Question

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.

Literacy rate (in %)	$45 - 55$	$55 - 65$	$65 - 75$	$75 - 85$	$85 - 95$
Number of cities	$3$	$10$	$11$	$8$	$3$

Answer 1

Hint: Here we will discuss the concept of the mean; we create a table by adding the columns to the given data, first we have to find the class mark by using the relation. Then we make the table for our clearance and use the mean formula. Finally we get the mean literacy rate.

Formula used: $({x_i}) = \dfrac{{Upperclass\lim it + Lowerclass\lim it}}{2}$
\[\overline X = a + \dfrac{{\sum\limits_{i = 1}^n {{f_i}{u_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }} \times h\]

Complete step-by-step solution:
To find the class mark $({x_i})$ for each interval, using the following relation:
$({x_i}) = \dfrac{{Upperclass\lim it + Lowerclass\lim it}}{2}$
Hence we get the values for ${x_1} = \dfrac{{45 + 55}}{2} = \dfrac{{100}}{2}$
$ = 50$
Similarly, we can do the rest of the given interval. So we will get the column of ${x_i}$
Also, taking $70$ as assumed mean $(a)$ (because it is the mid value of the given interval)
Now we have to calculating ${d_i},{u_i},{f_i}{u_i}.$
So, we need to make table, and we can write it as:

Literacy rate (%)	Number of cities $({f_i})$	${x_i}$	${d_i} = {x_i} - 70$	${u_i} = \dfrac{{{d_i}}}{{10}}$	${f_i}{u_i}$
$45 - 55$	$3$	$50$	$ - 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	$35$

From the table, we can find the sum of the all ${f_i}'s$
$ \Rightarrow \sum\limits_{i = 1}^n {{f_i} = 35} $
Now we added all the ${f_i}{u_i}'s$
$ \Rightarrow \sum\limits_{i = 1}^n {{f_i}{u_i} = - 6 - 10 + 0 + 8 + 6 = - 2} $
Also, the class size $(h)$for this data $ = 10$(because in the class interval $55 - 45 = 10,65 - 55 = 10)$
By applying mean formula
$\overline X = a + \left( {\dfrac{{\sum\limits_{i = 1}^n {{f_i}{u_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}} \right) \times h$
Now we just substituted all the values in the formula above
$ = 70 + \left( {\dfrac{{ - 2}}{{35}}} \right) \times 10$
First we multiply the terms,
$ = 70 - \dfrac{{20}}{{35}}$
By cancelling the fraction terms by$5$,
$ = 70 - \dfrac{4}{7}$
Now dividing the fraction term to reduce the term
$ = 70 - 0.57$
Now subtraction is needed and we done, we get
$ = 69.43$

Therefore the mean literacy rate is $69.43\% $

Note: In this question we have an alternative method,
Here we have to use the formula, Mean,\[\overline X = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
Now students might use the formula, where product of ${f_i}$ and ${x_i}$ for each class will be as:

Class	Frequency $({f_i})$	${x_i}$	${f_i}{x_i}$
$45 - 55$	$3$	$50$	$150$
$55 - 65$	$10$	$60$	$600$
$65 - 75$	$11$	$70$	$770$
$75 - 85$	$8$	$80$	$640$
$85 - 95$	$3$	$90$	$270$
Total	$35$		$2430$

Now put the values in the above mean formula and get,
$\overline X = \dfrac{{2430}}{{35}}$
On dividing the term and we get,
$ \Rightarrow 69.428$
We can round of the last digit hence we get,
$ \Rightarrow 69.43\% $
Therefore we get the mean literacy rate.
Thus we get the same answer in both the methods.