 61.25 50 36 5.25 4.5 29.25 62.5 49 60.75 Σf = 100 ΣfX = 550 Σfx2 = 399

Thus, mean and variance are 5.5 and 3.99 respectively.

#### Question 31:

Calculate  Mean, Standard Deviation and Variance:

 Variable : 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

 Variable Mid Value (m) Frequency (f) fm Deviation Mean $x=\left(m-\overline{)\mathrm{X}}\right)\phantom{\rule{0ex}{0ex}}\overline{)X}=21.9$ x2 fx2 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 2 5 7 13 21 16 8 3 5 37.5 87.5 227.5 472.5 440 260 112.5 19.4 14.4 9.4 4.4 0.6 5.6 10.6 15.6 376.36 207.36 88.36 19.36 0.36 31.36 112.36 243.36 752.72 1036.8 618.52 251.68 7.56 501.76 898.88 730.08 Σf = 75 Σfm = 1642.5 Σfx2 = 4798

#### Question 32:

Calculate the Coefficient of Variation for the following distribution of the wages of 200 workers in a factory:

 Wages ( in ₹) : 40-49 50-59 60-69 70-79 80-89 90-99 100-109 110-119 No. of workers : 11 23 40 60 35 16 9 6

In order to calculate coefficient of variation, first we will convert the given inclusive series into exclusive series as follows:

 Wages Mid Point (m) Frequency (f) fm Deviation from $x=\left(m-\overline{)\mathrm{X}}\right)\phantom{\rule{0ex}{0ex}}\overline{)\mathrm{X}}=74.45$ x2 fx2 39.5 − 49.5 49.5 − 59.5 59.5 − 69.5 69.5 − 79.5 79.5 − 89.5 89.5 − 99.5 99.5 − 109.5 109.5 − 119.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5 114.5 11 23 40 60 35 16 9 6 489.5 1253.5 2580 4470 2957.5 1512 940.5 687 29.95 19.95 9.95 .05 10.05 20.05 30.05 40.05 897 398 99 .002 101 402 903 1604 9867 9154 3960 0.12 3535 6432 8127 9624 Σf = 200 Σfm = 14890 Σfx2 = 50699.12

#### Question 34:

The index number of prices of cotton and coal shares in 1998 were as under:
Index Number of Prices:

 Cotton : 188 178 173 164 172 183 184 185 211 217 232 240 Coal : 131 130 130 129 129 129 127 127 130 137 140 142

Which of these two shares do you consider more variable in prices:

 S. No. Price X Deviation from mean $\left(x=X-\overline{)\mathit{X}}\right)\phantom{\rule{0ex}{0ex}}\overline{)\mathit{X}}=193.9$ x2 1 2 3 4 5 6 7 8 9 10 11 12 188 178 173 164 172 183 184 185 211 217 232 240 5.9 15.9 20.9 29.9 21.9 10.9 9.9 8.9 17.1 23.1 38.1 46.1 34.81 252.81 436.81 894.01 479.61 118.81 98.01 79.21 292.41 533.61 1451.61 2125.61 ΣX = 2327 Σx2 = 6796.92

 S. No. Price X Deviation from mean $x=X\mathit{-}\overline{)\mathit{X}}$ x2 1 2 3 4 5 6 7 8 9 10 11 12 131 130 130 129 129 129 127 127 130 137 140 142 −.75 −1.75 −1.75 −2.75 −2.75 −2.75 −4.75 −4.75 −1.75 5.25 8.25 10.25 .56 3.06 3.06 7.56 7.56 7.56 22.56 22.56 3.06 27.56 68.06 105.06 ΣX = 1581 Σx2= 278.22

Since coefficient of variance is more of cotton than that of coal. So, prices of cotton shares are more variable.

#### Question 35:

Calculate the arithmetic mean, standard deviation and variance from the following distribution:

 Class : 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) (f) fm Deviation from mean x x2 fx2 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 2 5 7 13 21 16 8 3 5 37.5 87.5 227.5 472.5 440 260 112.5 19.4 14.4 9.4 4.4 .6 5.6 10.6 15.6 376.36 207.36 88.36 19.36 .36 31.36 112.36 243.36 752.72 1036.8 618.52 251.68 7.56 501.76 898.88 730.08 Σf = 75 Σfm = 1642.5 Σfx2 = 4798

#### Question 36:

Calculate arithmetic mean, standard deviation and variance from the following series:

 Marks : 70-80 60-70 50-60 40-50 30-40 20-30 No of students : 7 11 22 0 15 5

 Marks Mid Value (m) f fm Deviation from mean $x=\left(m-\overline{)\mathit{X}}\right)\phantom{\rule{0ex}{0ex}}\overline{)\mathit{X}}=51.67$ x2 fx2 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 25 35 45 55 65 75 5 15 0 22 11 7 125 525 0 1210 715 525 26.67 16.67 6.67 3.33 13.33 23.33 711.29 277.89 44.49 11.09 177.69 544.29 3556.45 4168.35 0 243.98 1954.59 3810.03 Σf = 60 Σfm = 3100 Σfx2 = 13733.4

