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Question-1

Find the mean deviation from the mean for the following data: 4, 7, 8, 9, 10, 12, 13, 17

Solution:
= = = 10

= 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24

M.D.() = 24/8 = 3

Question-2

Find the mean deviation from the mean for the following data:38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solution:
=== 50

= 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84

M.D.() = 84/10 = 8.4

Question-3

Find the mean deviation from the mean for the following data:13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Solution:
=== 14
= 1 + 3 + 2 + 0 + 3 + 1 + 4 + 2 + 3 + 4 + 2 + 3 = 28

M.D.() = 28/12 = 2.33

Question-4

Find the mean deviation from the mean for the following data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Solution:
=== 50

= 14 + 22 + 4 + 8 + 10 + 5 + 3 + 4 + 1 + 1 = 72

M.D.() = 72/10 = 7.2

Question-5

 


Solution:

xi

fi

f ixi

|xi - |

fi|xi - |

  5

7

35

7

49

10

4

40

2

  8

15

6

90

3

18

20

3

60

8

24

25

5

75

13  

65

Total

25   

300  

 

164  

= 300/25 = 12
M.D () = 164/25 = 6.56

 

Question-6

 


Solution:

xi

fi

f ixi

|xi - |

fi|xi - |

10

  4

    40

40

160

30

24

  720

20

480

50

28

1400

  0

   0

70

16

1120

20

320

90

  8

  720

40

320

Total

80

4000

 

1280  

= 4000/80 = 50
M.D () = 1280/80 = 16

 

Question-7

 


Solution:

xi

fi

f ixi

|xi -|

fi|xi -|

10

  4

    40

40

160

30

24

  720

20

480

50

28

1400

  0

   0

70

16

1120

20

320

90

  8

  720

40

320

Total

80

4000

 

1280  

= 234/26 = 9
M.D () = 88/26 = 3.39

 

Question-8

fi: 8

 


Solution:

xi

fi

c.f.

|xi - |

fi|xi - |

15

3

3

|15-30|=15

45

21

5

8

|21-30|=9

45

27

6

14

|27-30|=3

 18

30

7

21

|30-30|=0

 0

35

8

29

|35-30|=5

 40

Total

29  

   

 

148

N = 29(odd), which is an odd number, therefore

Median observation =

Hence, Median = 30

Mean Deviation (M) =

Question-9

 


Solution:

Classes

xi

fi

f ixi

|xi - |

fi|xi - |

    0-100

  50

4

  200

308

1232

100-200

150

8

1200

208

1664

200-300

250

9

2250

108

972

300-400

350

10   

3500

   8

  80

400-500

450

7

3150

  92

644

500-600

550

5

2750

192

960

600-700

650

4

2600

292

1168

700-800

750

3

2250

392

1176

Total

 

50  

17900   

 

7896

=17900/50 = 358
M.D () = 7896/50 = 157.92

 

Question-10

 


Solution:

Classes

xi

fi

f ixi

|xi -|

fi|xi - |

  95-105

100

  9

900

25.3    

227.7

105-115

110

13

 1430   

15.3    

198.9

115-125

120

26

 3120   

5.3  

137.8

125-135

130

30

 3900   

4.7  

 141    

135-145

140

12

 1680   

14.7   

176.4

145-155

150

10

 1500   

24.7   

 247    

Total

 

100  

12530   

 

1128.8    

=12530/100 = 125.3
M.D () = 1128.8/100 = 11.282

 

Question-11

 


Solution:

Classes

xi

fi

f ixi

|xi - |

fi|xi - |

  0-10

 5

 6

    30

 8.5

 51

10-20

15

 8

  120

 1.5

 12

20-30

25

14

  350

11.5

161

30-40

35

16

  560

21.5

344

40-50

45

 4

  180

31.5

126

50-60

55

 2

  110

41.5

  83

Total

 

100   

 1350

 

777

=1350/100 = 13.5
M.D () = 777/100 = 7.77

 

Question-12

 


Solution:

Classes

xi

fi

c.f.

|xi - |

fi|xi - |

  15.5-20.5

18

  5

  5

    

100

20.5-25.5

23

6

  11

90

25.5-30.5

28

12

23   

  

120

30.5-35.5

33

14

37   

70  

35.5-40.5

26

63

63  

0

40.5-45.5

43

12

75   

60   

45.5-50.5

  48

16

91   

 

160    

50.5-55.5

53

9

100

135

 

     

 

35.5 – 40.5

Median = l +

Median Deviation(M) =

 

Question-13

Find the mean and variance for the following data: 6, 7, 10, 12, 13, 4, 8, 12

Solution:
==== 9.

The respective (xi - )2 are 32, 22, 12, 32, 42, 52, 12, 32.

(xi - )2 = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74

Hence variance (σ 2) = 74/8 = 9.25

Question-14

Find the mean and variance for first n natural numbers.

Solution:
Mean of first ‘n’ natural numbers :

Variance :

                                =

                                =

                                =

                                =

                                =

Question-15

Find the mean and variance of first 10 multiples of 3.

