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Page No 32.11:

Question 1:

Calculate the mean deviation from the median of the following frequency distribution:

Heights in inches 58 59 60 61 62 63 64 65 66
No. of students 15 20 32 35 35 22 20 10 8

Answer:

In order to find the mean deviation from the median, we will first have to calculate the median.
M is the value of x corresponding to the cumulative frequency just greater than or equal to N2.
 

xi fi Cumulative Frequency di = xi-61 fidi
58 15 15 3 45
59 20 35 2 40
60 32 67 1 32
61 35 102 0 0
62 35 137 1 35
63 22 159 2 44
64 20 179 3 60
65 10 189 4 40
66 8 197 5 40
  N=Σfi=197     i=1nfidi=336

Here,
N2=1972=98.5

The cumulative frequency just greater than 98.5 is 102. The corresponding value of x is 61.
Therefore, the median is 61.

MD = 1Ni=1nfidi = 1197×336 = 1.7055

Page No 32.11:

Question 2:

The number of telephone calls received at an exchange in 245 successive one-minute intervals are shown in the following frequency distribution:

Number of calls 0 1 2 3 4 5 6 7
Frequency 14 21 25 43 51 40 39 12
Compute the mean deviation about median.

Answer:

We will first calculate the median.

xi fi Cumulative Frequency di=xi-4 fidi
0 14 14 4 56
1 21 35 3 63
2 25 60 2 50
3 43 103 1 43
4 51 154 0 0
5 40 194 1 40
6 39 233 2 78
7 12 245 3 36
  N=Σfi=245     i=1nfidi=366

Here,
N2=2452=122.5

The cumulative frequency just greater than 122.5 is 154 and the corresponding value of x is 4.

∴ Median, M = 4

MD = 1Ni=1nfidi = 1245×366 = 1.493

Page No 32.11:

Question 3:

Calculate the mean deviation about the median of the following frequency distribution:

xi 5 7 9 11 13 15 17
fi 2 4 6 8 10 12 8

Answer:

We will first calculate the median for the data.
 

xi fi Cumulative Frequency di = xi-13 fidi
5 2 2 8 16
7 4 6 6 24
9 6 12 4 24
11 8 20 2 16
13 10 30 0 0
15 12 42 2 24
17 8 50 4 32
  N=Σfi=50     i=1nfidi=136

Here,
N2=502=25
The cumulative frequency just greater than 25 is 30 and the corresponding value of x is 13.

  Median, M=13

MD=1Ni=1nfidi          =150×136          =2.72

Page No 32.11:

Question 4:

Find the mean deviation from the mean for the following data:
(i)

xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

(ii)
xi 5 10 15 20 25
fi 7 4 6 3 5

(iii)
xi 10 30 50 70 90
fi 4 24 28 16 8

(iv)
Size 20 21 22 23 24
Frequency 6 4 5 1 4
                                                                                      
                                                                                       [NCERT EXEMPLAR]
(v)
Size 1 3 5 7 9 11 13 15
Frequency 3 3 4 14 7 4 3 4
                                                                                      
                                                                                       [NCERT EXEMPLAR]

Answer:

i)

xi fi fixi
xi-x¯
fixi-9
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
  N=Σfi=26 i=1nfixi = 234   i=1nfixi-9=88
x=fii=1nxiN=23426=9
M.D.=1Ni=1nfixi-x=126×88=3.39


ii)
xi fi fixi xi-x¯ fixi-14
5 7 35 9 63
10 4 40 4 16
15 6 90 1 6
20 3 60 6 18
25 5 125 11 55
  N=25 i=1nfixi=350   i=1nfixi-14=158
x=fii=1nxiN=35025=14
MD=1Ni=1nfixi-x=125×158=6.32


iii)
xi fi fi​xi xi-x¯ fixi-50
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
  N=80 i=1nfixi=4000   i=1nfixi-50=1280
x=fii=1nxiN=400080=50
MD=1Ni=1nfixi-x=180×1280=16

(iv)
Size(xi) Frequency (fi) fi​xi xi-x¯=xi-21.65 fixi-x=fixi-21.65
20 6 120 1.65 9.9
21 4 84 0.65 2.6
22 5 110 0.35 1.75
23 1 23 1.35 1.35
24 4 96 2.35 9.4
  N=20 i=1nfixi=433   i=1nfixi-x=25
x=fii=1nxiN=43320=21.65
MD=1Ni=1nfixi-x=120×25=1.25

(v)
Size(xi) Frequency (fi) fi​xi xi-x¯=xi-8 fixi-x=fixi-8
1 3 3 7 21
3 3 9 5 15
5 4 20 3 12
7 14 98 1 14
9 7 63 1 7
11 4 44 3 12
13 3 39 5 15
15 4 60 7 28
  N=42 i=1nfixi=336   i=1nfixi-x=124
x=fii=1nxiN=33642=8
MD=1Ni=1nfixi-x=142×124=2.95

Page No 32.11:

Question 5:

Find the mean deviation from the median for the following data:
(i)

xi 15 21 27 30 35
fi 3 5 6 7 8

(ii)
xi 74 89 42 54 91 94 35
fi 20 12 2 4 5 3 4

Answer:

​i) 

xi fi Cumulative Frequency
xi-30
fixi-30
15 3 3 15 45
21 5 8 9 45
27 6 14 3 18
30 7 21 0 0
35 8 29 5 40
  N=Σfi=29     i=1nfixi-30=148

Here, N2=292=14.5

The cumulative frequency greater than 14.5 is 21 and the corresponding value of x is 30.
Median, M=30
MD=1Ni=1nfixi-M=129×148=5.10

ii)
xi fi Cumulative Frequency xi-74 fixi-74
35 4 4 39 156
42 2 6 32 64
54 4 10 20 80
74 20 30 0 0
89 12 42 15 180
91 5 47 17 85
94 3 50 20 60
  N=Σfi=50     i=1nfixi-74=625

Here, N2=502=25

The cumulative frequency greater than 25 is 30 and the corresponding value of x is 74.

Median, M=89

MD=1Ni=1nfixi-M=150×775=12.5



Page No 32.16:

Question 1:

Compute the mean deviation from the median of the following distribution:

Class 0-10 10-20 20-30 30-40 40-50
Frequency 5 10 20 5 10

Answer:

                         

Class   Frequency
fi
Cumulative frequency Mid-values
xi
di=xi-25 fidi
0−10 5 5 5 20 100
10−20 10 15 15 10 100
20−30 20 35 25 0 0
30−40 5 40 35 10 50
40−50 10 50 45 20 200
   N=fii=15=50           fidii=15=450












Here, N =50   N2=25
The cumulative frequency greater than N2=25  is 35 and the corresponding class is 20−30.
Therefore, the median class is  20−30.

   l=20, f=20, F=15, N=50, h =10 Median =l+N2-Ff×h                       = 20+502-1520×10                =20+25-1520×10                 =25 
Mean deviation from the median =i=15fi diN  = 45050=9

Page No 32.16:

Question 2:

Find the mean deviation from the mean for the following data:
(i)

Classes 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Frequencies 4 8 9 10 7 5 4 3

(ii)
Classes 95-105 105-115 115-125 125-135 135-145 145-155
Frequencies 9 13 16 26 30 12

(iii)
Classes 0-10 10-20 20-30 30-40 40-50 50-60
Frequencies 6 8 14 16 4 2

Answer:

(i)  We will compute the mean deviation from the mean in the following way:
 

Classes  fi Midpoints
xi
fixi  xi -X 
=  xi-358 
fi  xi - X 
0−100 4 50 200 308 1232
100−200 8 150 1200 208 1664
200−300 9 250 2250 108 972
300−400 10 350 3500 8 80
400−500 7 450 3150 92 644
500−600 5 550 2750 192 960
600−700 4 650 2600 292 1168
700−800 3 750 2250 392 1176
  i=18fi=50
 
   i=18fi xi=17900   i=18fi  xi-X =7896


















 N=i=16fi=50 and  i=16fixi=17900

 X=i=18fixi fii=18=1790050=358 



  Mean deviation=1Ni=18fi  xi-X                            =789650                           =157.92 
    
              
(ii) We will compute the mean deviation from the mean in the following way:
 
     
Classes  Frequency fi Midpoints
xi
fixi  xi -X 
= xi-128.58 
fi  xi - X 
95−105 9 100 900 28.58 257.22
105−115 13 110 1430 18.58 241.54
115−125 16 120 1920 8.58 137.28
125−135 26 130 3380 1.42 36.92
135−145 30 140 4200 11.42 342.6
145−155 12 150 1800 21.42 257.04
   i=16fi=106     i=16fixi=13630   i=18fi xi-X =1272.6 













 N=i=16fi=106 and  i=16fixi=13630 

   X=i=16fixii=16fi           =13630106          =128.58


  Mean deviation=1Ni=18fi  xi-X                                 =1272.6106                                =12.005 



(iii) We will compute the mean deviation from the mean in the following way:
 
 
Classes  Frequency fi Midpoints
xi
fixi  xi -X 
=  xi-358 

fi  xi - X 
0−10 6 5 30 22 132
10−20 8 15 120 12 96
20−30 14 25 350 2 28
30−40 16 35 560 8 128
40−50 4 45 180 18 72
50−60 2 55 110 28 56
  i=16fi=50

 
   i=16fixi=1350

 
  i=18fi  xi-X =512

 















 
 N=i=16fi=50 and  i=16fixi=1350 ​

 X=i=16fixii=16fi   =135050   =27 


 Mean deviation=1Ni=18fi  xi-X                                =51250                               =10.24

Page No 32.16:

Question 3:

Compute mean deviation from mean of the following distribution:

Mark 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
No. of students 8 10 15 25 20 18 9 5

Answer:

Computation of mean deviation from the mean:
 

Marks  Number of Students
fi
Midpoints
xi
fixi  xi-X 
= xi-49 
fi xi-X 
10−20 8 15 120 34 272
20−30 10 25 250 24 240
30−40 15 35 525 14 210
40−50 25 45 1125 4 100
50−60 20 55 1100 6 120
60−70 18 65 1170 16 288
70−80 9 75 675 26 234
80−90 5 85 425 36 180
   N=i=18fi=110     i=18fixi=5390    i=18fi xi-X =1644
















 
 N=i=18fi=110 


and  i=18fixi=5390

X =i=18fixiN   =5390110   =49
 

Mean deviation=i=18fi xi-X  N                          =1644110                          =14.945                          14.95 

Page No 32.16:

Question 4:

The age distribution of 100 life-insurance policy holders is as follows:

Age (on nearest birth day) 17-19.5 20-25.5 26-35.5 36-40.5 41-50.5 51-55.5 56-60.5 61-70.5
No. of persons 5 16 12 26 14 12 6 5
Calculate the mean deviation from the median age.

