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#### Question 1:

Calculate figures and coefficient  of range of the following series, which gives the monthly expenditure (in ₹) of seven students: 22, 35, 32, 45, 42, 48, 39

Given:
Highest Value (H) =48
Lowest Value (L) =22

Range = Highest Value − Lowest Value

i.e R= H−  L

Substituting the given values in the formula.

R= 48 − 22 = 26

Thus, range is 26 and coefficient of range is 0.37

#### Question 2:

Find range and coefficient of range from the weekly wage (in ₹) of  10 workers of a factory: 310, 350, 420, 105, 115, 290, 245, 450, 300, 375.

Given:
Highest Value (H) = 450
Lowest Value (L) = 105

Range = Highest Value − Lowest Value

i.e R= H−  L

Substituting the given values in the formula.

R= 450 − 105 = 345

Thus, range is 345 and coefficient of range is 0.62

#### Question 3:

Find the range and coefficient of range of the following:

 Size of shoes 6 7 8 9 10 11 12 13 No. of shoes sold 8 12 14 20 15 8 7 10

 Size of Shoes No. of Shoes Sold 6 8 7 12 8 14 9 20 10 15 11 8 12 7 13 10

Highest value (H) = 13
Lowest value (L) = 6
Range = Highest value − Lowest value
i.e. R = HL
or, R = 13 − 6=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 4:

From the following data calculate range and coefficient of range:

 Marks 10 20 30 40 50 60 70 No. of Studends 8 12 7 30 10 5 2

 Marks No. of Students 10 8 20 12 30 7 40 30 50 10 60 5 70 2

Highest value (H) = 70
Lowest value (L) = 10

Range = Highest value − Lowest value

i.e. R = H−L

or, R = 70 − 10= 60 marks

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 5:

Find out range and coefficient range.

 Marks 10−20 20−30 30−40 40−50 50−60 No. of students 12 18 14 63 19

 Marks No. of Students 10 − 20 12 20 − 30 18 30 − 40 14 40 − 50 63 50 − 60 19

Hence, the range is 50 marks and coefficient of range is 0.715

#### Question 6:

Following are the marks obtained by 25 students of class XI in an exam. Find out range and coefficient of range of the marks.

 Marks 5−9 10−14 15−19 20−24 25−29 30−34 No. Students 4 6 3 2 6 4

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 5 − 9 10 −14 15 −19 20 −24 25 −29 30− 34 4.5 − 9.5 9.5 − 14.5 14.5 − 19.5 19.5 − 24.5 24.5 − 29.5 29.5 − 34.5 4 6 3 2 6 4

#### Question 7:

Calculate interquartile range, quartile deviation and coefficient of quartile deviation from the following data:

 Family A B C D E F G Income (in ₹) 1,200 1,400 1,500 1,700 2,000 2,100 2,200

 Family S. No. Income Rs A 1 1200 B 2 1400 C 3 1500 D 4 1700 E 5 2000 F 6 2100 G 7 2200 N = 7

#### Question 8:

find out the value of quartile deviation and its coefficient from the following data:

 Roll No. 1 2 3 4 5 6 7 Marks 20 28 40 12 30 15 50

 Roll No. Marks in Ascending Order 1 12 2 15 3 20 4 28 5 30 6 40 7 50 N = 7

#### Question 9:

Form the following figures, find the quartile deviation and its coefficient:

 Height (cm) 150 151 152 153 154 155 156 157 158 No. of students 15 20 32 35 33 22 20 12 10

 Height (c.m) No. of Students Cumulative Frequency 150 15 15 151 20 15 + 20 = 35 152 32 35 + 32 = 67 153 35 67 + 35 = 102 154 33 102 + 33 = 135 155 22 135 + 22 = 157 156 20 157 + 20 = 177 157 12 177 + 12 = 189 158 10 189 + 10 = 199 N = 199

#### Question 10:

Compute coefficient of Quartile deviation form the following data:

 Marks 10 20 30 40 50 60 No. of students 4 7 15 8 7 2

 Marks No. of Students Cumulative Frequency 10 4 4 20 7 7 + 4 = 11 30 15 11 + 15 = 26 40 8 26 + 8 = 34 50 7 34 + 7 = 41 60 2 41 + 2 = 43 N = 43

#### Question 11:

Calculate Quartile Deviation and its Coefficient from the following data:

 Size 5−10 10−15 15−20 20−25 25−30 30−35 35−40 40−45 45−50 Frequency 6 10 18 30 15 12 10 6 4

 Size Frequency Cumulative Frequency 5−10 6 6 10−15 10 16 15−20 18 34 20−25 30 64 25−30 15 79 30−35 12 91 35−40 10 101 40−45 6 107 45−50 4 111 Σf=N = 111

Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{111}{4}\right)\mathrm{th}$ item
= Size of 27.75th item

27.75 item lies in group 15−20 and falls within 34th c.f. of the series

Like wise,
Q3 = Size of $3\left(\frac{\mathrm{N}}{4}\right)\mathrm{th}$ item
= Size of 3
= Size of 83.25th  item

83.25th  item lies in the group 30-35 within the 33rd c.f. of the series.

