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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:
6.5, 5, 5.25, 5.5, 4.75, 4.5, 6.25, 7.75, 8.5

Solution:
= = = 6

= 0.5 + 1 + 0.75 + 0.5 + 1.25 + 1.5 + 0.25 + 1.75 + 2.5 = 10

M.D.() = 10/9 = 1.1

Question-3

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-4

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-5

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-6

 


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-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  

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

Question-8

 


Solution:

xi

fi

f ixi

|xi - |

fi|xi - |

5

8

40

4

32

7

6

42

2

12

9

2

18

0

 0

10

2

20

1

 2

12

2

24

3

 6

15

6

90

6

36

Total

26  

234   

 

88

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

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

Find the mean deviation from the median for the following data:
34, 66, 30, 38, 44, 50, 40, 60, 42, 51

Solution:
No of observations n = 10

Arrangement in ascending order are as follows:

30, 34, 38, 40, 42, 44, 50, 51, 60, 66.

Median is 5th and 6th term i.e 42 and 44.

Therefore the median is (42 + 44)/2 = 43

|xi - Median| = 13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 17 + 23

Hence M.D (Median) = |xi - Median|/n = 87/10 = 8.7

Question-13

Find the mean deviation from the median for the following data:
22, 24, 30, 27, 29, 31, 25, 28, 41, 42

Solution:
No of observations n = 10

Arrangement in ascending order are as follows:

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Median is 5th and 6th term i.e 28 and 29.

Therefore the median is (28 + 29)/2 = 28.5

|xi - Median| = 6.5 + 4.5 + 3.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 + 12.5 + 13.5

Hence M.D (Median) = |xi - Median|/n = 47/10 = 4.7

Question-14

Find the mean deviation from the median for the following data:
38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Solution:
No of observations n = 10

Arrangement in ascending order are as follows:

34, 38, 42, 44, 47, 48, 55, 53, 63, 70,

Median is 5th and 6th term i.e 47 and 48.

Therefore the median is (47 + 48)/2 = 47.5

|xi - Median| = 13.5 + 9.5 + 5.5 + 3.5 + 0.5 + 0.5 + 7.5 + 5.5 + 15.5 + 22.5

Hence M.D (Median) = |xi - Median|/n = 84/10 = 8.4

Question-15

 


Solution:
xi

15

21

27

30

35

fi

3

5

6

7

8

cfi

3

8

14

21

29

No of observations n = 29

Median is 15th term i.e 30

Therefore the median is 30.

|xi - Median| has respective values 15, 9, 3, 0, 5.

Therefore fi |xi - Median| = 3× 15 + 5× 9 + 6× 3 + 7× 0 + 8× 5

= 45 + 45 + 18 + 0 + 40 = 358

Hence M.D (Median) = fi |xi - Median|/n = 148/29 = 5.1

Question-16

 


Solution:
xi

74

89

42

54

91

94

35

 fi

20

12

2

4

5

3

4

cfi

20

32

34

38

43

46

50

No of observations n = 50

Median is 25th and 26th term i.e 89

Therefore the median is 89.

|xi - Median| has respective values 15, 0, 47, 35, 2, 5, 54.

Therefore fi |xi - Median| = 20× 15 + 12× 0 + 2× 47 + 4× 35 + 5× 2 + 3× 5 + 4× 54

                                      = 300 + 0 + 94 + 140 + 10 + 15 + 216 = 775

Hence M.D (Median) = fi |xi - Median|/n = 775/50 = 15.5

Question-17

Find the arithmetic mean of the series 1, 2, 22, ……., 2n –1.

Solution:
= 1 + 2 + 22 + …….. + 2n-1

Sum are in G.P

== 2n - 1

A.M = /n = (2n-1)/ n

Question-18

 


Solution:
Let the missing frequencies of the classes 375-400 and 425 – 450 be a and b respectively.
 

Value

frequency

c.f.