Answer:

To make this function continuous, we need to subtract 0.25 from the lower limit and add 0.25 to the upper limit of the class.

 

Age  Number of People
fi
Cumulative Frequency   Midpoints
xi
di=xi-38.63  fidi
16.7519.75
 
5 5 18.25 20.38 101.9
19.7525.75 16 21 22.75 15.88 254.08
 
25.7535.75 12 33 30.75 7.88 94.56
 
35.7540.75 26 59 38.25 0.38 9.38
 
40.7550.75 14 73 45.75 7.12 99.68
 
50.7555.75 12 85 53.25 14.62 175.44
 
55.7560.75 6 91 58.25 19.62 117.72
 
60.7570.75 5 96 65.75 27.12 135.6
 
   N=fii=18=96       i=18fidi=988.36
 

 

                                                                                                                                                                               

N=96N2=48The cumulative frequency just greater than  N2=38 is 59 and the corresponding class is  35.75-40.75.Thus, the median class is 35.75-40.75l=35.75, f= 26, F=33, h=5Median =l+N2-Ff×h             = 35.75+48-3326×5                =35.75+2.88             = 38.63Mean deviation from the median =i=18fidiN                                                   =988.3696                                                  =10.29

Page No 32.16:

Question 5:

Find the mean deviation from the mean and from median of the following distribution:

Marks 0-10 10-20 20-30 30-40 40-50
No. of students 5 8 15 16 6

Answer:

Computation of mean distribution from the median:
 

Marks  Number of Students
fi
Cumulative Frequency  Midpoints
xi
di=xi-28 fidi fixi xi-27
fixi-27
0−10 5 5 5 23 115 25 22 110
10−20 8 13 15 13 104 120 12 96
20−30 15 28 25 3 45 375 2 30
30−40 16 44 35 7 112 560 8 128
40−50 6 50 45 17 102 270 18 108
  N=50        i=15fidi=478            1350   i=15fixi-27=472














 N=50  , N2=25

The cumulative frequency just greater than N2=25  is 28 and the corresponding class is 20−30.
Thus, the median class is 20−30.
 
Using formula:

  l=20, F=13, f=15, h=10    Median  =l+N2-Ff × h      Substituting the values:Median =20+25-1315×10                =20 +8               =28        
 Mean distribution from the median =i=15fi diN                                            = 47850                                           =9.56

    Mean (X)=i=15fixiN                  =135050                  =27Mean deviation from the mean  =150×i=15fixi-27                                                  =47250                                                  =9.44 

 Mean deviation from the median and the mean are 9.56 and 9.44, respectively.

Page No 32.16:

Question 6:

Calculate mean deviation about median age for the age distribution of 100 persons given below:

Age: 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55
Number of persons 5 6 12 14 26 12 16 9

Answer:

Since the function is not continuous, we subtract 0.5 from the lower limit of the class and add 0.5 to the upper limit of the class so that the class interval remains same, while the function becomes continuous.

 Thus, the mean distribution table will be as follows:
 

Age   Number of Persons
fi
Midpoint
xi
Cumulative Frequency di=xi-38 fidi
15.5−20.5 5 18 5 20 100
20.5−25.5 6 23 11 15 90
25.5−30.5 12 28 23 10 120
30.5−35.5 14 33 37 5 70
35.5−40.5 26 38 63 0 0
40.5−45.5 12 43 75 5 60
45.5−50.5 16 48 91 10 160
50.5−55.5 9 53 100 15 135
  N=i=18fi=100       i=18fidi=735

















    
N=100   N2=50 
Thus, the cumulative frequency slightly greater than 50 is 63 and falls in the median class 35.5−40.5.

 l=35.5  , F=37 , f=26 , h=5Median  = l+N2- Ff× h                               = 35.5+50-3726×5                      =35.5+2.5                           =   38 Mean deviation about the median age = i=18fidiN                                                    =735100                                                    =7.35         


 Thus, the mean deviation from the median age is 7.35 years.      



Page No 32.17:

Question 7:

Calculate the mean deviation about the mean for the following frequency distribution:
 

Class interval: 0–4 4–8 8–12 12–16 16–20
Frequency 4 6 8 5 2

Answer:

Let the assumed mean A = 10 and h = 4.
 

Class Interval Mid-Value(xi) Frequency(fi) di=xi-104 fidi xi-X=xi-9.2 fixi-X
0–4 2 4 −2 −8 7.2 28.8
4–8 6 6 −1 −6 3.2 19.2
8–12 10 8 0 0 0.8 6.4
12–16 14 5 1 5 4.8 24
16–20 18 2 2 4 8.8 17.6
    N = 25   fidi = −5   fixi-X=96

Here, N = 25 and fidi = −5

Mean,
X=A+h1Nfidi   =10+4125×-5   =10-2025   =10-0.8   =9.2

∴ Mean deviation about mean=1Nfixi-X=125×96=3.84

Page No 32.17:

Question 8:

Calculate mean deviation from the median of the following data:                    [NCERT EXEMPLAR]
 

Class interval: 0–6 6–12 12–18 18–24 24–30
Frequency 4 5 3 6 2

Answer:

Calculation of mean deviation about the median.
 

Class Interval Mid-Values
(xi)
Frequency
(fi)
Cummulative
Frequency (c.f.)
xi-14 fixi-14
0–6 3 4 4 11 44
6–12 9 5 9 5 25
12–18 15 3 12 1 3
18–24 21 6 18 7 42
24–30 27 2 20 13 26
    N = 20     fixi-14=140

Here, N = 20. So, N2=10

The cummulative frequency just greater than N2 is 12. Thus, 12–18 is the median class.

Now, l = 12, h = 6, f = 3 and F = 9

Median=l+N2-Ff×h=12+10-93×6=14

Now,

Mean deviation about median = 1Nfixi-14=120×140=7



Page No 32.28:

Question 1:

Find the mean, variance and standard deviation for the following data:
(i) 2, 4, 5, 6, 8, 17.
(ii) 6, 7, 10, 12, 13, 4, 8, 12.
(iii) 227, 235, 255, 269, 292, 299, 312, 321, 333, 348.
(iv) 15, 22, 27, 11, 9, 21, 14, 9.

Answer:

(i)    2,4,5,6,8,17

Mean =X=2+4+5+6+8+176 =426= 7
 

xi xi-X  =xi-7 xi-72
2 -5 25
4 -3 9
5 -2 4
6 -1 1
8 1 1
17 10 100
     i=16xi-X2=140












n=6

Variance (X) =i=16xi-X2n                     =1406                     =23.33


Standard deviation =Variance (X)                                     =23.33                                   =4.83



(ii) 6,7,10,12,13,4,8,12

Mean =6+7+10+12+13+4+8+128            =728            =9

 
xi xi-X =xi-9 xi-X2
6 -3 9
7 -2 4
10 1 1
12 3 9
13 4 16
4 -5 25
8 -1 1
12 3 9
    i=18xi-X2 =74















 n=8
 Variance X =i=18xi-X2n                     =748                      =9.25Standard deviation =Variance X                                     =9.25                                    =3.04


(iii)      227,235,255,269,292,299,312,321,333,348,


Mean = 227+235+255+269+292+299+312+321+333+34810            =289110 =289.1

 
xi xi-X =xi-289.1 xi-X2
227 − 62.1 3856.41
235 − 54.1 2926.81
255 − 34.1 1162.81
269 − 20.1 404.01
292 2.9 8.41
299 9.9 98.01
312 22.9 524.41
321 31.9 1017.61
333 43.9 1927.21
348 58.9 3469.21
    i=110xi-x¯2 =15394.9

















n=10 Variance X =i=110xi-X¯2n                       =15394.910                        =1539.49Standard deviation   =Variance X                                     =1539.49                                      =39.24



(iv)   15,22,27,11,9,21,14,9

                  Mean=15+22+27+11+9+21+14+98 =1288=16
 
 
xi xi-X=xi-16 xi-X¯2
15 −1 1
22 6 36
27 11 121
11 5 25
9 −7 49
21 5 25
14 −2 4
9 −7 49
    i=18xi-X¯2 =310















 n=8Variance X =i=18xi-x2n                      =3108                     =38.75Standard deviation=Variance X                             =38.75                             =6.22

Page No 32.28:

Question 2:

The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.

Answer:

Let x1, x2, x3, ..., x20  be  the 20 given observations.

Variance (X)=5

Variance (X) =120×xi-X2  =5                  (Here, X is the mean of the given observations.)                                                                                                


Let u1,u2,,u3, ..., u20 be the new observations, such that
ui=2xi       (for i=1,2,3, ..., 20)                   ...(1)

Mean= U ¯ =i=120uin                              =i=1202xi20           substituting ui from eq(1) and taking n as 20            = 2×xii=12020                =2 X¯


 ui-U¯  =2xi-2X¯         (for i=1,2, ..., 20)             =2xi-X¯                     ui-U¯2=2xi-X¯2                                squaring both the sides              =4xi-X¯2  i=120ui-U¯2   =4i=120xi-X¯2      i=120ui-U¯220  =4i=120xi-X¯2  20                          =4 i=120xi-X¯2  20Variance (U)=4×Variance (X)                     =4×5                     =20


 Thus, variance of the new observations is 20.