$\overline{)\mathit{X}}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{3100}{60}=51.67\phantom{\rule{0ex}{0ex}}SD=\sqrt{\frac{\mathrm{\Sigma }f{x}^{2}}{\mathrm{N}}}=\sqrt{\frac{13733.4}{60}}=\sqrt{228.89}=15.13\phantom{\rule{0ex}{0ex}}\mathrm{Variance}={\left(SD\right)}^{2}=228.89$

#### Question 37:

You are given the following data about height of boys and girls:

 Boys Girls Number 72 38 Average height( in inches) 68 61 Variance of distribution (in inches) 9 4

(a) Calculate Coefficient of Variation.
(b) Decide whose height is more variable.

(a)
For Boys

For Girls

(b) Since coefficient of variance of boys is more than that of girls, thus the height of boys is more variable.

#### Question 38:

Lives of two models of refrigerators in a recent survey are as under:

 Life No fo Years : 0-2 2-4 4-6 6-8 8-10 10-12 No of Refrigerators : Model A : 5 16 13 7 5 4 Model B : 2 7 12 19 9 1

What is the average life of each model of refrigerators? Which model has more uniformity?

Model A:

 Years Mid Value (m) f fm Deviation from mean ${x}_{\mathrm{A}}=\left(m-\overline{)\mathit{X}}\right)\phantom{\rule{0ex}{0ex}}\left(\overline{)\mathit{X}}=5.12\right)$ ${x}_{\mathrm{A}}^{2}$ $f{x}_{\mathrm{A}}^{2}$ 0 − 2 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 1 3 5 7 9 11 5 16 13 7 5 4 5 48 65 49 45 44 −4.12 −2.12 −.12 1.88 3.88 5.88 16.97 4.49 0.01 3.53 15.05 34.57 84.85 71.84 0.13 24.71 75.25 138.28 Σf = 50 Σfm = 256 $\sum f{x}_{\mathrm{A}}^{2}=395.06$

Average life of model A is 5.12 year.

Model B:
 Years Mid Value (m) f fm Deviation from mean ${x}_{B}=\left(m-\overline{)\mathit{X}}\right)\phantom{\rule{0ex}{0ex}}\left(\overline{)\mathit{X}}=6.16\right)$ ${x}_{B}^{2}$ $f{x}_{B}^{2}$ 0 − 2 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 1 3 5 7 9 11 2 7 12 19 9 1 2 21 60 133 81 11 −5.16 −3.16 −1.16 0.84 2.84 4.84 26.62 9.98 1.34 0.70 8.06 23,42 53.24 69.86 16.08 13.3 72.54 23.42 Σf = 50 Σfm = 308 $\sum f{x}_{B}^{2}=248.44$

Average life of model B is 6.16 year.

Since the coefficient of variation of model B is less than that of A, thus model B has more uniformity.

#### Question 39:

Two brands of Tyres are tested with the following results:

 Life ( '000 miles) : 20-25 25-30 30-35 35-40 45-50 50-55 Brand X 8 15 12 18 13 9 Brand Y 6 20 32 30 12 0
Which brand has a greater variation?

For Brand 'X'

 Life Mid value m f fm Deviation from mean $x=m-\overline{)\mathit{X}}$ $\overline{)\mathit{X}}=35.17$ x2 fx2 20−25 22.5 8 180 −12.67 160.53 1284.24 25−30 27.5 15 412.5 −7.67 58.83 882.45 30−35 32.5 12 390 −2.67 7.13 85.56 35−40 37.5 18 675 2.33 5.43 97.74 40−45 42.5 13 552.5 7.33 53.73 698.49 45−50 47.5 9 427.5 12.33 152.03 1368.27 Σf =75 Σfm =2637.5 Σfx2= 4416.75

For Brand 'Y'
 Life Mid value m f fm Deviation from mean $x=m\mathit{-}\overline{)\mathit{X}}$ $\overline{)\mathit{X}}=33.60$ x2 fx 20−25 22.5 6 135 −11.1 123.21 739.26 25−30 27.5 20 550 −6.1 37.21 744.2 30−35 32.5 32 1040 −1.1 1.21 38.72 35−40 37.5 30 1125 3.9 15.21 456.3 40−45 42.5 12 510 8.9 79.21 950.52 45−50 47.5 0 0 13.9 193.21 0 Σf=100 Σfm=3360 $\mathrm{\Sigma }f{x}_{}^{2}=2929$

As the coefficient of variation is more for brand X, it is said to be more variable than brand Y.

