Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75 -85 | 85-98 |
Number of cities | 3 | 10 | 11 | 8 | 3 |
Find the midpoint of the given interval using the formula.
Midpoint $(x_i)$ = $\dfrac{(upper limit + lower limit)}{2}$ |
In this case, the value of Mid-point $(x_i)$ is very large, so let us assume the mean value, a = 70.
Class interval (h) = 10
So, $u_i$ = $\dfrac{(x_i – a)}{h}$
$u_i$ = $\dfrac{(x_i – 70)}{10}$
Substitute and find the values as follows:
Class Interval | Frequency $(f_i)$ | $(x_i)$ | $u_i = \dfrac{(x_i – 70)} {10}$ | $f_i u_i$ |
---|---|---|---|---|
45-55 | 3 | 50 | -2 | - 6 |
55-65 | 10 | 60 | -1 | -10 |
65-75 | 11 | 70 = a | 0 | 0 |
75-85 | 8 | 80 | 1 | 8 |
85- 95 | 3 | 90 | 2 | 6 |
Sum $f_i$ = 35 | Sum $f_i u_i$ = - 2 |
So,
Mean = $\overline{x} = a + ( \dfrac{∑f_iu_i }{∑f_i}) × h$ |
= 70 + ($\dfrac{-2}{35}$) × 10
= 69.43
Therefore, the mean literacy part = 69.43%