#### Question 12:

Calculate coefficient of quartile deviation from the following data:

 X (Less than) 200 300 400 500 600 Frequency 8 20 40 46 50

Converting less than cumulative frequency into normal frequency distribution:

 X c.f. Frequency 100−200 8 8 200−300 20 20−8=12 300−400 40 40−20=20 400−500 46 46−40=6 500−600 50 50−46=4 N=50

Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{50}{4}\right)\mathrm{th}$ item
= Size of 12.5th item

12.5th item lies in group  200−300 and falls within 20th c.f. of the series.

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 $3\left(\frac{50}{4}\right)\mathrm{th}$ item
= Size of 37.5th item

37.5th item lies in group 300−400 and fall within the 40th c.f. of the series

#### Question 13:

Estimate an appropriate measure of the following data:

 Wages (₹) Less than 25 25−30 30−35 35−40 Above 40 No. of workers 2 10 26 16 7

 Wages No. of worker (Frequency) Cumulative Frequency Less than 25 2 2 25−30 10 12 30−35 26 38 35−40 16 54 Above 40 7 61 N=61

Q1 = Size of $\left(\frac{N}{4}\right)\mathrm{th}$ item
= Size of $\left(\frac{61}{4}\right)\mathrm{th}$ item
= Size of 15.25th item

15.25th item lies in group 30−35 and falls within 38th c.f. of the series.

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 $3\left(\frac{61}{4}\right)\mathrm{th}$ item
= Size of 45.75th item

47.75th item lies in group 35−40 and fall within the 54th c.f. of the series

#### Question 14:

Calculate the mean deviaiton from median and its coefficient from the data: 100, 150, 80, 90, 160, 200, 140

 Sr. No Values     (X) Deviation from Median              M= 140 1 2 3 4 5 6 7 80 90 100 140 150 160 200 60 50 40 0 10 20 60 N=7 $\Sigma \left|{d}_{m}\right|=240$

(M) Median = Size of
= Size of
= Size of 4th item
= 140

#### Question 15:

Compute Mean deviation and its coefficient by mean from the data given below:

 X 210 220 225 225 225 240 250 270 280

 Sr. No Values (X) Deviation from Mean $\left|{d}_{\overline{X}}\right|=\left|X-\overline{X}\right|$ 1 210 28 2 220 18 3 225 13 4 225 13 5 225 13 6 235 3 7 240 2 8 250 12 9 270 32 10 280 42 N=10 ΣX = 2380 Σ$d\overline{{}_{X}}$ = 176

#### Question 16:

Following are the marks of the students. Find mean deviation and coefficient mean deviation from mean.

 Marks 5 10 15 20 25 30 35 40 No. of students 16 32 36 44 28 18 12 14

 Marks (X) Frequency (f) fX Deviation from Mean $\left|{d}_{\overline{X}}\right|=\left|X-\overline{X}\right|\phantom{\rule{0ex}{0ex}}\left|X-20.2\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$ 5 16 80 15.2 243.2 10 32 320 10.2 326.4 15 36 540 5.2 187.2 20 44 880 0.2 8.8 25 28 700 4.8 134.4 30 18 540 9.8 176.4 35 12 420 14.8 177.6 40 14 560 19.8 277.2 ΣX = 180 Σf = 200 ΣfX=4040 Σ$f\left|d\overline{){}_{X}}\right|$=1531.2

#### Question 17:

Find out the mean deviation from the median and its coefficient.

 Marks 10 11 12 13 14 No. of students 3 12 18 12 3

 Marks (X) Frequency (f) Cumulative frequency (c.f.) Deviation from Median $\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{X}\mathbf{-}\mathbf{M}\right|\phantom{\rule{0ex}{0ex}}\mathbit{M}\mathbit{e}\mathbit{d}\mathbit{i}\mathbit{a}\mathbit{n}\mathbf{=}\mathbf{12}$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$ 10 3 3 2 6 11 12 15 1 12 12 18 33 0 0 13 12 45 1 12 14 3 48 2 6 N=Σf = 48 Σf|dM| =36

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 24.5 th item is 33 (in the c.f. column), which is corresponding to 12.

Hence, median is 12.
Thus we calculate  the deviation of the values from 12.

