No of Students per teacher | Number of states / U.T |
15-20 | 3 |
20-25 | 8 |
25-30 | 9 |
30-35 | 10 |
35-40 | 3 |
40-45 | 0 |
45-50 | 0 |
50-55 | 2 |
Solution:
Given information:
Modular class = 30 – 35,
l = 30,
Class width (h) = 5,
fm = 10, f1 = 9 and f2 = 3
Mode Formula:
\[\begin{array}{*{35}{l}}
~ \\
Mode\text{ }=\text{ }l+\text{ }\left[ \left( fm-f1 \right)/\left( 2fm-f1-f2 \right) \right]\times h \\
\end{array}\]
Substitute the qualities in the given recipe
\[\begin{array}{*{35}{l}}
~ \\
Mode\text{ }=\text{ }30+\left( \left( 10-9 \right)/\left( 20-9-3 \right) \right)\times 5 \\
~ \\
Mode\text{ }=\text{ }30+\left( 5/8 \right)\text{ }=\text{ }30+0.625 \\
~ \\
Mode\text{ }=\text{ }30.625 \\
\end{array}\]
Accordingly, the method of the given information = 30.625
Computation of mean:
Discover the midpoint utilizing the recipe, xi =(upper limit +lower limit)/2
Class Interval Frequency (fi) Mid-point (xi) fixi
15-20 3 17.5 52.5
20-25 8 22.5 180.0
25-30 9 27.5 247.5
30-35 10 32.5 325.0
35-40 3 37.5 112.5
40-45 0 42.5 0
45-50 0 47.5 0
50-55 2 52.5 105.5
Total fi = 35 Sum fixi = 1022.5
\[\begin{array}{*{35}{l}}
Mean\text{ }=\text{ }x\text{ }=\text{ }\sum fixi/\sum fi \\
~ \\
\end{array}\]
= 1022.5/35
= 29.2 In this manner, mean = 29.2