Page No 32.28:

Question 3:

The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Answer:

Let x1,,x2,,x3 , ..., x15 be the given observations.

Variance X  is given as 4.
If  X¯ is the mean of the given observations, then we get:
 Variance X=115i=115xi-X2                         =4

Let u1,u2,u3 ... u15 be the new observations such that
ui=xi+9      for i=1,2 ,3, ...,15           ....(1) U¯ =1ni=115ui   =115i=115xi+9   =115xii=115 +9×1515                      as  i=1159 =9×15  =X  +9                            ...(2)ui-U¯ =xi+9-9+X¯               from eq (1) and eq (2)           =xi-X¯115×ui -U¯2   =115xi-X¯2         squaring both the sides and then dividing by 15115×i=115ui -U¯2   =115×i=115xi-X¯2       115×i=115ui -U¯2   = 4Variance U =4 

Thus, variance of the new observation is 4.

Page No 32.28:

Question 4:

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Answer:

Let x and y be the other two observations.

Mean is 4.4.

1+2+6+x+y5=4.4

   9+x+y=22   x+y =13            ... (1)

Let Var (X) be the variance of these observations, which is given to be 8.24.

If X¯ is the mean, then we have:

8.24 =1512+22+62+x2+y2  -x2         =151+4+36+x2+y2 -   4.42         =1541+x2+y2 -19.36x2+y2   = 97               ...(2) x+y2+x-y2=2x2+y2  132+x-y2=2×97                 using eq(1) and eq(2)x-y2=194-169 =25x-y =±5                    ....(3) Solving eq(1) and eq (3)  for  x-y= -5 and  x+ y= 13 2x=18 x=9y=4

Thus, the other two observations are 9 and 4.

Page No 32.28:

Question 5:

The mean and standard deviation 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.

Answer:

 Mean=X¯ =8       n=6     σ =S.D =4If x1,x2....x6  are the given observations  X=1n×i=16xi 8=16×i=16xiLet  u1,u2.....u6 be the new  observations  ui=3xi                for i=1,2,3...6 Mean of new observations= U¯ =1n×uii=16                                                             =16×3xii=16                                                             =3×16×i=16xi                                                                = 3 X¯                                                               =3×8                                                              =24

Thus, mean of the new observations is 24.


SD=σx =4 σx2 =Variance  X Variance X=1616i=16xi-X¯2=16         ...1Variance U=σu2=16i=16ui-U¯2                                  =16×i=16 3xi -3X¯ 2                                           =32×16i=16xi-X¯2                                   =9×16σu =Variance U     =9×16     =12

Thus, standard deviation of the new observations is 12.

Page No 32.28:

Question 6:

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.

Answer:

 Let x and y be the remaining two observations.

n=8    Variance=9.25X¯=Mean =9   6+7+10+12+12+13+x+y8=9  60+x+y =72  x+y =12          ...(1) 
 
Variance X=1ni=18xi2  -X¯29.25  =18×62+72+102+122+122+132+x2+y2-929.25=18642+x2+y2 -81 9.25×8 =642+x2+y2-648  x2+y2=80             ....(2) We know,  x+y2+x-y2=2x2+y2122+x-y2 =2×80              using equations (1) and (2)144+x-y2=160 x-y2=16 x-y=±4             If x-y=4, then x+y=12 and x-y=4 give x=8 and y=4If x-y=-4, then x+y=12 and x-y=4 give x=4 and y=8

Thus, the remaining two observations are 8 and 4.

Page No 32.28:

Question 7:

For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

Answer:

We have:
   n=200,  X¯=40, σ =15

1nxi =X¯1200xi =40 xi =40×200 =8000  Since the score was misread, this sum is incorrect. Corrected xi =8000 -34-53+43+35                                  = 8000-7                                    =7993 Corrected mean  =Correctedxi200                                        =7993200                                       =39.955  


SD =σ = 15     Variance= 152 =225According to the formula,Variance =1nxi2-1nxi21200xi2-402 =2251200xi2-1600 =225xi2=200×1825 =365000  This is an incorrect reading.Corrected xi2  =365000-342-532+432+352                                 =365000-1156-2809+1849+1225                                 =364109 


Corrected variance=1n× Correctedxi -Corrected mean2                                 =1200×364109 -39.9552                                       = 1820.545 -1596.402                                 = 224.14 Corrected  SD  =Corrected variance                              = 224.14                              =14.97                     

Page No 32.28:

Question 8:

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

Answer:

n=100      Mean=X¯ = 40        σ = SD =5.11nxi =X¯ xi=100×40=4000          This is an incorrect reading due to misread values.Corrected sum, xi =4000-50+40                                      =  3990 Corrected mean  = Corrected sum 100                                    =3990100                                    =39.9              ...(1)

To find the corrected SD:

Variance = σ  σ2 =5.12= VarianceAccording to the formula,1nxi2  -X¯2   = Variance  1100xi2-402  =26.011100xi2 -1600 =26.011100xi 2 =1626.01xi2 =162601     But, this is incorrect due to misread values  Corrected  xi2  =162601-502+402                                       =  161701      ....(2)Corrected variance =1100Correctedxi2  -Corrected mean2                                    = 161701100-39.92                 using equations (1) and (2)                                     =1617.01-1592. 01                                     =  25 Corrected SD =Corrected variance                            =25                             =5

 Corrected mean = 39.9
 Corrected standard deviation = 5

Page No 32.28:

Question 9:

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.

Answer:

n=20            Mean=X  ¯               SD= σ =2      1nxi  =X  ¯ 120xi  =10 xi   =200       This is incorrect due to misread values.        ...(1)  Variance = σ2 =41nxi2  -X  ¯ 2 =4  120xi2-102 =4  120xi2 =104 xi2=104×20 =2080                  This is incorrect due to misread values.      ...(2)


(i)     If  observation 8 is omitted, then total 19 observations are left.

         Incorrected xi=200
       
        Corrected xi+8=200

Corrected xi=192Corrected mean =Corrected xi19                                     =19219                                     =10.10Using equation (2), we get:Corrected  xi2+82=2080Corrected  xi2 =2080-64                                     =2016 119Corrected  xi2 -Corrected mean 2  =Corrected variance  Corrected variance= 119×2016 -192192   Corrected variance= 2016×19-1922192    Corrected variance= 38304-36864192   Corrected variance=1440192  Corrected SD =Corrected variance                            =1440192                           =121019                          =1.997

Thus, if  8 is omitted, then the mean is 10.10 and SD is 1.997.



(ii)  When incorrect observation 8 is replaced by 12:

         
 From equation (1):Incorrected xi= 200Corrected xi=200-8+12  =204 Corrected X¯ =20420=10.2Incorrected  xi2=2080           from (2)Corrected xi2 =2080-82+122                           = 2160Corrected variance=120×Corrected xi2 -Corrected X¯2                                = 120×2160 -204202                                  =2160×20-2042202                                   =43200-41616400                                      =1584400 Corrected SD = Corrected variance                              =1584400                               =39610                              =19.89910                              =1.9899
 
If 8 is replaced by 12, then the mean  is 10.2 and SD is 1.9899.

Page No 32.28:

Question 10:

The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations were omitted.

Answer:

 n= 100      Mean=X¯ =20               SD=  σ   =3             Misread values are 21, 21 and18.1nxi=X¯ 1100xi =20xi =20 ×100=2000     This sum is incorrect due to misread values.       ....(1) If three misread values are to be omitted, the total number of enteries will be 97.Also,   xi=2000-21-21-18 =1940   Corrected  X¯ =194097 =20       ....(2)

σ =3     Variance =σ2  =9=Variance=1nxi2-X¯21100xi2-202  =91100xi2 = 9+400 1100xi2 =409xi2=409 ×100 =40900     This is an incorrect sum due to misread values.   ...(2)Corrected xi2 =40900 -212+212+182                             =40900 -441-441-324                             =39694                 ....(3) From equations (2) and (3), we get:Corrected variance =1nxi2-X¯2                              =197×39694  -202                              =409.216-400                                  = 9.216 Corrected SD= Corrected variance                           = 9.216                          = 3.0357 

 Thus, after omitting three values, the mean would be 20 and SD would be 3.0357.

Page No 32.28:

Question 11:

Show that the two formulae for the standard deviation of ungrouped data

σ=1nxi-X2 and σ'=1nxi2-X2 are equivalent, where X=1nxi

Answer:


σ=1nxi-X2=1nxi2-2xiX+X2=1nxi2-1n2xiX+1nX2=1nxi2-1n×2Xxi+1n×X21=1nxi2-1n×2X×nX+1n×X2×n                   X=1nxi
=1nxi2-2X2+X2=1nxi2-X2=σ'

Hence, the formulae σ=1nxi-X2 and σ'=1nxi2-X2 are equivalent, where X=1nxi.



