Since, the given frequency distribution is not continuous, so to make it continuous, we will subtract 0.5 from lower limit and add 0.5 to the upper limit of each class.
Class | fi | xi | ui=xi−Ah | u2i | fiui | fiu2i |
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 | 10 0 | | | | 25 | 199 |
Here, N=100,h=4
Let the assumed mean A=42.5
Mean ¯x=A+∑5i=1fiuiN×h=42.5+25100×4=43.5
Variance (σ2)=h2N2⎡⎣N5∑i=1fiu2i−(5∑i=1fiui)2⎤⎦
=1610000[100×199−(25)2]
=1610000[19900−625]
=1610000×19275
=30.84
∴ Standard deviation (σ)=5.55