#### Question 18:

Compute the mean deviation from the median and from the mean for the following distribution of the scores of 50 college students:

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

 Scores (X) Frequency (f) Cumulative Frequency (c.f.) Mid -Values (m) Deviation from Median $\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{M}\right|\phantom{\rule{0ex}{0ex}}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{172}\mathbf{.}\mathbf{78}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$ 140−150 4 4 145 27.78 111.12 150−160 6 10 155 17.78 106.68 160−170 10 $\overline{)20}$ 165 7.78 77.8 $\overline{)170}$−180 $\overline{)18}$ 38 175 2.22 39.96 180−190 9 47 185 12.22 109.98 190−200 3 50 195 22.22 66.66 N=Σf = 50 Σf|dM|=512.2

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$item, i.e. ${\left(\frac{50}{2}\right)}^{th}$   item, which is 25th item.

This corresponds to the class interval of 170-180, so this is the median class.

 Scores (X) Frequency (f) Mid -Values (m) fm Deviation from Mean $\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\overline{)\mathbf{X}}\right|\phantom{\rule{0ex}{0ex}}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{171}\mathbf{.}\mathbf{2}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$ 140−150 4 145 580 26.2 104.8 150−160 6 155 930 16.2 97.2 160−170 10 165 1650 6.2 62 170−180 18 175 3150 3.8 68.4 180−190 9 185 1665 13.8 124.2 190−200 3 195 585 23.8 71.4 Σf = 50 Σfm=8560 Σ$\mathbit{f}\left|{\mathbf{d}}_{\overline{)\mathbf{X}}}\right|$=528

#### Question 19:

Find the mean deviation from mean and its coefficient for the given data:

 X 0−10 10−20 20−30 30−40 40−50 50−60 F 3 5 7 2 9 4

 Values (X) Frequency (f) Mid-Values (m) fm $\left|{\mathbit{d}}_{\overline{X}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\overline{\mathbit{X}}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbit{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|\mathbit{d}\overline{{}_{\mathbf{X}}}\right|$ 0−10 3 5 15 27 81 10−20 5 15 75 17 85 20−30 7 25 175 7 49 30−40 2 35 70 3 6 40−50 9 45 405 13 117 50−60 4 55 220 23 92 Σf = 30 Σfm=960 Σ$f\left|{d}_{\overline{X}}\right|$=430

#### Question 20:

Calculate mean deviation and its coefficient from the following figures:

 Class-Interval Frequency Less than 10 5 Less than 20 12 Less than 30 20 Less than 40 35 Less than 50 54 Less than 60 60

Converting less than cumulative frequency distribution into simple frequency distribution:

 Class Interval Cumulative Frequency (c.f.) Frequency (f) Mid-Values (m) $\left|{\mathbit{d}}_{\mathbit{M}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\mathbit{M}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\mathbf{36}\mathbf{.}\mathbf{67}\right|$ $\mathbit{f}\mathbf{×}\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathbit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$ 0−10 5 5 5 31.67 158.35 10−20 12 7 15 21.67 151.69 20−30 20 8 25 11.67 93.36 30−40 35 15 35 1.67 25.65 40−50 54 19 45 8.33 158.27 50−60 60 6 55 18.33 109.98 Σf=N=60 Σf|dM|=696.7

Median class is given by the size of  , i.e. , which is the 30th item.

This corresponds to the class interval of 30-40, so this is the median class.

#### Question 21:

Calculate the Standard deviation from the following values:8 , 9, 15, 23, 5, 11, 19, 8, 10, 12.

 Values (X) $\mathbit{x}\mathbf{=}\left|\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right|$ x2 8 −4 16 9 −3 9 15 3 9 23 11 121 5 −7 49 11 −1 1 19 7 49 8 −4 16 10 −2 4 12 0 0 ΣX=120 Σx2=274

Hence, the standard deviation of the above series is 5.23

#### Question 22:

Find the standard deviation for the following data: 3, 5, 6, 7, 10, 12, 15, 18.

 Values (X) $\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{\mathbf{X}}$ x2 3 −6.5 42.25 5 −4.5 20.25 6 −3.5 12.25 7 −2.5 6.25 10 .5 .25 12 2.5 6.25 15 5.5 30.25 18 8.5 72.25 ΣX=76 Σx2=190

Hence, standard deviation of the above series is 4.87

#### Question 23:

Find out the standard deviation of the height of 10 men given below: 160, 160, 161, 162, 163, 163, 163, 164, 164, 170.

 Values (X) $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right)$ x2 160 −3 9 160 −3 9 161 −2 4 162 −1 1 163 0 0 163 0 0 163 0 0 164 1 1 164 1 1 170 7 49 ΣX=1630 Σx2=74

Hence, standard deviation of the above series is 2.72

#### Question 24:

Calculate standard deviation of the given below:

 Size 3 4 5 6 7 8 9 Frequency 3 7 22 60 85 32 8

 Size (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right)$ x2 fx2 3 3 9 3.59 12.89 38.67 4 7 28 2.59 6.71 46.97 5 22 110 1.59 2.53 55.66 6 60 360 0.59 0.35 21 7 85 595 0.41 0 .17 14.45 8 32 256 1.41 1.99 63.68 9 8 72 2.41 5.81 46.48 N=Σf=217 ΣfX=1430 Σfx2=286.91

Hence, standard deviation of the above series is 1.149

#### Question 25:

Find the value of standard deviation and coefficient of variation from the following:

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

 Size (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbf{X}\mathbf{-}\overline{\mathbf{X}}\right)$ x2 fx2 10 6 60 −34 1156 6936 20 8 160 −24 576 4608 30 16 480 −14 196 3136 40 15 600 −4 16 240 50 32 1600 +6 36 1152 60 11 660 16 256 2816 70 12 840 26 676 8112 Σf=100 ΣfX=4400 Σfx2=27000

#### Question 26:

Measurements are made to the nearest cm. of the heights of 10 children. Calculate mean and standard deviation.