Page No 32.37:

Question 1:

Find the standard deviation for the following distribution:

x : 4.5 14.5 24.5 34.5 44.5 54.5 64.5
f : 1 5 12 22 17 9 4

Answer:

x: 4.5 14.5   24.5 34.5 44.5 54.5 64.5
f: 1 5 12 22 17 9 4

Median value of x is 34.5.
 
xi fi di=xi-34.5 ui=xi-34.510 fiui ui2 fiui2
4.5 1 -30 -3 -3 9 9
14.5 5 -20 -2 -10 4 20
24.5 12 -10 -1 -12 1 12
34.5 22 0 0 0 0 0
44.5 17 10 1 17 1 17
54.5 9 20 2 18 4 36
64.5 4 30 3 12 9 36
  N=fi=70     fiui=22   fiui2=130

Var(X)=h21Ni=1nfiui2-1Ni=1nfiui2

We have
N=70, i=1nfiui=22, i=1n fiui2=130, h=10

Plugging all the values in the formula of variance:
Var(X)=102170×130-170×222           =10013070-22702           =100137-1211225           =1001.857-0.0987           = 1001.7583             =175.83

Standard deviation, SD=Var(X)
SD=Var(X)       =175.83       =13.26



Page No 32.38:

Question 2:

Table below shows the frequency f with which 'x' alpha particles were radiated from a diskette

x : 0 1 2 3 4 5 6 7 8 9 10 11 12
f : 51 203 383 525 532 408 273 139 43 27 10 4 2
Calculate the mean and variance.

Answer:

Mean, X¯=fixifi=100782600=3.88

xi fi fixi xi-X¯ xi-X¯2 fixi-X¯2
0 51 0 −3.88 15.05 767.55
1 203 203 −2.88 8.29 1682.87
2 383 766 −1.88 3.53 1351.99
3 525 1575 −0.88 0.77 404.25
4 532 2128 0.12 0.014 7.448
5 408 2040 1.12 1.25 510
6 273 1638 2.12 4.49 1225.77
7 139 973 3.12 9.73 1352.47
8 43 344 4.12 16.97 729.71
9 27 243 5.12 26.21 707.67
10 10 100 6.12 37.45 374.5
11 4 44 7.12 50.69 202.76
12 2 24 8.12 65.93 131.86
  fi=N=2600 fixi=10078     fixi-X¯2=9448.848

Variance, σ2=fixi-X¯2N=9448.8482600=3.63

Page No 32.38:

Question 3:

Find the mean, mode, S.D. and coefficient of skewness for the following data:

Year render: 10 20 30 40 50 60
No. of persons (cumulative): 15 32 51 78 97 109

Answer:

Year render No. of persons (cumulative) fi ui=xi-3510 fiui ui2 fiui2
10 15 15 -2.5 -37.5 6.25 93.75
20 32 17 -1.5 -25.5 2.25 38.25
30 51 19 -0.5 -9.5 0.25 4.75
40 78 27 0.5 13.5 0.25 6.75
50 97 19 1.5 28.5 2.25 42.75
60 109 12 2.5 30 6.25 75
    fi=N=109   fiui=-0.5   fiui2=261.25

X¯=A+hfiuiN  =35+10-0.5109  =34.96 years
σ2=h2fiui2N-fiuiN2  =100261.25109-0.2511881  =100×2.396   =239.6

σ=239.6  =15.47 years

Coefficient of skewness = Mean - Mode
                                    = 34.96 - 40
                                    = -5.04

Page No 32.38:

Question 4:

Find the standard deviation for the following data:

x : 3 8 13 18 23
f : 7 10 15 10 6

Answer:

xi fi fixi xi-X¯ xi-X¯2 fixi-X¯2
3 7 21 9.79 95.84 670.88
8 10 80 4.79 22.94 229.4
13 15 195 0.21 0.04 0.6
18 10 180 5.21 27.14 271.4
23 6 138 10.21 104.24 625.44
  fi=48 fixi=614     fixi-X¯2=1797.32


Variance, σ2=fixi-X¯2fi=1797.3248=37.44

SD, σ = 37.44 = 6.12



Page No 32.41:

Question 1:

Calculate the mean and S.D. for the following data:

Expenditure in Rs: 0-10 10-20 20-30 30-40 40-50
Frequency: 14 13 27 21 15

Answer:

Expenditure (Rs) fi Midpoint xi fixi xi-X¯ xi-X¯2 fixi-X¯2
0−10 14 5 70 -21.1 445.21 6233.94
10−20 13 15 195 -11.1 123.21 1601.73
20−30 27 25 675 -1.1 1.21 34.67
30−40 21 35 735 8.9 79.21 1663.41
40−50 15 45 675 18.9 357.21 5358.15
  fi=90   fixi=2350     fixi-X¯2=14891.9

Mean, X¯ = 235090 = 26.11
Variance, σ2 = 14891.990 = 165.47

 SD, σ = 165.47 = 12.86

Page No 32.41:

Question 2:

Calculate the standard deviation for the following data:

Class: 0-30 30-60 60-90 90-120 120-150 150-180 180-210
Frequency: 9 17 43 82 81 44 24

Answer:

Class fi Midpointxi yi=xi-10530 yi2 fiyi fiyi2
030 9 15 3 9 27 81
3060 17 45 2 4 34 68
6090 43 75 1 1 43 43
90120 82 105 0 0 0 0
120150 81 135 1 1 81 81
150180 44 165 2 4 88 176
180210 24 195 3 9 72 216
  fi=N=300       fiyi=137 fiyi2=665
 
Mean, x¯=a+h1NΣfiyi=105+30137300=118.7

Variance:
 σ2=h2N2Nfiyi2-fiyi2=90090000300×665-18769=1100199500-18769=180731100=1807.31SD, σ = 1807.31 = 42.51



Page No 32.42:

Question 3:

Calculate the A.M. and S.D. for the following distribution:

Class: 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency: 18 16 15 12 10 5 2 1

Answer:

Class fi Midpointxi ui=xi-3510 fiui fiui2
0−10 18 5 -3 -54 162
10−20 16 15 -2 -32 64
20−30 15 25 -1 -15 15
30−40 12 35 0 0 0
40−50 10 45 1 10 10
50−60 5 55 2 10 20
60−70 2 65 3 6 18
70−80 1 75 4 4 16
  fi=79     fiui=-71 fiui2=305

X¯=a+hfiuiN=35+10-7179=26.01

AM = 26.01

σ2=h2fiui2N-fiuiN2=10030579-50416241=305.20σ=305.20=17.47

Page No 32.42:

Question 4:

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, the correct figure being 40. Find the correct mean and S.D.

Answer:

x¯o = 40, σo = 5.1, and n=100Σxo=4000Corrected sum of observations, Σxn=4000-50+40=3990n=100x¯n=Σxnn=39.90Also, SD, σo=5.1Σxi-x¯o2=2601Σxi-x¯n2=2601-50-402+40-39.902=2601-100+0.01=2501.01σn=Σxi-x¯n2n = 2501.01100=5

Page No 32.42:

Question 5:

Calculate the mean, median and standard deviation of the following distribution:

Class-interval: 31-35 36-40 41-45 46-50 51-55 56-60 61-65 66-70
Frequency: 2 3 8 12 16 5 2 3

Answer:

Class Interval fi Midpoint xi ui=xi-534 ui 2 fiui fiui2
3135 2 33 -5 25 -10 50
3640 3 38 -3.75 14.06 -11.25 42.18
4145 8 43 -2.5 6.25 -20 50
4650 12 48 -1.25 1.56 -15 18.72
5155 16 53 0 0 0 0
5660 5 58 1.25 1.56 6.25 7.8
6165 2 63 2.5 6.25 5 12.5
6670 3 68 3.75 14.06 11.25 42.18
  N = 51       i=1nfiui=-33.75 i=1nfiui2=223.38

X=a+hi=1nfiuiN  =53+4-33.7551   =50.36

σ2=h2i=1nfiui2N-i=1nfiuiN2   =16223.3851-1139.062601   =63.07σ=63.07  =7.94

 
fi CF
(Cumulative frequency)
2 2
3 5
8 13
12 25
16 41
5 46
2 48
3 51

fi=51=NN2=25.5

Median class interval is
5155.
L=51F=25f=16h=4

Median=L+N2-Ff×h

         =51+25.5-2516×4=51+0.54=51.125

Page No 32.42:

Question 6:

Find the mean and variance of frequency distribution given below:
 

xi: 1 ≤ x < 3 3 ≤ x < 5 5 ≤ x < 7 7 ≤ x < 10
fi: 6 4 5 1

Answer:

xi Mid-Values(yi) yi2 fi fi yi fi yi2
1–3 2 4 6 12 24
3–5 4 16 4 16 64
5–7 6 36 5 30 180
7–10 8.5 72.25 1 8.5 72.25
      N = fi= 16 fiyi=66.5 fiyi2=340.25

Therefore,

Mean = fiyifi=66.516=4.16

Variance = 1Nfiyi2-1Nfiyi2=116×340.25-116×66.52=21.26-17.22=4.04

Page No 32.42:

Question 7:

The weight of coffee in 70 jars is shown in the following table:                                                  
 

Weight (in grams): 200–201 201–202 202–203 203–204 204–205 205–206
Frequency: 13 27 18 10 1 1

Determine the variance and standard deviation of the above distribution.                                          [NCERT EXEMPLAR]

Answer:

 

Weight (in grams) Mid-Valuesxi Frequencyfi di=xi-202.5 di2 fidi fidi2
200–201 200.5 13 −2 4 −26 52
201–202 201.5 27 −1 1 −27 27
202–203 202.5 18 0 0 0 0
203–204 203.5 10 1 1 10 10
204–205 204.5 1 2 4 2 4
205–206 205.5 1 3 9 3 9
    N = fi=70     fidi=-38 fidi2=102

Now,

Variance, σ2
=1Nfidi2-1Nfidi2=170×102-170×-382=1.457-0.295=1.162 gm

Standard deviation, σ = Variance=1.162=1.08 gm

Page No 32.42:

Question 8:

Mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.                                  
                                                                                                                                                                                             [NCERT EXEMPLAR]

Answer:

Given:

Number of observations, n = 100

Mean, x = 40

Standard deviation, σ = 10

We know that

x=xi100xi100=40xi=4000

∴ Correct xi=4000-30+70+3+27=3930

Correct mean = Correct xi100=3930100=39.3

Now,

Incorrect variance, σ2=xi2100-402
xi2100=100+1600=1700xi2=170000

Correct xi2=170000-302-702+32+272=164939

∴ Correct standard deviation

=164939100-39.32=1649.39-1544.49=104.9=10.24

Page No 32.42:

Question 9:

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
                                                                                                                                                                                            [NCERT EXEMPLAR]

Answer:

Given:

Number of observations, n = 10

Mean, x = 45

Variance, σ2 = 16

Now,

Incorrect mean, x = 45

Incorrect xi10=45Incorrect xi=450

∴ Correct xi=450-52+25=423

⇒ Correct mean = Correct xi10=42310=42.3

Incorrect variance, σ2 = 16

16=Incorrect xi210-452Incorrect xi2=1016+2025=20410

∴ Correct xi2=20410-522+252=20410-2704+625=18331

Now,

Correct variance = 1833110-42.32=1833.1-1789.29=43.81

Page No 32.42:

Question 10:

Calculate the mean, variance and standard deviation of the following frequency distribution.
 

