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Chapter 32: Statistics – Exercise 32.3

Statistics – Exercise – 32.3 – Q.1

We have to calculate mean deviation from the median. So, first we calculate the median.

CI x F cf d = (x – med) fd
0 – 10 5 5 5 20 100
10 – 20 15 10 15 10 100
20 – 30 25 20 35 0 0
30 – 40 35 5 91 10 50
40 – 50 45 10 101 20 200
    50     450

 

Statistics – Exercise – 32.3 – Q.2(i)

CI x f xf d = (x-mean) fd
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
    50 17900   7896

 

Statistics – Exercise – 32.3 – Q.2(ii)

Classes fi xi di fidi    
95 – 105 9 100 -3 -27 28.58 257.22
105 – 115 13 110 -2 -26 18.58 241.54
115 – 125 16 120 -1 -16 8.58 137.28
125 – 135 26 130 0 0 1.42 36.92
135 – 145 30 140 1 30 11.42 342.6
145 – 155 12 150 2 24 21.42 257.04
  N = 106     Total = -15   Total = 1272.60

 

Statistics – Exercise – 32.3 – Q.2(iii)

CI x f xf d = (x - mean) fd
0 – 10 5 6 30 22 132
10 – 20 15 8 120 12 96
20 – 30 25 14 350 2 28
30 – 40 35 16 560 8 128
40 – 50 45 4 180 18 72
50 – 60 55 2 110 28 56
           
    50 1350   512
  Mean 27    
Deviation 10.24

 

Statistics – Exercise – 32.3 – Q.3

Find the mean deviation from the mean for the data:

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

 

Statistics – Exercise – 32.3 – Q.4

We have to calculate mean deviation from the median. So, first we calculate the median.

CI x f cf D = (x – med) fd
17 – 19.5 18.25 5 5 20 100
20 – 25.5 22.75 16 21 15.5 248
26 – 35.5 30.75 12 33 7.5 90
36 – 40.5 38.25 26 59 0 0
41 – 50.5 45.75 14 73 7.5 105
51 – 55.5 53.25 12 85 15 180
56 – 60.5 58.25 6 91 20 120
61 – 70.5 65.75 5 96 27.5 137.5
    96     980.5

We have N = 96 ⟹ N/2 = 48

The cumulative frequency just greater than N/2 is 59 and the corresponding value of x is 38.25.

Hence, median = 38.25

 

Statistics – Exercise – 32.3 – Q.5

M.D from Median

Marks Students xi Cum. Freq fidi
0 – 10 5 5 5 55/3 275/3
10 – 20 8 15 13 25/3 200/3
20 – 30 15 25 28 3-May 75/3
30 – 40 16 35 44 35/3 560/3
40 – 50 6 45 50 65/3 390/3
  N = 50       Total = 500

M.D from mean

Marks Students xi   fidi |xi - 27| fi|xi - 27|
0 – 10 5 5 -3 -15 22 110
10 – 20 8 15 -2 -16 12 96
20 – 30 15 25 -1 -15 2 30
30 – 40 16 35 0 0 8 128
40 – 50 6 45 1 6 18 108
  N = 50     Total = 40   Total = 472

 

Statistics – Exercise – 32.3 – Q.6

Converting the given data into continuous frequency distribution by sub tracing 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Age x: f: Cumulative frequency |d:|=|x:-38| f:|d:|
15.5 – 20.5 18 5 5 20 100
20.5 – 25.5 23 6 11 15 90
25.5 – 30.5 28 12 23 10 120
30.5 – 35.5 33 14 37 5 70
35.5 – 40.5 38 26 63 0 0
40.5 – 45.5 43 12 75 5 60
45.5 – 50.5 48 16 91 10 160
50.5 – 55.5 53 9 100 15 135
    N = Σf: = 100     Σf:|d:| = 735

Clearly, N = 100 ⟹ N/2 = 50.

Cumulative frequency is just greater than N/2 is 63 and the corresponding class is 35.5 - 40.5.

I = 35.5, f = 26, h = 5, F = 37

 

Statistics – Exercise – 32.3 – Q.7

Classes fi xi fixi |xi - 9.2| fi|xi - 9.2|
0 – 4 4 2 8 7.2 28.8
4 – 8 6 6 35 3.2 19.2
8 – 12 8 10 80 0.8 6.4
12 – 16 5 14 70 4.8 24
16 – 20 2 18 36 8.8 17.6
  N = 25   Total = 230   Total = 96.0

 

Statistics – Exercise – 32.3 – Q.8

Classes fi xi fixi |xi - 14.1| fi|xi - 14.1|
0 – 6 4 3 12 11.1 44.4
6 – 12 5 9 45 5.1 25.5
12 – 18 3 15 45 0.9 2.7
18 – 24 6 21 126 6.9 41.4
24 – 30 2 27 54 12.9 25.8
  N = 20   Total = 282   Total = 139.8


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