#### Question 40:

From the data given below state which series is more consistent:

 Variable : 10-20 20-30 30-40 40-50 50-60 60-70 Series A : 10 18 32 40 22 18 Series B : 18 22 40 32 18 10

For Series A

 Variable Mid value (m) f fm Deviation from mean $x=m-{\overline{)\mathit{X}}}_{\mathrm{A}}$ ${\overline{)\mathit{X}}}_{\mathrm{A}}=42.14$ ${\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ $\mathbit{f}{\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ 10−20 15 10 150 −27.14 736.58 7365.8 20−30 25 18 450 −17.14 293.78 5288.04 30−40 35 32 1120 −7.14 50.98 1631.36 40−50 45 40 1800 2.86 8.18 327.2 50−60 55 22 1210 12.86 165.38 3638.36 60−70 65 18 1170 22.86 522.58 9406.44 Σf=140 Σfm=5900 $\sum f{x}_{A}^{2}=27657.2$

For Series B
 Variable Mid value (m) f fm Deviation from mean $x=m-{\overline{)\mathrm{X}}}_{\mathrm{B}}$ ${\overline{)\mathrm{X}}}_{\mathrm{B}}=37.86$ ${\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ $\mathbit{f}{\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ 10−20 15 18 270 −22.86 522.58 9406.44 20−30 25 22 550 −12.86 165.38 3638.36 30−40 35 40 1400 −2.86 8.18 327.2 40−50 45 32 1440 7.14 50.98 1631.36 50−60 55 18 990 17.14 293.78 5288.04 60−70 65 10 650 27.14 736.58 7365.8 Σf=140 Σfm=5300 $\mathrm{\Sigma }f{x}_{\mathrm{B}}^{2}=27657.2$

As coefficient of variation for series 'A' is less than series 'B', therefore, series A is more consistent.

#### Question 41:

The following table gives the distribution of wages in the two branches of a factory:

 Monthly wages ( in ₹) : 100-150 150-200 200-250 250-300 300-350 Total Number of workers : Branch A : 167 207 253 205 168 1000 Branch B : 63 93 157 105 82 500
Find the mean and standard deviation for the two branches for the wages separately.
(a) Which branch pays higher average wages?
(b) Which branch has greater variability in wages in relation to the average wages?
(c) What is the average monthly wage for the factory as a whole?
(d) What is the variance of wages of all the workers in the two branches$—$A and B taken together?

For Branch A

 Monthly Wages Mid Value (m) f fm Deviation from mean ${\mathbit{x}}_{\mathbf{A}}\mathbf{=}\mathbit{m}\mathbf{-}{\overline{)\mathit{X}}}_{\mathbf{A}}\phantom{\rule{0ex}{0ex}}{\overline{)\mathit{X}}}_{\mathbf{A}}\mathbf{=}\mathbf{225}$ ${\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ $\mathbit{f}{\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350 125 175 225 275 325 167 207 253 205 168 20875 36225 56925 56375 54600 −100 −50 0 50 100 10000 2500 0 2500 10000 1670000 517500 0 512500 1680000 Σf = 1000 Σfm = 225000 $\mathrm{\Sigma }f{x}_{\mathrm{A}}^{2}=4380000$

For Branch B
 Monthly wage Mid value (m) f fm Deviation from mean ${\mathbit{x}}_{\mathbf{B}}\mathbf{=}\mathbit{m}\mathbf{-}{\overline{)\mathbf{X}}}_{\mathbf{B}}\phantom{\rule{0ex}{0ex}}{\overline{)\mathbf{X}}}_{\mathbf{B}}\mathbf{=}\mathbf{230}$ ${\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ $\mathbit{f}{\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350 125 175 225 275 325 63 93 157 105 82 7875 16275 35325 28875 26650 −105 −55 −5 45 95 11025 3025 25 2025 9025 694575 281325 3925 212625 740050 Σf = 500 Σfm = 115000 $\mathrm{\Sigma }f{x}_{\mathrm{B}}^{2}=1932500$

(a) Branch B pays higher average monthly wages i.e. Rs 230.
(b) Branch A has greater variability since its coefficient of variation is higher than that of B.
(c)

Thus, the average monthly wage for the factory as a whole is Rs 226.67.

(d)

Thus, the variance of wages of all the workers in the two branches- A and B taken together is Rs 4217.95.

#### Question 42:

Goals scored by two teams A and B in football matches were as follows:

 No. of Goals in a match : 0 1 2 3 4 No.of Matches:     A : 27 9 8 5 4 B : 17 9 6 5 3
Find the team which is more consistent in its performance.