Class: 1–10 10–20 20–30 30–40 40–50 50–60
Frequency: 11 29 18 4 5 3

Answer:

Let the assumed mean A = 25.
 

Class Mid-Valuesxi di=xi-A   =xi-25 di2 Frequency fi fidi fidi2
1–10 5.5 −19.5 380.25 11 −214.5 4182.75
10–20 15 −10 100 29 −290 2900
20–30 25 0 0 18 0 0
30–40 35 10 100 4 40 400
40–50 45 20 400 5 100 2000
50–60 55 30 900 3 90 2700
        N = fi = 70 fidi = −274.5 fidi2=12182.75

Mean = A+fidifi=25+-274.570=25-3.92=21.08

Variance = σ2=1Nfidi2-1Nfidi2=12182.7570--274.5702=174.02-15.37=158.65

Standard deviation = σ=158.65=12.6



Page No 32.47:

Question 1:

Two plants A and B of a factory show following results about the number of workers and the wages paid to them

  Plant A Plant B
No. of workers 5000 6000
Average monthly wages Rs 2500 Rs 2500
Variance of distribution of wages 81 100
In which plant A or B is there greater variability in individual wages?

Answer:

Variance of the distribution of wages in plant Aσ2=81
Standard deviation of the distribution of wages in plant Aσ=9

Variance of the distribution of wages in plant Bσ2=100
Standard deviation of the distribution of wages in plant Bσ=10
Average monthly wages in both the plants are Rs 2500.
Thus, the plant with greater value of SD will have more variability in salary.
Plant B has more variability in individual wages than plant A.

Page No 32.47:

Question 2:

The means and standard deviations of heights ans weights of 50 students of a class are as follows:

  Weights Heights
Mean 63.2 kg 63.2 inch
Standard deviation 5.6 kg 11.5 inch
Which shows more variability, heights or weights?

Answer:

 Coeffient of variations (CV) in weights =SDMean×100                                                          =5.663.2×100                                                          =8.86

Coefficient of variations (CV) in heights =11.563.2×100                                                             =18.19

CV in heights is greater than CV in weights.
Thus, heights will show more variability than weights.



Page No 32.48:

Question 3:

Coefficient of variation of two distributions are 60% and 70% and their standard deviations are 21 and 16 respectively. What are their arithmetic means?

Answer:

The coefficient of variation (CV) for the first distribution is 60.
The coefficient of variation (CV) for the second distribution is 70.
SDσ1=21
SDσ2=16
We know:
CV=σX¯×100

From the above formula, we get:
X¯=σCV×100

X1¯=2160×100=35X2¯=1670×100=22.86

Page No 32.48:

Question 4:

Calculate coefficient of variation from the following data:

Income (in Rs): 1000-1700 1700-2400 2400-3100 3100-3800 3800-4500 4500-5200
No. of families: 12 18 20 25 35 10

Answer:

Income (Rs) fi Midpointxi ui=xi-3450700 fiui fiui2
1000−1700 12 1350 −3 −36 108
1700−2400 18 2050 −2 −36 72
2400−3100 20 2750 −1 −20 20
3100−3800 25 3450 0 0 0
3800−4500 35 4150 1 35 35
4500−5200 10 4850 2 20 40
  fi=120    
fiui=-37
 
fiui2=275

​​X¯=a+hfiuiN=3450+700-37120=3234.17

σ2=h2fiui2N-fiuiN2=490000275120-136914400=1076332.64σ=1076332.64=1037.46CV=σX¯×100=1037.463234.17×100=32.08 

Page No 32.48:

Question 5:

An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:

  Firm A Firm B
No. of wage earners 586 648
Average weekly wages Rs 52.5 Rs. 47.5
Variance of the
distribution of wages
100
 
121
 
(i) Which firm A or B pays out larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?

Answer:

Average weekly wages=Total weekly wagesNumber of workers
Total weekly wages = (Average weekly wages) (Numbers of workers)
Total weekly wages for firm A = Rs 52.5×586 = Rs 30765
Total weekly wages for firm B = Rs 47.5×648 = Rs 30780
(i) Firm B pays a larger amount as weekly wages.

(ii)  SD (firm A) = 10
SD (firm B) = 11

CV (firm A)=1052.5×100                    =19.04CV (firm B)=1147.5×100                   =23.15

Since CV of firm B is greater than that of firm A, firm B has greater variability in individual wages.

Page No 32.48:

Question 6:

The following are some particulars of the distribution of weights of boys and girls in a class:

Number Boys Girls
  100 50
Mean weight 60 kg 45 kg
Variance 9 4
Which of the distributions is more variable?

Answer:

​We know:
SD (boys) is 3 and SD (girls) is 2.

CV=σX¯×100

  CV (boys)=360×100                      =5

CV (girls)=245×100                 =4.44

Since CV for boys is greater than that of  girls, distribution of the weights of boys is more variable than that of girls.

Page No 32.48:

Question 7:

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:

Subject Mathematics Physics Chemistry
Mean 42 32 40.9
Standard 12 15 20
Deviation      
Which of the three subjects shows the highest variability in marks and which shows the lowest?

Answer:

We know:
CV = σX¯×100

Xm¯=42, σm=12Xp¯=32, σp=15Xc¯=40.9, σc=20
CV of mathematics marks =1242×100=120042=28.57

CV of physics marks ​=1532×100=150032=46.87

CV of chemistry marks =2040.9×100=200040.9=48.89

Since CV of chemistry is the greatest, the variability of marks in chemistry is the highest and that of mathematics is the lowest.

Page No 32.48:

Question 8:

From the data given below state which group is more variable, G1 or G2?

Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Group G1 9 17 32 33 40 10 9
Group G2 10 20 30 25 43 15 7

Answer:

Marks fi Midpoint xi ui=xi-4510 fiui fiui2
10−20 9 15 −3 −27 81
20−30 17 25 −2 −34 68
30−40 32 35 −1 −32 32
40−50 33 45 0 0 0
50−60 40 55 1 40 40
60−70 10 65 2 20 40
70−80 9 75 3 27 81
  N=150     fiui=-6 fiui2=342

h=5, a=45
For group 1:
X¯=a+hfiuiN=45+10-6150=44.6
σ2=h2fiui2N-fiuiN2=100342150-3622500=227.84σ=227.84=15.09
 
Marks
fi
Midpointxi ui=xi-4510 fiui fiui2
1020 10 15 −3 −30 90
20−30 20 25 −2 −40 80
30−40 30 35 −1 −30 30
40−50 25 45 0 0 0
50−60 43 55 1 43 43
60−70 15 65 2 30 60
70−80 7 75 3 21 63
  fi=150     fiui=-6 fiui2=366

For group 2:
X¯=a+hfiuiN=45+10-6150=44.6
σ2=h2fiui2N-fiuiN2=100366150-3622500=243.84σ=243.84=15.62
Mean of both the groups are same and SD of group 2 is greater than that of group 1.
So, group 2 will be more variable.

Page No 32.48:

Question 9:

Find the coefficient of variation for the following data:

Size (in cms): 10-15 15-20 20-25 25-30 30-35 35-40
No. of items: 2 8 20 35 20 15

Answer:

Size (cm) fi Midpoint xi ui=xi-27.55 fiui fiui2
10−15 2 12.5 -3 -6 18
15−20 8 17.5 -2 -16 32
20−25 20 22.5 -1 -20 20
25−30 35 27.5 0 0 0
30−35 20 32.5 1 20 20
35−40 15 37.5 2 30 60
  fi=N=100     fiui=8 fiui2=150

Here,
h=5, a=27.5
X¯=a+hfiuiN=27.5+58100=27.9
σ2=h2fiui2N-fiuiN2=25150100-6410000=37.34

σ=37.34=6.11

We know that
CV=σX¯×100=6.1127.9×100=21.9

Page No 32.48:

Question 10:

From the prices of shares X and Y given below: find out which is more stable in value:

X: 35 54 52 53 56 58 52 50 51 49
Y: 108 107 105 105 106 107 104 103 104 101

Answer:

Let Ax = 51

xi
di=xi-51 di2
35 -16 256
54 3 9
52 1 1
53 2 4
56 5 25
58 7 49
52 1 1
50 -1 1
51 0 0
49 -2 4
  di=0 di2=350

Here, we have
n=10, X¯=51σ2=di2n-din2   =35010-0102   =35-0   =35

σ=35=5.91

CVx=5.9151×100        =11.58


Let Ay =105
xi
di=xi-105 di2
108 3 9
107 2 4
105 0 0
105 0 0
106 1 1
107 2 4
104 -1 1
103 -2 4
104 -1 1
101 -4 16
  di=0 di2=40

n=10, Y¯=105σ2=di2n-din2    =4010-0102    =4-0     =4

σ=4=2

CVy=2105×100        =1.90

Since CV of prices of share Y is lesser than that of X, prices of shares Y are more stable.