 Heights (cms) 60 61 62 63 64 65 66 67 68 No. of children 2 0 15 29 25 12 10 4 3

 Height (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{\mathbit{X}}\right)$ x2 fx2 60 2 120 −3.89 15.13 30.26 61 0 0 −2.89 8.35 0 62 15 930 −1.89 3.57 53.55 63 29 1827 −0.89 0.79 22.91 64 25 1600 0.11 0.01 0.25 65 12 780 1.11 1.23 14.76 66 10 660 2.11 4.45 44.5 67 4 268 3.11 9.67 38.68 68 3 204 4.11 16.89 50.67 Σf=100 ΣfX=6389 Σfx2=255.58

Hence mean of the above series is 63.89 cms and standard deviation is 1.6 cms.

#### Question 27:

Calculate standard deviation for the given data:

 Age (in yrs) 20−25 25−30 30−35 35−40 40−45 45−50 50−55 No. of workers 17 11 8 5 4 3 2

 Age (X) No. of Workers (f) Mid-Values (m) fm m2 $\mathbit{f}\mathbf{×}{\mathbit{m}}^{\mathbf{2}}\mathbf{=}\mathbit{f}{\mathbit{m}}^{\mathbf{2}}$ 20−25 17 22.5 382.5 506.25 8606.25 25−30 11 27.5 302.5 756.25 8318.75 30−35 8 32.5 260 1056.25 8450 35−40 5 37.5 187.5 1406.25 7031.25 40−45 4 42.5 170 1806.25 7225 45−50 3 47.5 142.5 2256.25 6768.75 50−55 2 52.5 105 2756.25 5512.5 Σf=50 Σfm=1550 Σfm2=51912.5

Hence, standard deviation of the above series is 8.79 years.

#### Question 28:

Calculate Standard deviation from the following series:

 Class 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Frequency 2 4 6 8 6 4 2

 Class Interval (X) Frequency (f) Mid-Values (m) fm m2 $\mathbit{f}\mathbf{×}{\mathbit{m}}^{\mathbf{2}}\mathbf{=}\mathbit{f}{\mathbit{m}}^{\mathbf{2}}$ 0−10 2 5 10 25 50 10−20 4 15 60 225 900 20−30 6 25 150 625 3750 30−40 8 35 280 1225 9800 40−50 6 45 270 2025 12150 50−60 4 55 220 3025 12100 60−70 2 65 130 4225 8450 Σf=32 Σfm=1120 Σfm2=47200

Hence, standard deviation of the above series is 15.81

#### Question 29:

From the following figures, find the standard deviation and the coefficient of variation:

 Marks 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80 No. of persons 5 10 20 40 30 20 10 4

 Marks (X) No. of Persons (f) Mid-Values (m) fm m2 fm2 0−10 5 5 25 25 125 10−20 10 15 150 225 2250 20−30 20 25 500 625 12500 30−40 40 35 1400 1225 49000 40−50 30 45 1350 2025 60750 50−60 20 55 1100 3025 60500 60−70 10 65 650 4225 42250 70−80 4 75 300 5625 22500 Σf=139 Σfm=5475 Σfm2=249875

#### Question 31:

The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3 and the standard deviations are 8 and 7. Find the mean and the standard deviation of the sample of size 150 obtained by combining the two samples.

Calculating combined standard deviation,

To calculate combined standard deviation, we need to find the combined mean of the observations.

Thus, Combined mean is

Hence, the combined standard deviation is 7.56

#### Question 32:

For a group of 100 males, mean and standard deviation of their daily wages are ₹ 36 and ₹ 9 respectively. For a group of 50 females, it is ₹ 45 and ₹ 6. Find the standard deviation for the whole group.

Calculating combined standard deviation,

To calculate combined standard deviation, we need to find the combined mean of the observations.

Thus, Combined mean is

Hence, the combined standard deviation is Rs 9.16

#### Question 33:

The profit of two business concerns for 5 years are as given below. Draw Lorenz Curves to show the distribution.