 X f fX Deviation from mean ${\mathbit{x}}_{\mathbf{A}}\mathbf{=}\left(\mathbit{X}\mathbf{-}{\overline{)\mathit{X}}}_{\mathbf{A}}\right)$ ${\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ $\mathbit{f}{\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ 0 1 2 3 4 27 9 8 5 4 0 9 16 15 16 −1.06 −0.06 0.94 1.94 2.94 1.12 0.0036 0.88 3.76 8.64 30.24 0.03 7.04 18.8 34.56 Σf = 53 ΣfX = 56 $\mathrm{\Sigma }f{x}_{\mathrm{A}}^{2}=90.68$

 X f fX Deviation from mean ${\mathbit{x}}_{\mathbf{B}}\mathbf{=}\left(\mathbit{X}\mathbf{-}{\overline{)\mathit{X}}}_{\mathbf{B}}\right)$ ${\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ $\mathbit{f}{\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ 0 1 2 3 4 17 9 6 5 3 0 9 12 15 12 −1.2 −0.2 0.8 1.8 2.8 1.44 0.04 0.64 3.24 7.84 24.48 0.36 3.84 16.2 23.52 Σf = 40 ΣfX= 48 $\mathrm{\Sigma }f{x}_{\mathrm{B}}^{2}=68.4$

Since, coefficient of variation is lesser for team B, it is more consistent.

#### Question 43:

Find mean and the standard deviation of the following two groups taken together:

 Group Number Mean S.D. A 113 159 22.4 B 121 149 20.0

Now, to calculate combined standard deviation, we need to find,

Combined Standard Deviation

Thus, mean and standard deviation of the two groups-A and B taken together are 23.33 and 21.8 respectively.

#### Question 44:

The number examined , the mean weight and standard deviations in each group of examination by two medical examiners are given below . Calculate mean and standard deviation of both the groups taken together.

 Medical Examiner Number Examined Mean Weight (lbs) Standard deviation (lbs) A 50 113 6.5 B 60 120 8.2

Combined Standard Deviation

Thus, mean and standard deviation of both the groups taken together are 116.82 and 8.25 respectively.

#### Question 45:

The following data gives arithmetic means, standard deviations of three sub-groups . Calculate arithmetic mean  and standard deviation of the whole group.

 Sub-group No. of Men Average Wage (in ₹) Standard deviation  (in ₹) A 50 61.0 8.0 B 100 70.0 9.0 C 120 80.5 10.0

Combined Standard Deviation

Thus, arithmetic mean and standard deviation of the whole group are 73 and 11.9 respectively.

#### Question 46:

A sample of 35 values has mean 80 and Standard Deviation 4 . A second sample of 65 values from same population has mean 70 and standard deviation 3. Find the mean and standard deviation of combined sample of 100 values.

Given,

Combined Standard Deviation

Thus, mean and standard deviation of combined sample of 100 values are 73.5 and 5.84 respectively.

#### Question 47:

For a group of 50 male workers , the mean and standard deviation of their weekly wages are ₹ 63 and ₹ 9 respectively. For a group of 40 female workers , these are ₹ 54 and ₹ 6 respectively. Find mean and standard deviation for a combined group of 90 workers.

Given,

Combined Standard Deviation

Thus, mean and standard deviation for a combined group of 90 workers are 59 and 9 respectively.

#### Question 48:

Coefficient of variation of two series are 58% and 69% and their standard deviation are 21.2 and 15.6 . What are their means?

Let the first series be A.
Now,
Coefficient of variation of A = 58%
σA = 21.2

Let the second series B.
Now,
Coefficient of variation of B= 69%
σB = 21.2

Thus, means of series A and series B are 36.55 and 22.608 respectively.

#### Question 49:

If the coefficient of variation of X series is 14.6%  and that of Y series is 36.9% and their means are 101.2 and 101.25 respectively. Find their standard deviation.

For first series,
Coefficient of variation of A = 14.6%

For second series,
Coefficient of variation of B= 36.9%

Thus, standard deviations of first and second series are 14.775 and 37.36 respectively.

#### Question 50:

Draw a Lorenz curve from the following:

 Income (₹ '000) : 20 40 60 100 160 No of Persons ('000) Class A : 10 20 40 50 80 Class B : 16 14 10 6 4

 Income Cumulative Income Cumulative Percentage Person of Class A Cumulative Cumulative Percentage Person of Class B Cumulative Persons Cumulative Percentage 20 40 60 100 160 20 60 120 220 380 5.26 15.78 31.57 57.89 100 10 20 40 50 80 10 30 70 120 200 5 15 35 60 100 16 14 10 6 4 16 30 40 46 50 32 60 80 92 100 #### Question 51:

Following is the frequency distribution of marks obtained by students in Economics and Statistics . Analyse the data by drawing a Lorenz Curve.

 Marks (mid-values) : 5 15 25 35 45 55 65 75 85 95 No.of students (Economics) : 10 12 13 14 22 27 20 12 11 9 No. of students (Statistics) : 1 2 26 50 59 40 10 8 3 1 