Page No 32.49:

Question 11:

Life of bulbs produced by two factories A and B are given below:
 

Length of life
(in hours):
550–650 650–750 750–850 850–950 950–1050

Factory A:
(Number of bulbs)
 
10 22 52 20 16

Factory B:
(Number of bulbs)
 
8 60 24 16 12

The bulbs of which factory are more consistent from the point of view of length of life?                             [NCERT EXEMPLAR]

Answer:


For factory A

Let the assumed mean A = 800 and h = 100.
 

Length of Life
(in hours)
Mid-Valuesxi ui=xi-800100 ui2 Number of bulbsfi fiui fiui2
550–650 600 −2 4 10 −20 40
650–750 700 −1 1 22 −22 22
750–850 800 0 0 52 0 0
850–950 900 1 1 20 20 20
950–1050 1000 2 4 16 32 64
        fi=120 fiui=10 fiui2=146

Mean, XA=A+hfiuifi=800+100×10120=808.33

Standard deviation, σA=hfiui2fi-fiuifi2=100146120-101202=100×1.0998=109.98

∴ Coefficient of variation = σAXA×100=109.98808.33×100=13.61

For factory B

Let the assumed mean A = 800 and h = 100.
 
Length of Life
(in hours)
Mid-Valuesxi ui=xi-800100 ui2 Number of bulbsfi fiui fiui2
550–650 600 −2 4 8 −16 32
650–750 700 −1 1 60 −60 60
750–850 800 0 0 24 0 0
850–950 900 1 1 16 16 16
950–1050 1000 2 4 12 24 48
        fi=120 fiui=-36 fiui2=156

Mean, XB=A+hfiuifi=800+100×-36120=800-30=770

Standard deviation, σB=hfiui2fi-fiuifi2=100156120--361202=100×1.1=110

∴ Coefficient of variation = σBXB×100=110770×100=14.29

Since the coefficient of variation of factory B is greater than the coefficient of variation of factory A, therefore, factory B has more variability than factory A. This means bulbs of factory A are more consistent from the point of view of length of life.

Page No 32.49:

Question 12:

Following are the marks obtained,out of 100 by two students Ravi and Hashina in 10 tests:
 

Ravi: 25 50 45 30 70 42 36 48 35 60
Hashina: 10 70 50 20 95 55 42 60 48 80

Who is more intelligent and who is more consistent?                                                               [NCERT EXEMPLAR]

Answer:


For Ravi
 

Marks xi di=xi-45 di2
25 −20 400
50 5 25
45 0 0
30 −15 225
70 25 625
42 −3 9
36 −9 81
48 3 9
35 −10 100
60 15 225
  di=-9 di2=1699

Mean, XR=A+di10=45+-910=44.1

Standard deviation, σR=di210-di102=169910--9102=169.09=13.003

Coefficicent of variation = σRXR×100=13.00344.1×100=29.49

For Hashina
 
Marks xi di=xi-55 di2
10 −45 2025
70 15 625
50 −5 25
20 −35 1225
95 40 1600
55 0 0
42 −13 169
60 5 25
48 −7 49
80 25 625
  di=-20 di2=6368

Mean, XH=A+di10=55+-2010=53

Standard deviation, σH=di210-di102=636810--20102=632.8=25.16

Coefficicent of variation = σHXH×100=25.1653×100=47.47

Since the coefficient of variation in mark obtained by Hashima is greater than the coefficient of variation in mark obtained by Ravi, so Hashina is more consistent and intelligent.

Page No 32.49:

Question 1:

Write the variance of first n natural numbers.

Answer:

​Sum of first n natural numbers =nn+12


Mean, X¯=Sum of all the observationsTotal number of observations
=nn+12n=n+12



 σ2=xi-X¯2n=xi-n+122n=1nxi2-xin+1+n+122=nn+12n+16n-nn+12n+1n+n+124n×n=n+12n+16-n+122+n+124=n+1n-112=n2-112

Page No 32.49:

Question 2:

If the sum of the squares of deviations for 10 observations taken from their mean is 2.5, then write the value of standard deviation.

Answer:

The sum of the squares of deviations for 10 observations, taken from their mean, is 2.5.
Square of each deviation = 2.510=0.25
Standard deviation = 0.25=0.5

Page No 32.49:

Question 3:

If x1, x2, ..., xn are n values of a variable X and y1, y2, ..., yn are n values of variable Y such that yi = axi + b; i = 1, 2, ..., n, then write Var(Y) in terms of Var(X).

Answer:

VarX=xi-X¯2nVarY=yi-Y¯2n

We have:
yi=axi+by¯=yin=axi+nbn=aX¯+b
Var Y=axi+b-aX¯-b2n=axi-aX¯2n=a2xi-X¯2n=a2VarX

Page No 32.49:

Question 4:

If X and Y are two variates connected by the relation Y=aX+bc and Var (X) = σ2, then write the expression for the standard deviation of Y.

Answer:

​​Y=aX+bcY=yin=aX+nbcn                  =aXnc+nbnc                  =aX¯c+bc

We know:

VarX=xi-X¯2n          =σ2VarY=yi-Y¯2n=aXc+bc-acX¯-bc2n=aXc-acX¯2n=ac2xi-X¯2n=ac2σ2SD σ=ac2σ2            =acσ

Page No 32.49:

Question 5:

In a series of 20 observations, 10 observations are each equal to k and each of the remaining half is equal to − k. If the standard deviation of the observations is 2, then write the value of k.

Answer:

n=20di=xi-a  =xi-xi20   =xi-0   =xidi=xi=0di2=20k2σ2=di2n-din2        =20k220-0        =k2σ=2=k2k=±2

Page No 32.49:

Question 6:

If each observation of a raw data whose standard deviation is σ is multiplied by a, then write the S.D. of the new set of observations.

Answer:

Standard deviation, σ=ixi-x2n

Here, x represents the arithmetic mean.

Multiplying each xi by a:
xnew=1nia.xi        =a×1nixi         =a.xold
 

New standard deviation, σnew=ia.xi-a.x2n        =ia2.xi-x2n        =aixi-x2n        =a.σ

Page No 32.49:

Question 7:

If a variable X takes values 0, 1, 2,..., n with frequencies nC0, nC1, nC2 , ... , nCn, then write variance X.

Answer:

x=i=0nxifii=0nfi=0×Con+1×C1n+...+n×CnnCon+C1n+...+Cnnx=n×2n-12nn+1      =nn+12Var(X)=σ2           =1ni=0nxi-x2           =1n0+1+2+....+n-nx2σ2=1nnn+12-n×nn+122        =1nnn+121-n2        =n24nn+12n-12



Page No 32.50:

Question 1:

For a frequency distribution mean deviation from mean is computed by
(a) M.D. = ΣfΣf d

(b) M.D. = ΣdΣf

(c) M.D. = ΣfdΣf

(d) M.D. = Σf dΣf

Answer:

(d) MD=fdf

Page No 32.50:

Question 2:

For a frequency distribution standard deviation is computed by applying the formula
(a) σ=Σfd2Σf-ΣfdΣf2

(b) σ=ΣfdΣf2-Σfd2Σf

(c) σ=Σfd2Σf-ΣfdΣf

(d) ΣfdΣf2-Σfd2Σf

Answer:

(a) σ=fd2f-fdf2

Page No 32.50:

Question 3:

If v is the variance and σ is the standard deviation, then
(a) v=1σ2

(b) v=1σ

(c) v = σ2

(d) v2 = σ

Answer:

(c) v=σ2

The variance is the square of the standard deviation.

Page No 32.50:

Question 4:

The mean deviation from the median is
(a) equal to that measured from another value
(b) maximum if all observations are positive
(c) greater than that measured from any other value.
(d) less than that measured from any other value.

Answer:

(d) less than that measured from any other value.

In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.

Page No 32.50:

Question 5:

If n = 10, X=12  and Σxi2=1530, then the coefficient of variation is
(a) 36%
(b) 41%
(c) 25%
(d) none of these

Answer:

(c) 25%

Standard deviation is expressed in the following manner:
 σ=1nixi2-X2  =153010-122  =9  =3

CV=σX×100     =312×100     =25%

Page No 32.50:

Question 6:

The standard deviation of the data:

x: 1 a a2 .... an
f: nC0 nC1 nC2 .... nCn
is
(a) 1+a22n-1+a2n

(b) 1+a222n-1+a2n

(c) 1+a22n-1+a22n

(d) none of these

Answer:

(d)   None of these
 

xi fi fixi xi2 fixi2
1 C0n C0n 1 1
a C1n C1n a2 a2  C1n
a2 C2n a2 C2n a4   a4  C2n
a3 C3n  a3 C3n a6  a6   C3n
:
:
:
:
:
:
:
:
:
:
:
:
:
 
:
:
:
:
an Cnn an  Cnn a2n a2n Cnn
 
i=1nfi=2n
 
i=1nfixi=1+an   i=1nfi xi2 =1+a2  n

 

















 Number of terms, N= i=1nfi=2ni=1nfixi  =C0n+ a C1n+ a2 C2n+...+ an  Cnn=  1+an        X =i=1nfixiN    =1+an  2ni=1nfixi2=1+a2nσ2 =Variance X=1Ni=1nfixi2 -i=1nfixiN2    =1+a2n2n-1+an  2n2     =1+a22n-1+a22nσ =Variance X    =1+a22n-1+a22n

Page No 32.50:

Question 7:

The mean deviation of the series a, a + d, a + 2d, ..., a + 2n from its mean is
(a) (n+1) d2n+1

(b) nd2n+1

(c) n (n+1) d2n+1

(d) (2n+1) dn (n+1)

Answer:

(c) n(n+1)d2n+1
 

xi xi-X=xi-a+nd
a nd
a + d (n-1)d
a + 2d (n-2)d
a + 3d (n-3)d
: :
: :
a + (n - 1)d d
a + nd 0
a + (n+1)d d
: :
: :
a + 2nd nd
xi=2n+1a+nd xi-X =nn+1d


















There are 2n+1 terms .  N=2n+1  xi=a+a+d+a+2d+a+3d+... +a+2nd         = (2n+1)a + d (1+2+3+... +2n)               a+a+a + ...(2n+1) times =2n+1 a          =(2n+1)a+2n2n+1d2                              Sum of the first n natural numbers is  n (n+1)2, but here we are considering the first 2n numbers.          =2n+1a+2n+1nd            =2n+1a+ndX =2n+1a+nd2n+1     =  a+nd xi-X  =nd+(n-1)d +(n-2)d +...+d+0+d+2 d+3d +...+nd                    = dn+n-1+n-2 +...+1  +0+ d1+2+3+...+n                   =dnn+12+dnn+12                       1+2+3+... +n=nn+12                   = n(n+1)dMean deviation about the mean = xi-X N                                                   =n(n+1)d2n+1

Page No 32.50:

Question 8:

A batsman scores runs in 10 innings as 38, 70, 48, 34, 42, 55, 63, 46, 54 and 44. The mean deviation about mean is
(a) 8.6
(b) 6.4
(c) 10.6
(d) 7.6

Answer:

Disclaimer: No option is matching the answer.