 Year 2001 2002 2003 2004 2005 Firm A 15 30 45 60 50 Firm B 20 30 45 60 45

 Years Firm A Cumulative Profits of Firm A Cumulative Percentage Profits of Firm A    (x) Firm B Cumulative Profits of Firm B Cumulative Percentage Profits of Firm B     (y) Coordinates of Lorenz Curve    (x,y) 2001 15 15 $\frac{15}{200}×100=7.5$ 20 20 $\frac{20}{200}×100=10$ (7.5,10) 2002 30 45 $\frac{45}{200}×100=22.5$ 30 50 $\frac{50}{200}×100=25$ (22.5, 25) 2003 45 90 $\frac{90}{200}×100=45$ 45 95 $\frac{95}{200}×100=47.5$ (45, 47.5) 2004 60 150 $\frac{150}{200}×100=75$ 60 155 $\frac{155}{200}×100=77.5$ (75,77.5) 2005 50 200 $\frac{200}{200}×100=100$ 45 200 $\frac{200}{200}×100=100$ (100,100) $\Sigma {N}_{1}$=200 $\Sigma {N}_{2}$=200

#### Question 34:

The given table shows the daily income of workers of two factories. Draw the Lorenz Curves for both the factories.

 Daily Income (₹) 0−100 100−200 200−300 300−400 400−500 Factory A 8 7 5 3 2 Factory B 15 6 2 1 1

 Daily Income Mid Value Cumulative Mid Value % Cumulative Mid Value No. of worker (f) (c.f.) % Cumulative Frequency 0−100 50 50 4 8 8 32 100−200 150 200 16 7 15 60 200−300 250 450 36 5 20 80 300−400 350 800 64 3 23 92 400−500 450 1250 100 2 25 100

 Daily Income Mid Value Cumulative Mid Value % Cumulative Mid Value No. of worker (f) (c.f.) % Cumulative Frequency 0−100 50 50 4 15 15 60 100−200 150 200 16 6 21 84 200−300 250 450 36 2 23 92 300−400 350 800 64 1 24 96 400−500 450 1250 100 1 25 100

#### Question 35:

Find the mean deviation from mean and its coefficient for the given data:

 Marks (more than) 0 10 20 30 40 50 60 No. of students 200 180 150 100 40 15 5

 Marks (X) Frequency (f) Mid-Values (m) fm $\left|{\mathbit{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\overline{\mathbit{X}}\right|\mathbf{=}\left|\mathbit{m}\mathbf{-}\mathbf{29}\mathbf{.}\mathbf{5}\right|$ $\mathbit{f}\mathbf{×}\left|\mathbit{d}\overline{{}_{\mathbf{X}}}\right|\mathbf{=}\mathbit{f}\left|{\mathbit{d}}_{\overline{\mathbf{X}}}\right|$ 0−10 20 5 100 24.5 490 10−20 30 15 450 14.5 435 20−30 50 25 1250 4.5 225 30−40 60 35 2100 5.5 330 40−50 25 45 1125 15.5 387.5 50−60 10 55 550 25.5 255 60−70 5 65 325 35.5 177.5 Σf=200 Σfm=5900 Σ$f\left|{d}_{\overline{X}}\right|$=2300

#### Question 36:

Calculate the Mean and Standard Deviation from the following distribution.

 Age (years) 15−19 20−24 25−29 30−34 35−39 40−44 No. of Persons 4 20 38 24 10 4

In order to calculate the mean and standard deviation, we first need to convert the inclusive series into exclusive series as given below:

 Age (X) Frequency (f) Mid-Values (m) fm m2 fm2 14.5−19.5 4 17 68 289 1156 19.5−24.5 20 22 440 484 9680 24.5−29.5 38 27 1026 729 27702 29.5−34.5 24 32 768 1024 24576 34.5−39.5 10 37 370 1369 13690 39.5−44.5 4 42 168 1764 7056 Σf=100 Σfm=2840 Σfm2=83860

Hence, mean age of the persons is 28.4 years and standard deviation is 5.66 years.

#### Question 37:

The following table gives the weights of one hundred persons. Copute the coefficient of dispersion by the method of limits.

 Class-interval 40−45 45−50 50−55 55−60 60−65 65−70 70−75 75−80 80−85 85−90 No. of persons 4 13 8 14 9 16 17 9 8 2

Given:

Upper limit of the Highest Class Interval (H) = 90
Lower limit of the Lowest Class Interval (L) = 40

Range = Highest Value − Lowest Value

i.e R= H−  L

Substituting the given values in the formula.

R= 90 − 40= 50

Hence, range of the above series is 50 kg and coefficient of range is 0.384

#### Question 38:

Calculate standard deviation of the following data:

 Age in years (below) 10 20 30 40 50 60 70 80 No. of persons 15 30 53 75 100 110 115 125

 Age (X) Cumulative Frequency (c.f.) Frequency (f) Mid-Values (m) fm m2 fm2 0−10 15 15-0=15 5 75 25 375 10−20 30 30-15=15 15 225 225 3375 20−30 53 53-15=23 25 575 625 14375 30−40 75 75-23=22 35 770 1225 26950 40−50 100 100-75=25 45 1125 2025 50625 50−60 110 110-100=10 55 550 3025 30250 60−70 115 115-110=5 65 325 4225 21125 70−80 125 125-115=10 75 750 5625 56250 Σf=125 Σfm=4395 Σfm2=203325

Hence, standard deviation of the above series is 19.76 years.