Solution:
The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

Mean =

Now, 32 + 62+92+122+152+182+212+ 242+ 272+302. = 3465

 

Variance =

Question-16

 


Solution:

xi

fi

f ixi

xi -

(xi - )2

f i(xi - )2

6

  2

  12

-13

169

338

10

  4

  40

-9

81

324

14

  7

  98

-5

25

175

18

12

216

-1

1

12

24

  8

192

5

25

200

28

  4

112

9

81

324

30

  3

  90

11

121

363

Total

40

760  

 

 

1736   

==760/40 = 19

Variance (σ 2) = 1736/40 = 43.4

Therefore Standard Deviation (σ ) = =6.58

Question-17

 


Solution:

xi

fi

f ixi

xi -

(xi - )2

f i(xi - )2

  92

3

276

-8

64

192

  93

2

186

-7

49

  98

  97

3

291

-3

  9

  27

  98

2

196

-2

  4

   8

102

6

612

  2

  4

 24

104

3

312

  4

16

 48

109

3

327

  9

81

243

Total

22  

2200

 

 

640

==2200/22 = 100

Variance (σ 2) = 640/22 = 29.09

Therefore Standard Deviation (σ ) = =5.39

Question-18

 


Solution:

xi

yi = xi - 64

f i

f i y i

f i y i2

60

-4

  2

 -8

32

61

-3

  1

 -3

  9

62

-2

12

-24

48

63

-1

29

-29

29

64

0

25

  0

  0

65

1

12

12

12

66

2

10

20

40

67

3

  4

12

36

68

4

  5

20

80

Total

 

100  

 0

286  

== 64

Variance (σ 2) = [(1)2/100][0× 286 -0] = 2.86

Therefore Standard Deviation (σ ) = =1.691

Question-19

 


Solution:

Classes

xi

yi = (xi – 105)/30

f i

f i y i

f i y i2

   0-30

  15

-3

  2

-6

18

 30-60

  45

-2

  3

-6

12

 60-90

  75

-1

  5

-5

  5

  90-120

105

0

10

 0

  0

120-150

135

1

  3

 3

  3

150-180

165

2

  5

10

20

180-210

195

3

  2

 6

18

Total

 

 

30

 2

76


== 107

Hence Variance (σ 2) = [(30)2/30][76 – 4/30] = 2276

Question-20

 


Solution:

Classes

xi

yi = (xi - 25)/10

f i

f i y i

f i y i2

  0-10

  5

-2

  5

-10

20

10-20

15

-1

  8

 -8

  8

20-30

25

 0

15

  0

  0

30-40

35

 1

16

16

16

40-50

45

 2

  6

12

24

Total

 

 

50

10

68

== 27

Hence Variance (σ 2) = [(10)2/50][68 - 100/50] = 132

Question-21

Calculate the standard deviation and mean diameter of the circles.

 


Solution:

Classes

xi

yi = (xi – 42.5)/4

f i

f i y i

f i y i2

32.5-36.5

34.5

-2

15

-30

60

36.5-40.5

38.5

-1

17

-17

17

40.5-44.5

42.5

0

21

  0

  0

44.5-48.5

46.5

1

22

 22

 22

48.5-52.5

50.5

2

25

 50

100

Total

 

 

100  

 25

199

Mean diameter of the circles = = = 43.5

Variance (σ 2) = [(4)2/100][199 – 625/100] = 30.84
Hence the Standard Deviation is (
σ ) = = 5.55

Question-22

 


Solution:

Group A

Group B

f i

f i y i

f i y i2

f i

f i y i

f i y i2

10-20

15

-3

  9

-27

81

10

-54  

162

20-30

25

-2

17

-34

68

20

-44  

  88

30-40

35

-1

32

-32

32

30

-40  

  40

40-50

45 =A

0

33

   0

0

25

0

   0

50-60

55

1

40

 40

40

43

32

  32

60-70

65

2

10

20

40

 15

16

  32

70-80

75

3

 9

   7

  81

  7

  6

  18

Total

 

 

150  

-6

342

150  

-6

366

 Group A
Variance (
σ 2) = h2[ - ] = (10)2[] = 100[ 2.28 – 0.0016] = 100 × 2.2784 = 227.84

Group B
Variance (
σ 2) = h2[ - ] = (10)2[] = 100[ 2.44 – 0.0016] = 243.84

Since variance for Group B is more than Group A. Therefore group B is more variable than A.

Question-23

 


Solution:

X

X2

y

Y2

 35

  1225

108

11664

  54

  2916

107

11449

52

2704

105

11025

53

2809

105

11025

56

3136

106

11236

58

3364

107

11449

50

2500

103

10609

51

2601

104

10816

49

2401

101

10201

X = 458

  X2 = 25456

  X = 946

Y2 = 99474  

σx 2 = = = 2828.44 – 2589.68 = 238.76

σY 2 = = = 11052.66 –11048.35 = 4.31

Since the variance for x is greater than variance of y , therefore y is more stable.

Question-24

Which firm A or B, shows greater variability in individual wages?

 


Solution:
Mean of monthly waqes for firm A and firm B are same i.e.,Rs.5253 .Since the wage earners of firm B are more than that of A,hence firm B pays out larger amount as monthly wages.

Variance for firm A = 100

Variance for firm B = 121

Since variance for firm B is greater than that of A, therefore firmB shows greater variability than that of A.

Question-25

For the team B , mean number of goals scored goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

 


Solution:

No. of goals scored xi

fi

fi xi

fi xi2 

0

1

0

0

1

9

9

9

2

7

14

28

3

5

15

 45

4

3

12

48

Total

n = 25

= 50

2 = 130

       

 












= = 2

S.D = = = = = 1.09

For team B, = 2, S.D = 1.25

Since their means are same , therefore σΑ is less than that σΒ, therefore team A is more consistent than team B.

 

Question-26

The sum and sum of squares corresponding to length x (in cm )and weight y
(in gm ) of 50 plant products are given below:
= 212 , = 902.8, = 261,
= 1457.6 which is more varying , the length or weight ?

Solution:
n =10

= 212 , = 902.8 , = 261 , = 1457.6

σ2length = = - = 18.056 – 17.9776 = 1.0790

σ2weigth = = - = 29.152 – 27.2484 = 1.9036.
As variance for weights is more than that for length , therefore weights are more varying than length.





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