 N=10X=38+70+48+34+42+55+63+46+54+4410        =49410      =49.4                                                        
 

xi   di =xi-49.4
34 15.4
38 11.4
42 7.4
44 5.4
46 3.4
48 1.4
54 4.6
55 5.6
63 13.6
70 20.6
  i=ndi=88.8


Mean deviation from the mean =88.810                                               = 8.88



Page No 32.51:

Question 9:

The mean deviation of the numbers 3, 4, 5, 6, 7 from the mean is
(a) 25
(b) 5
(c) 1.2
(d) 0

Answer:

(c) 1.2

Mean X  =3+4+5+6+75                  =255                  =5
Taking the absolute value of deviation of each term from the mean, we get:

MD = (3-5)+(4-5)+(5-5)+(6-5)+(7-5)5          =2+1+0+1+25         =65         =1.2 

Page No 32.51:

Question 10:

The sum of the squares deviations for 10 observations taken from their mean 50 is 250. The coefficient of variation is
(a) 10 %
(b) 40 %
(c) 50 %
(d) none of these

Answer:

(a) 10%

We have: X  = 50, n=10      i=110xi-X 2 =250 SD=Variance of X                   =      i=110xi-X 2n                          =25010                 =5

Using  CV=σX×100    
 
  CV =550×100                 = 10%

Page No 32.51:

Question 11:

Let x1, x2, ..., xn be values taken by a variable X and y1, y2, ..., yn be the values taken by a variable Y such that yi = axi + b, i = 1, 2,..., n. Then,
(a) Var (Y) = a2 Var (X)
(b) Var (X) = a2 Var (Y)
(c) Var (X) = Var (X) + b
(d) none of these

Answer:

(a) Var (Y)=a2Var(X)

Var(X) = i=1n(xi-X¯)2n    where   Mean  X =i=1nxinVar(Y) =i=1n(yi-Y )2n and    Y =i=1nyinWe have,yi=axi+bY = i=1nyin   =i=1naxi+bn   =    ai=1nxin  +   nbn       =      aX+bVar(Y)=i=1nyi-Y2n                         =    i=1naxi+b-aX+b2n               =i=1n(axi-aX)2n               =a2i=1n(xi-X)2n               =a2Var(X)

Page No 32.51:

Question 12:

If the standard deviation of a variable X is σ, then the standard deviation of variable a X+bc is
(a) a σ

(b) acσ

(c) ac σ

(d) aσ+bc

Answer:

(c) acσ


Y=aX+bcY=yin=aX+nbcn                  =aXnc+nbnc                  =aX¯c+bc


VarX=xi-X¯2n          =σ2VarY=yi-Y¯2n=aXc+bc-acX¯-bc2n=aXc-acX¯2n=ac2xi-X¯2n=ac2σ2SD σ=ac2σ2            =acσ

Page No 32.51:

Question 13:

If the S.D. of a set of observations is 8 and if each observation is divided by −2, the S.D. of the new set of observations will be
(a) −4
(b) −8
(c) 8
(d) 4

Answer:

(d) 4

If a set of observations, with SD σ, are multiplied with a non-zero real number a, then SD of the new observations will be a σ.
Dividing the set of observations by − 2 is same as multiplying the observations by 1-2.

New S.D.=-12×8                 =82                 =4

Page No 32.51:

Question 14:

If two variates X and Y are connected by the relation Y=a X+bc, where a, b, c are constants such that ac < 0, then
(a) σY=acσX

(b) σY=-acσX

(c) σY=acσX+b

(d) none of these

Answer:

(b) σY=-acσX

Y=aX+bc Y=i=1naX+bcn    =ai=1nX+nbcn    =aci=1nXn+bc    =aXc+bcWe know:Var (X)=i=1nxi-X2n             =    σ2Var(Y) = i=1n(yi-Y)2n           =i=1naXc+bc-acX-bc2n             =i=1naXc-acX2n        =ac2i=1nxi-X2n           =ac2σ2SD of Y (σy)=   ac2σ2                         =acσ


 ac<0 a<0 or c<0  ac=  -ac

   σY=-acσX

Page No 32.51:

Question 15:

If for a sample of size 60, we have the following information xi2=18000 and xi=960, then the variance is

(a) 6.63                                  (b) 16                                   (c) 22                                  (d) 44                               

Answer:


Given: xi2=18000, xi=960 and n = 60

∴ Variance

=xi2n-xin2=1800060-960602=300-256=44

Hence, the correct answer is option (d).

Page No 32.51:

Question 16:

Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is

(a) s                                  (b) ks                                  (c) s + k                                  (d) sk                              

Answer:

The given observations are a, b, c, d, e.

Mean = m = a+b+c+d+e5

xi=a+b+c+d+e=5m            .....(1)

Standard deviation, s = xi25-m2          

Now, consider the observations a + k, b + k, c + k, d + k, e + k.

New mean =a+k+b+k+c+k+d+k+e+k5

                  =a+b+c+d+e+5k5=5m+5k5=m+k

∴ New standard deviation

=xi+k25-m+k2=xi2+k2+2xik5-m2+k2+2mk=xi25+k25+2xik5-m2+k2+2mk=xi25-m2+5k25-k2+2kxi5-2mk
=xi25-m2+2k×5m5-2mk               Using 1=xi25-m2=s

Hence, the correct answer is option (a).

Page No 32.51:

Question 17:

The standard deviation of first 10 natural numbers is

(a) 5.5                                  (b) 3.87                                   (c) 2.97                                  (d) 2.87                            

Answer:

We know that the standard deviation of first n natural number is n2-112.

∴ Standard deviation of first 10 natural numbers

=102-112=9912=8.25=2.87

Hence, the correct answer is option (d).

Page No 32.51:

Question 18:

Consider the first 10 positive integers. If we multiply each number by −1 and then add 1 to each number, the variance of the numbers so obtained is

(a) 8.25                                  (b) 6.5                                   (c) 3.87                                  (d) 2.87                            

Answer:

The first 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Multiplying each number by −1, we get

−1, −2, −3, −4, −5, −6, −7, −8, −9, −10

Adding 1 to each of these numbers, we get

0, −1, −2, −3, −4, −5, −6, −7, −8, −9

Now,

xi=0+-1+-2+-3+-4+-5+-6+-7+-8+-9=-45

xi2=0+1+4+9+16+25+36+49+64+81=285

∴ Variance of the obtained numbers

=xi210-xi102=28510--45102=28.5-20.25=8.25

Hence, the correct answer is option (a).

Page No 32.51:

Question 19:

Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is

(a) 6.5                                  (b) 2.87                                   (c) 3.87                                 (d) 8.25                          

Answer:

The given numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

If 1 is added to each number, then the new numbers obtained are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

Now,

xi=2+3+4+5+6+7+8+9+10+11=65

xi2=4+9+16+25+36+49+64+81+100+121=505

∴ Variance of the numbers so obtained

=xi210-xi102=50510-65102=50.5-42.25=8.25

Hence, the correct answer is option (d).

Page No 32.51:

Question 20:

The mean of 100 observations is 50 and their standard deviation is 5. The sum of all squares of all the observations is

(a) 50,000                                  (b) 250,000                                   (c) 252500                                  (d) 255000                           

Answer:

Let x and σ be the mean and standard deviation of 100 observations, respectively.

 x=50, σ=5 and n = 100

Mean, x=50

xi100=50xi=5000                     .....1

Now,

Standard deviation, σ=5

xi2100-xi1002=5xi2100-50001002=25                    From 1xi2100=25+2500=2525xi2=252500

Thus, the sum of all squares of all the observations is 252500.

Hence, the correct answer is option (c).

Page No 32.51:

Question 21:

Let x1, x2, ..., xn be n observations. Let yi=axi+b for i = 1, 2, 3, ..., n, where a and b are constants. If the mean of xi's is 48 and their standard deviation is 12, the mean of yi's is 55 and standard deviation of yi's is 15, the values of a and b are

(a) a = 1.25, b = −5                 (b) a = −1.25, b = 5                 (c) a = 2.5, b = −5                 (d) a = 2.5, b = 5                            

Answer:

It is given that yi=axi+b for i = 1, 2, 3, ..., n, where a and b are constants.

xi = 48 and σxi=12

yi=55 and σyi=15

yi=axi+byin=axi+bnyin=axin+bnyi=axi+b   55=48a+b           .....1                      

Now,

Standard deviation of yi = Standard deviation of axi+b

σyi=a×σxi15=12aa=1512=1.25

Putting a = 1.25 in (1), we get

b=55-48×1.25=55-60=-5

Thus, the values of a and b are 1.25 and −5, respectively.