#### Question 39:

Price of a particular item in 10 years in two cities are given below, which city has more stable prices?

 City A 55 54 52 53 56 58 52 50 51 49 City B 108 107 105 105 106 107 104 103 104 101

 City A City B XA ${\mathbit{x}}_{\mathbit{A}}\mathbf{=}{\mathbit{X}}_{\mathbit{A}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{A}}$ ${\mathbit{x}}_{\mathbf{A}}^{\mathbf{2}}$ XB ${\mathbit{x}}_{\mathbf{B}}\mathbf{=}{\mathbit{X}}_{\mathbf{B}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{B}}$ ${\mathbit{x}}_{\mathbf{B}}^{\mathbf{2}}$ 55 2 4 108 3 9 54 1 1 107 2 4 52 −1 1 105 0 0 53 0 0 105 0 0 56 3 9 106 1 1 58 5 25 107 2 4 52 −1 1 104 −1 1 50 −3 9 103 −2 4 51 −2 4 104 −1 1 49 −4 16 101 −4 16 ΣXA=530 $\mathrm{\Sigma }{x}_{\mathrm{A}}^{2}=70$ ΣXB=1050 $\mathrm{\Sigma }{x}_{\mathrm{B}}^{2}=40$

Since C.V. of city B is less. Therefore, prices are more stable in city B.

#### Question 40:

For a distribution, the coefficient of variation is 22.5% and mean is 7.5. Calculate standard deviation.

Given,

Coefficient of variation= 22.5%
Mean, $\overline{)\mathit{X}}$= 7.5

Hence, standard deviation is 1.69

#### Question 41:

Find out the arithmetic mean and standard deviation from the following data:

 Variable 5−10 10−15 15−20 20−25 25−30 30−35 Frequency 2 9 29 54 11 5

 Variable (X) Frequency (f) Mid-Values (m) fm m2 fm2 5−10 2 7.5 15 56.25 112.5 10−15 9 12.5 112.5 156.25 1406.25 15−20 29 17.5 507.5 306.25 8881.25 20−25 54 22.5 1215.0 506.25 27337.5 25−30 11 27.5 302.5 756.25 8318.75 30−35 5 32.5 162.5 1056.25 5281.25 Σf=110 Σfm=2315 Σfm2=51337.5

#### Question 42:

The mean and standard deviation of a series of 20 items are 20 and 5 respectively. While calculating these measures, an item of 13 was wrongly read as 30. Find out the correct mean and standard deviation.

Calculating correct Mean, by using the following observations:

Squaring both sides

$25=\frac{\mathrm{\Sigma }{x}^{2}}{20}-400$

425 × 20 = Σx2

$\Sigma {{x}^{2}}_{wrong}$= 8500

$\Sigma {{x}^{2}}_{correct}$ = $\Sigma {{x}^{2}}_{wrong}$ − (Incorrect item)2 + (Correct item)2

$\Sigma {{x}^{2}}_{correct}$= 8500 − (30)2 + (13)2
$\Sigma {{x}^{2}}_{correct}$= 8500 − 900 + 169 = 7769

Hence, correct mean and standard deviation are 19.15 and 4.66 respectively.

#### Question 43:

Following are the marks obtained by two students: Mollie and Isha, in 10 sets of examinations:

 Marks of obtained by Mollie 44 80 76 48 52 72 68 56 60 54 Marks of obtained by Isha 48 75 54 60 63 69 72 51 57 66
Out of Mollie and Isha, who is more consistent?

 Mollie Isha XM ${\mathbit{x}}_{\mathbf{M}}\mathbf{=}{\mathbit{X}}_{\mathbf{M}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{M}}$ ${\mathbit{x}}_{\mathbf{M}}^{\mathbf{2}}$ XI ${\mathbit{x}}_{\mathbf{I}}\mathbf{=}{\mathbit{X}}_{\mathbf{I}}\mathbf{-}{\overline{\mathbit{X}}}_{\mathbit{I}}$ ${\mathbit{x}}_{\mathbit{I}}^{\mathbf{2}}$ 44 −17 289 48 −13.5 182.25 80 19 361 75 13.5 182.25 76 15 225 54 −7.5 56.25 48 −13 169 60 −1.5 2.25 52 −9 81 63 1.5 2.25 72 11 121 69 7.5 56.25 68 7 49 72 10.5 110.25 56 −5 25 51 −10.5 110.25 60 −1 1 57 −4.5 20.25 54 −7 49 66 4.5 20.25 ΣXM=610 $\mathrm{\Sigma }{x}_{\mathrm{M}}^{2}=1370$ ΣXI=615 $\mathrm{\Sigma }{x}_{\mathrm{I}}^{2}=742.5$

