| Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-98 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Solution:
Discover the midpoint of the given stretch utilizing the recipe.
Midpoint (xi) = (maximum cutoff + lower limit)/2
For this situation, the worth of mid-point (xi) is exceptionally huge, so let us expect the mean worth, A = 70 and class stretch is h = 10.
Thus, ui = (xi-A)/h = ui = (xi-70)/10
Substitute and discover the qualities as follows:
Class Interval Frequency (fi) (xi)
![\[di\text{ }=\text{ }xi\text{ }\text{ }a~~~~~~~~~~~~ui\text{ }=\text{ }di/h\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0d5bfdfd7c90e8706d1999f953aad727_l3.png "Rendered by QuickLaTeX.com")
45-55 3 50 -20 -2 -6
55-65 10 60 -10 -1 -10
65-75 11 70 0 0 0
75-85 8 80 10 1 8
85-95 3 90 20 2 6
Total fi = 35 Sum fiui = – 2
![\[Thus,\text{ }Mean\text{ }=\text{ }x\text{ }=\text{ }A+\left( \sum fiui/\sum fi \right)\times h\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-671b3c41e006b48208cc1977ba76ddc4_l3.png "Rendered by QuickLaTeX.com")
= 70+(- 2/35)×10
= 69.42 In this manner, the mean education part = 69.42