Q.
Given below are the diameters of circles (in mm) drawn in a design.
Diameter
33 - 36
37 - 40
41 - 44
45 -48
49 - 52
Number of circles
15
17
21
22
25
find the mean diameter of the circles.
| Diameter | 33 - 36 | 37 - 40 | 41 - 44 | 45 -48 | 49 - 52 |
| Number of circles | 15 | 17 | 21 | 22 | 25 |
Statistics
Solution:
Converting the given series into an exclusive series we prepare the table given below.
Class
Frequency$f_i$
Mid point$( x_i)$
$y_{i} = \frac{\left(x_{i}-42.5\right)}{4}$
$y_i^2$
$f_iy_i$
$f_i y_i^2$
32.5 - 36.5
15
34.5
-2
4
-30
60
36.5 - 40.5
17
38.5
-1
1
-17
17
40.5 - 44.5
21
42.5=A
0
0
0
0
44.5 - 48.5
22
46.5
1
1
22
22
48.5 - 52.5
25
50.5
2
4
50
100
N=100
25
199
$\therefore A = 42.5, h=4$,
$N= \sum f_{i} = 100, \sum f_{i} y_{i} = 25$
and $\sum f_{i} y_{i}^{2} = 199$.
$ \therefore \bar{x} = \left( A+ \frac{\sum f_{i}y_{i}}{N} \times h\right)$
$\Rightarrow \bar{x} = \left(42.5 +\frac{25}{100}\times4\right) = 43.5 $
Thus the mean $= 43.5$
| Class | Frequency$f_i$ | Mid point$( x_i)$ | $y_{i} = \frac{\left(x_{i}-42.5\right)}{4}$ | $y_i^2$ | $f_iy_i$ | $f_i y_i^2$ |
|---|---|---|---|---|---|---|
| 32.5 - 36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
| 36.5 - 40.5 | 17 | 38.5 | -1 | 1 | -17 | 17 |
| 40.5 - 44.5 | 21 | 42.5=A | 0 | 0 | 0 | 0 |
| 44.5 - 48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
| 48.5 - 52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
| N=100 | 25 | 199 |