Isha is more consistent in securing makes as her C.V. (14.01)% is less than that of Mollie's C.V. (19.18)%

#### Question 44:

Calculate coefficient of variation from the following data:

 Marks (more than) 0 10 20 30 40 50 60 70 No. of students 100 90 75 50 20 10 5 0

Converting more than cumulative frequency distribution into simple frequency distribution:

 Marks (X) No. of Students (f) Mid-Values (m) fm m2 fm2 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 −70 70 −80 100-90=10 90-75=15 75-50=25 50-20=30 20-10=10 10-5=5 5-0=5 0-0=0 5 15 25 35 45 55 65 75 50 225 625 1050 450 275 325 0 25 225 625 1225 2025 3025 4225 5625 250 3375 15625 36750 20250 15125 21125 0 Σf =100 Σfm = 3000 Σfm2 =112500

#### Question 45:

In two Towns A and B, daily pocket money and the standard deviation are given below:

 Town Average Daily Pocket Standard Deviation No. of Teenagers A 34.5 5.0 476 B 28.5 4.5 524

(i) Which town, A or B, pays out the larger amount of daily pocket money?
(ii) What is the average daily pocket money of all teenagers taken together?
(iii) Calculate coefficient of variation of each town. Which town is more variable in terms of pocket money?

(i) Town A
Daily pocket money = Average Daily Pocket × No. of teenagers
= 34.5 × 476
= Rs 16422

Town B
Daily Pocket Money = 28.5 × 524
= Rs 14934

Town 'A' pays larger amount of daily pocket money.

(ii) Solution:

In order to find out average daily pocket money of all teenagers, we need find out the combined mean of their pocket money.

Thus,

(iii)

Town B is more variable in terms of pocket money as its C.V. is higher than that of C.V. of Town A.

#### Question 46:

The prices of share of Company X and Company Y are given below. State, which company is more stable?

 Company X 25 50 45 30 70 42 36 48 34 60 Company Y 10 70 50 20 95 55 42 60 48 80

 Company X Company Y X $\mathbit{x}\mathbf{=}\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}$ x2 Y $\mathbit{y}\mathbf{=}\mathbit{Y}\mathbf{-}\overline{)\mathbit{Y}}$ y2 25 50 45 30 70 42 36 48 34 60 −19 6 1 −14 26 −2 −8 4 −10 16 361 36 1 196 676 4 64 16 100 256 10 70 50 20 95 55 42 60 48 80 −43 17 −3 −33 42 2 −11 7 −5 27 1849 289 9 1089 1764 4 121 49 25 729 ΣX = 440 Σx2 = 1710 ΣY = 530 Σy2 = 5928

As C.V of prices of shares of Co. X is less than that of the prices of shares of Co. Y therefore,  price of share of  Co. X is more stable.

#### Question 47:

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50 instead of 40. Find the correct mean and standard deviation.

Calculating correct Mean, by using the following observations:

Squaring both sides

$26.01=\frac{\mathrm{\Sigma }{x}^{2}}{100}-1600\phantom{\rule{0ex}{0ex}}$

Hence, correct mean and standard deviation are 39.9 and 5 respectively.

#### Question 48:

From the following data, calculate standard deviation of the two groups A and B. Which group is more consistent?

 Class Interval Group A Group B 5−10 2 9 10−15 9 11 15−20 29 18 20−25 54 32 25−30 11 27 30−35 5 13

For group A

 Class Interval Frequency (f) Mid-Values (m) fm m2 fm2 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 2 9 29 54 11 5 7.5 12.5 17.5 22.5 27.5 32.5 15 112.5 507.5 1215 302.5 162.5 56.25 156.25 306.25 506.25 756.25 1056.25 112.5 1406.25 8881.25 27337.5 8318.75 5281.25 Σf = 110 Σfm = 2315 Σfm2 = 51337.5

For group B

 Class Interval Frequency (f) Mid-Values (m) fm m2 fm2 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 9 11 18 32 27 13 7.5 12.5 17.5 22.5 27.5 32.5 67.5 137.5 315 720 742.5 422.5 56.25 156.25 306.25 506.25 756.25 1056.25 506.25 1718.75 5512.5 16200 20418.75 13731.25 Σf = 110 Σfm = 2405 Σfm2 = 58087.5

Group A is more consistent as C.V. of group A is less than C.V. of group B.

#### Question 49:

From the following data of marks, calculate standard deviation. What will be the value of standard deviation, if marks obtained by each student is increased by one?