Hence, the correct answer is option (a).

Page No 32.51:

Question 22:

The mean deviation of the data 3, 10, 10, 4, 7, 10, 5 from the mean is

(a) 2                                  (b) 2.57                                   (c) 3                                 (d) 3.57                          

Answer:

The given observations are 3, 10, 10, 4, 7, 10, 5.

∴ Mean, x=3+10+10+4+7+10+57=497=7

Now,

Mean deviation from mean, MD

=xi-77


=3-7+10-7+10-7+4-7+7-7+10-7+5-77=4+3+3+3+0+3+27=187=2.57

Hence, the correct answer is (b).



Page No 32.52:

Question 23:

The mean deviation for n observations x1, x2, ...,xn from their mean X is given by

(a) i=1nxi-X                               (b) 1ni=1nxi-X                                   (c) i=1nxi-X2                                  (d) 1ni=1nxi-X2                            

Answer:

The mean deviation for n observations x1, x2, ...,xn from their mean X is 1ni=1nxi-X.


Disclaimer: There is some printing error in option (b) given in the question. The answer would be option (b) if it given as 1ni=1nxi-X.

Page No 32.52:

Question 24:

Let x1, x2, ...,xn be n observations and X be their arithmetic mean. The standard deviation is given by

(a) i=1nxi-X2                 (b) 1ni=1nxi-X2                 (c) 1ni=1nxi-X2                 (d) 1ni=1nxi2-X2                  

Answer:


It is given that x1, x2, ...,xn are n observations and X is their arithmetic mean.

The standard deviation of given observations is 1ni=1nxi-X2.

Also,

1ni=1nxi-X2 = 1ni=1nxi2-X2

Hence, the correct answers are options (c) and (d).


Disclaimer: For option (c) to be the only correct answer, option (d) should be different from the given value.

Page No 32.52:

Question 25:

The standard deviation of the observations 6, 5, 9, 13, 12, 8, 10 is

(a) 6                             (b) 6                             (c) 527                            (d) 527                           

Answer:

The given observations are 6, 5, 9, 13, 12, 8, 10.

Now,

xi=6+5+9+13+12+8+10=63

xi2=36+25+81+169+144+64+100=619

∴ Standard deviation of the observations, σ

=1Nxi2-1Nxi2=17×619-17×632=6197-81=619-5677=527

Hence, the correct answer is option (d).



Page No 32.6:

Question 1:

Calculate the mean deviation about the median of the following observations:
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Answer:

Formula used for mean deviation:
MD=1ni=1ndiHere, di=xi-M
 M = Median

i) Arranging the data in ascending order:
2354, 2780, 3011, 3020, 3541, 4150, 5000

Here, median (M)=3020 and n = 7.

xi di = xi - 3020
3011 9
2780 240
3020 0
2354 666
3541 521
4150 1130
5000 1980
Total 4546

MD=1ni=1ndi

 MD=17×4546=649.42

ii) Arranging the data in ascending order:
34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Here, n is equal to 10.
Median is the arithmetic mean of the fifth and the sixth observation.

Median, M = 46+482 = 47
 
xi di = xi-M
38 9
70 23
48 1
34 13
42 5
55 8
63 16
46 1
54 7
44 3
Total 86

MD=110×86=8.6

iii) Arranging the data in ascending order:
30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Here,
n=10
Also, median is the AM of the fifth and the sixth observation.

Median, M = 42+442 = 43
 
xi di = xi-M
34 9
66 23
30 13
38 5
44 1
50 7
40 3
60 17
42 1
51 8
Total 87

MD=110×87=8.7

iv) Arranging the data in ascending order.
22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Here, n=10.
Also, median is the AM of the fifth and the sixth observation.

Median, M = 28+292 = 28.5
 
xi di = xi-M
22 6.5
24 4.5
30 1.5
27 1.5
29 0.5
31 2.5
25 3.5
28 0.5
41 12.5
41 13.5
Total 47

MD=110×47=4.7

v) Arranging the data in ascending order:
34, 38, 42, 44, 47, 48, 53, 55, 63, 70

Here, n=10.
Also, median is the AM of the fifth and the sixth observation.

Median, M = 47+482 = 47.5
 
xi di = xi-M
38 9.5
70 22.5
48 0.5
34 13.5
63 15.5
42 5.5
55 7.5
44 3.5
53 5.5
47 0.5
Total 84

MD=110×84=8.4

Page No 32.6:

Question 2:

Calculate the mean deviation from the mean for the following data:
(i) 4, 7, 8, 9, 10, 12, 13, 17
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer:

Formula used for finding the mean deviation about the mean is given below:

MD = 1ni=1ndi , where di = xi-x


i)
Let x be the mean of the given data.

x= 4+7+8+9+10+12+13+178 = 10
 

xi di = xi - x¯
4 6
7 3
8 2
9 1
10 0
12 2
13 3
17 7
Total 24

MD = 1ni=1ndi
MD=18×24=3

ii)
Let x be the mean of the given data.

x=13+17+16+14+11+13+10+16+11+18+12+1712=14
 
xi di = xi - x¯
13 1
17 3
16 2
14 0
11 3
13 1
10 4
16 2
11 3
18 4
12 2
17 3
Total 28

MD = 112×28 = 2.33

iii)
Let x be the mean of the given data.

x=38+70+48+40+42+55+63+46+54+4410=50
 
xi di = xi - x¯
38 12
70 20
48 2
40 10
42 8
55 5
63 13
46 4
54 4
44 6
Total 84

MD = 110×84 = 8.4

iv)
Let x be the mean of the given data.

x=36+746+42+60+45+53+46+51+5910=50
 
xi di = xi - x¯
36 14
72 22
46 4
42 8
60 10
45 5
53 3
46 4
51 1
49 1
Total 72

MD = 110×72 = 7.2

Page No 32.6:

Question 3:

Calculate the mean deviation of the following income groups of five and seven members from their medians:

I
Income in Rs.
II
Income in Rs.
4000
4200
4400
4600
4800

 
 300
4000
4200
4400
4600
4800
5800

Answer:

Calculate the mean deviation for the first data set.
The data is already arranged in ascending order.
For this data set, n is equal to 5.
Also, median, M=4400
MD= 1ni=1ndi,  where di=xi-M
 

xi di = xi - M
4000 400
4200 200
4400 0
4600 200
4800 400
Total 1200

MD=15×1200=240
Therefore, for the income of families in the first group, the mean deviation from the median is Rs 240.

Now, consider the second data set. This is also arranged in ascending order.
Here, n=7.
Also, median, M=4400
 
xi di = xi - M
300 4100
4000 400
4200 200
4400 0
4600 200
4800 400
5800 1400
Total 6700

MD=17×6700=957.14
Therefore, for the income of families in the second group, the mean deviation from the median is Rs 957.14.

Page No 32.6:

Question 4:

The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find mean deviation from median
(ii) Find mean deviation from the mean also.

Answer:

i) Formula for the mean deviation from the median is as follows:
MD = 1ni=1ndi, where di = xi - M

Arranging the data in ascending order for finding the median:
15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.2, 72.9, 79

Here, n=10.
Therefore, median is the average of the fifth and the sixth observations.

M=40+52.32=46.15
 

xi di=xi-46.15
40 6.15
52.3 6.15
55.2 9.05
72.9 26.75
52.8 6.65
79 32.85
32.5 13.65
15.2 30.95
27.9 18.25
32 14.15
Total 164.6

MD=110×164.6=16.46

Mean deviation from median in 16.4 cm.

ii)
Let x¯ be the mean of the given data set.

x=40+52.3+55.2+72.9+52.8+79+32.5+15.2+27.9+30.210=45.98

 
xi di=xi-45.98
40 5.98
52.3 6.32
55.2 9.22
72.9 26.92
52.8 6.82
79 33.02
32.5 13.48
15.2 30.78
27.9 18.08
32 13.98
Total 164.6

MD = 110×164.6 = 16.46

Mean deviation from the mean is 16.4 cm.

Page No 32.6:

Question 5:

In question 1 (iii), (iv), (v) find the number of observations lying between X − M.D. and X + M.D, where M.D. is the mean deviation from the mean.

Answer:

​iii)
Let x¯ be the mean of the data set.

x=34+66+30+38+44+50+40+60+42+5110=45.5

MD = 1ni=1ndi, where di = xi - x
 

xi di = xi - 45.5
34 11.5
66 20.5
30 15.5
38 7.5
44 1.5
50 4.5
40 5.5
60 14.5
42 3.5
51 5.5
Total 90

MD = 110×90 = 9

x - M.D. = 45.5 - 9 = 36.5Also, x + M.D. = 45.5 + 9 = 54.5

Hence, there are 6 observations between 36.5 and 54.5.

iv)
Let x¯ be the mean of the data set.

x=22+24+30+27+29+31+25+28+41+4210=29.9
 
xi di = xi - 29.9
22 7.9
24 5.9
30 0.1
27 2.9
29 0.9
31 1.1
25 4.9
28 1.9
41 11.9
42 12.1
Total 48.8

MD = 110×48.8 = 4.88

x - M.D. = 29.9 - 4.88 = 25.02,and x + M.D. = 29.9 + 4.88 = 34.78

There are 5 observations between 25.02 and 34.78.

v)
Let x¯ be the mean of the data set.

x=38+70+48+34+63+42+55+44+53+4710=49.4
 
xi di = xi - 49.4
38 11.4
70 20.6
48 1.4
34 15.4
63 13.6
42 7.4
55 5.6
44 5.4
53 3.6
47 2.4
Total 86.8

MD = 110×86.6 = 8.68

x- MD = 49.4-8.68 = 40.72and  x+MD = 49.4+8.68 = 58.08

There are 6 observations between 40.72 and 58.08.



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