300-325

5

5

325-350

17

22

350-375

80

102

375-400

a

102+a

400-425

326

428+a

425-450

b

428+a+b

450-475

88

512+a+b

475-500

9

525+a+b

 

1000

N = 1000

Median is given to be 413.11

Median class is 400 – 425

Median = L +
L = 400

C = 102+a

h = 25

413.11 = 400 +

13.11 ×326 = (500-102-a) 25

4273.86 = (398-a) 25

a = 227.0456 = 227

Also 525 +a + b = 1000

b = 1000 – 525 – 227 = 248

Missing frequency are 227 and 248.

Question-19

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-20

Find the mean and variance for the following data:
2, 4, 5, 6, 8, 17

Solution:
= = = = 7.

The respective (xi - )2 are 52, 32, 22, 12, 12, 102.

(xi - )2 = 25 + 9 + 4 + 1 + 1 + 100 = 140

Hence variance (σ 2) = 140/6 = 23.33

Question-21

Find the mean for the following data:
First n natural numbers

Solution:
====

Question-22

 


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-23

 


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-24

 


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-25

 


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-26

 


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-27

 


Solution:

classes

xi

yi = (xi – 55)/10

f i

f i y i

f i y i2

10-20

15

-4

 3

-12

  48

20-30

25

-3

 4

-12

  36

30-40

35

-2

 7

-14

  28

40-50

45

-1

 7

 -7

   7

50-60

55

0

15

 0

   0

60-70

65

1

 9

 9

   9

70-80

75

2

 6

12

  24

80-90

85

3

 6

18

  54

90-100

95

4

 3

12

  48

Total

   

60

6

254

= = 56

Variance (σ 2) = [(10)2/60][254 –36/60] = 422.33

Therefore standard Deviation (σ ) = = 20.55

Question-28

[Hint: First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5,40.5-44.5,44.5-48.5, 48.5-52.5 and the proceed]

 


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-29

[Hint: Compare the variance of two groups. The group with greater variance is more variable]

 


Solution:

classes

xi

vi = (xi-45)/10
 

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

18

-54  

162

20-30

25

-2

17

-34

68

22

-44  

  88

30-40

35

-1

32

-32

32

40

-40  

  40

40-50

45

0

23

   0

0

18

0

   0

50-60

55

1

40

 40

40

32

32

  32

60-70

65

2

18

 36

72

  8

16

  32

70-80

75

3

  1

   3

  9

  2

  6

  18

Total

   

140  

-14

302

140  

-84

372

Group A

Variance (σ 2) = [(10)2/140][302 – 196/140] = 214.7

Group B

Variance (σ 2) = [(10)2/140][372 – 7056/140] = 229.7

The variance group B is more than group A. Therefore group B has more variable.

Question-30

 


Solution:

classes

xi

yi = (xi – 22.5)/5

f i

f i y i

f i y i2

 0-5

  2.5

-4

20

-80

320

  5-10

  7.5

-3

24

-72

216

10-15

12.5

-2

32

-64

128

15-20

17.5

-1

28

-28

  28

20-25

22.5

0

20

 0

   0

25-30

27.5

1

11

 11

  11

30-35

32.5

2

26

 52

104

35-40

37.5

3

15

 45

135

40-45

42.5

4

24

 96

384

Total

   

200  

-40

1326  

= = 21.5

Hence Variance (σ 2) = [(5)2/200][1326 – 1600/200] = 164.75

Question-31

The mean and variance of 8 observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Solution:
Let the remaining two observations be x and y.

Then mean = =9
60 + x + y = 72

x + y = 12                 ………………………(i)

Variance = = 9.25

(-3)2 + (-2)2 + (1)2 + (3)2 + (3)2 + (4)2 + x2 + y2 –18(x + y)+ 2× 92= 9.25× 8

x2 + y2 – 216 + 210 = 74

x2 + y2 = 80             ……………………….(ii)

But from (i)

x2 + y2 = 144 – 2xy ………………………(iii)

144 – 2xy = 80

2xy = 64                   ………………………(iv)

Subtracting (iv) from (ii)

x2 + y2 – 2xy = 80 – 64

(x – y)2 = 16

x – y = ± 4                ……………………….(v)

Hence solving (i) and (v)

x = 8, y = 4 and x = 4, y = 8

Therefore the remaining two observations are 4 and 8.