 Marks Obtained 1 2 3 4 5 6 7 8 9 No. of students 32 41 57 98 123 83 46 17 3

 Marks (X) Frequency (f) fX $\mathbit{x}\mathbf{=}\left(\mathbit{X}\mathbf{-}\overline{)\mathbit{X}}\right)$ x2 fx2 1 2 3 4 5 6 7 8 9 32 41 57 98 123 83 46 17 3 32 82 171 392 615 498 322 136 27 −3.55 −2.55 −1.55 −0.55 0.45 1.45 2.45 3.45 4.45 12.60 6.50 2.40 0.30 0.20 2.10 6.00 11.90 19.80 403.2 266.5 136.8 29.4 24.6 174.3 276 202.3 59.4 Σf = 500 Σfx = 2275 Σfx2 = 1572.5

As, standard deviation is independent of the change in origin, i.e. it will not get affected if the value of the series is increased or decreased by a constant quantity. Therefore it will remain the same.

#### Question 50:

From the following data of two workers, identify who is a more consistent worker?

 Worker A B Average Time in completing a job 40 42 Standard Deviation 8 6

Worker B is more consistent as his C.V. (14.29)% is less than that of C.V of worker A(20%).

#### Question 51:

Find the standard deviation and coefficient of standard deviation:

 X (less than) 10 20 30 40 50 60 70 80 Frequency 12 30 65 107 157 202 222 230

Converting less than frequency distribution into simple frequency distribution:

 X Frequency (f) Mid-Values (m) fm m2 fm2 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 12-0=12 30-12=18 65-30=35 107-65=42 157-107=50 202-157=45 222-202=20 230-222=8 5 15 25 35 45 55 65 75 60 270 875 1470 2250 2475 1300 600 25 225 625 1225 2025 3025 4225 5625 300 4050 21875 51450 101250 136125 84500 45000 Σf = 230 Σfm = 9300 Σfm2 = 444550

#### Question 52:

Find mean, standard deviation and coefficient of variation.

 Class-interval 0−4 4−8 8−12 12−16 16−20 20−24 Frequency 10 15 20 25 20 10

 Class Interval Frequency (f) Mid-Values (m) fm m2 fm2 0 − 4 4 − 8 8 − 12 12 − 16 16 − 18 20 − 24 10 15 20 25 20 10 2 6 10 14 18 22 20 90 200 350 360 220 4 36 100 196 324 484 40 540 2000 4900 6480 4840 Σf = 100 Σfm = 1240 Σfm2 = 18800

#### Question 53:

The following table shows the marks obtained by 60 students. Calculate mean and standard deviation.

 Marks (more than) 70 60 50 40 30 20 No. of students 7 18 40 40 55 60

Converting more than cumulative frequency into ordinary continuos series:

 Marks X Frequency (f) Mid-Values (m) fm m2 fm2 20 −30 30 −40 40 −50 50 −60 60 −70 70 −80 60-55=5 55-40=15 40-40=0 40-18=22 18-7=11 7-0=7 25 35 45 55 65 75 125 525 0 1210 715 525 625 1225 2025 3025 4225 5625 3125 18375 0 66550 46475 39375 Σf = 60 Σfm = 3100 Σfm2 = 173900

Hence, mean and standard deviation of the above series are 51.67 marks and 15.11 marks respectively.

#### Question 54:

Calculate standard deviation and coefficient of dispersion from the data below:

 Mid-Points 5 15 25 35 45 55 65 75 Frequency 5 8 7 12 28 20 10 10

 Mid-Values (m) Frequency (f) fm m2 fm2 5 15 25 35 45 55 65 75 5 8 7 12 28 20 10 10 25 120 175 420 1260 1100 650 750 25 225 625 1225 2025 3025 4225 5625 125 1800 4375 14700 56700 60500 42250 56250 Σf = 100 Σfm= 4500 Σfm2 = 236700

#### Question 55:

The following data shows the expected life of two models of T.V.: A and B:

 Life (no. of years) 0−2 2−4 4−6 6−8 8−10 10−12 Model A 5 16 13 7 5 4 Model B 2 7 12 19 9 1
Which model has greater uniformity.

 Life (No. of years) Frequency (f) Mid Values (m) fm m2 fm2 0 − 2 2 − 4 4 − 6 6 − 8 8 −10 10 − 12 5 16 13 7 5 4 1 3 5 7 9 11 5 48 65 49 45 44 1 9 25 49 81 121 5 144 325 343 405 484 Σf = 50 Σfm = 256 Σfm2 = 1706

 Life (No. of years) Frequency (f) Mid Values (m) fm m2 fm2 0 − 2 2 − 4 4 − 6 6 − 8 8 −10 10 − 12 2 7 12 19 9 1 1 3 5 7 9 11 2 21 60 133 81 11 1 9 25 49 81 121 2 63 300 931 729 121 Σf = 50 Σfm = 308 Σfm2 = 2146

As,  C.V of Model B is less than that of C.V of model A therefore model B has greater uniformity.

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