Q.
The diameters of circles (in mm) drawn in a design are given below.
Diameters
33- 36
37- 40
41- 44
45-48
49- 52
Number of circles
15
17
21
22
25
The standard deviation and mean diameter of the circles respectively are
| Diameters | 33- 36 | 37- 40 | 41- 44 | 45-48 | 49- 52 |
| Number of circles | 15 | 17 | 21 | 22 | 25 |
Statistics
Solution:
Class
$f_i$
Mid value$(x_i)$
Deviation from mean $d_i =\frac{ x_i - 42.5}{4} A =42.5, h = 4 $
$d_i^2$
$f_id_i$
$f_id_i^2$
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
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
Total
100
25
199
Mean $(\bar{x})=A+\frac{\Sigma f d_i}{\Sigma f_i} \times h$
$=42.5+\frac{25}{100} \times 4$
$=42.5+1=43.5$
Standard deviation $(\sigma)=\sqrt{\frac{\Sigma f d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f d_i}{\Sigma f_i}\right)^2} \times h$
$=\sqrt{\frac{199}{100}-\left(\frac{25}{100}\right)^2} \times 4$
$=\sqrt{\frac{199 \times 100-625}{100 \times 100}} \times 4$
$=\frac{4}{100} \sqrt{19900-625}$
$=\frac{1}{25} \times \sqrt{19275}=\frac{138.83}{25}=5.55$
| Class | $f_i$ | Mid value$(x_i)$ | Deviation from mean $d_i =\frac{ x_i - 42.5}{4} A =42.5, h = 4 $ | $d_i^2$ | $f_id_i$ | $f_id_i^2$ |
|---|---|---|---|---|---|---|
| 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 | 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 |
| Total | 100 | 25 | 199 |