Question-32

The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14, find the remaining two observations.

Solution:
Let the remaining two observations be x and y.

Then mean = =8
42 + x + y = 56

x + y = 14             ………………………(i)

Variance = = 16

(-6)2 + (-4)2 + (2)2 + (4)2 + (6)2 + x2 + y2 –16(x + y)+ 2× 82= 16× 7

x2 + y2 – 224 + 236 = 112

x2 + y2 = 100            …………………….(ii)

But from (i)
x2 + y2 = 196 – 2xy …………………….(iii)

196 – 2xy = 100

2xy = 96                 ……………………….(iv)

Subtracting (iv) from (ii)

x2 + y2 – 2xy = 100 – 96

(x – y)2 = 4

x – y = ± 2              ……………………….(v)

Hence solving (i) and (v)

x = 8, y = 6 and x = 6, y = 8

Therefore the remaining two observations are 8 and 6.

Question-33

The mean and variance of 6 observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Solution:
Let the observations be x1 , x2, x3, …., x20 and be their mean. Then

8 =

or

If each observation is multiplied by 3, the resulting observations are

3x1 , 3x2, 3x3, …., 3x20.


Their new mean == 3= 3× 8 = 24

and new variance === 3× 48 = 144

Therefore the new standard deviation is =12

Question-34

Given that is the mean and σ 2 is the variance of n observations x1, x2, x3,……xn. Prove that the mean and variance of the observations ax1, ax2, ax3,…..axn are a and a2 σ 2, respectively, (a 0).

Solution:
Let the observations be x1 , x2, x3, …., xn and be their mean. Thenσ 2 =

If each observation is multitplied by a, the resulting observations are

ax1, ax2, ax3,…..axn

Their new mean == a

And new variance === aσ 2

Hence proved.

Question-35

The mean of 20 observations are found to be 10. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean in each of the following cases:
(i)  If the wrong item is omitted.
(ii) If it is replaced by 12.

Solution:
Let the observations be x1 , x2, x3, …., x20 and be their mean. Then = 10
2 = or = 40

(i) Observation 8 is omitted.

New mean = = =10.11

(ii) Observation 8 is replaced by 12.

Difference = 12 – 8 = 4

New mean = = =10.2

Question-36

Prove that +…… where

Solution:
   =  = 0.

Question-37

Prove the identity  .

Solution:

     = (Since )
     = - n
     = -n(
     = -

Question-38

The mean of 9 items is 15. If one more item is added to this series, the mean becomes 16. Find the value of the 10th item.

Solution:
Let the value of 9 items be x1 , x , x2 …… x9

15 = x1+x2+…….x9 = 15 × 9 = 135

Let x10 be the 10th item

AM of x1,x2,……x9,x10 = 16

16 = x1+x2 ……….. x9+x10 = 160

135 + x10 = 160

Question-39

The average weight of a group of 25 items was calculated to be 78.4kg. It was later discovered that a weight was misread as 69kg instead of 96kg. Calculate correct average.

Solution:
No. of items = 25

Incorrect average = 78.4kg

Incorrect reading of weight of an item = 69kg

Correct reading of weight of an item = 96kg

Let the variable weight be denoted by ‘x’

Incorrect

78.4 =

Incorrect

New correct - incorrect weight of an item + correct weight of an item

Correct kg

Question-40

The mean of 9 items is 15. If one more item is added to this series, the mean becomes 16. Find the value of the 10th item.

Solution:
Let the value of 9 items be x1 , x , x2 …… x9

15 = x1+x2+…….x9 = 15 × 9 = 135

Let x10 be the 10th item


AM of x1,x2,……x9,x10 = 16

16 = x1+x2 ……….. x9+x10 = 160

135 + x10 = 160





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