The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section. $\begin{array}{|l|c|c|c|c|c|c|c|}\hline \text{Percentage of female teachers} & 15\text{-}25 & 25\text{-}35 & 35\text{-}45 & 45\text{-}55 & 55\text{-}65 & 65\text{-}75 & 75\text{-}85 \\ \hline \text{Number of States/U.T.} & 6 & 11 & 7 & 4 & 4 & 2 & 1 \\ \hline \end{array}$ (Source: Seventh All India School Education Survey conducted by NCERT)

Step 1: Calculate the Class Mark ($x_i$)

To find the mean using any method, we first need to determine the class mark ($x_i$) for each class interval. The class mark represents the midpoint of each interval and is calculated by averaging the lower and upper limits of the class. This value will be used in our calculations.

Step 2: Direct Method: Calculate $\Sigma f_i x_i$ and $\Sigma f_i$

The direct method for calculating the mean involves summing the products of each frequency ($f_i$) and its corresponding class mark ($x_i$), and then dividing by the sum of all frequencies. This formula provides a straightforward way to find the arithmetic mean of grouped data.

Step 3: Assumed Mean Method: Choose an Assumed Mean (A) and calculate $d_i$

In the assumed mean method, we choose an assumed mean (A) from the class marks (usually the middle one) to simplify calculations. We then calculate the deviation ($d_i$) of each class mark from this assumed mean. The mean is found by adding the assumed mean to the average of the deviations.

Step 4: Step Deviation Method: Calculate $u_i$

The step deviation method is an extension of the assumed mean method, used when the class intervals have a uniform width ($h$). We further simplify the deviations by dividing them by the class width to get $u_i$. This makes the numerical values smaller and easier to work with, especially for manual calculations. We then use the given formula to find the mean.

Step 5: Consolidate results in a table for all methods

We consolidate all necessary calculations into a single table. This table includes the class intervals, frequencies ($f_i$), class marks ($x_i$), product $f_i x_i$, deviations $d_i$, step deviations $u_i$, and product $f_i u_i$. The sums $\sum f_i$, $\sum f_i x_i$, and $\sum f_i u_i$ are then calculated for use in the mean formulas. We choose $A = 50$ (the class mark of the class with frequency 4) and $h = 10$ (the class width).

Step 6: Calculate Mean by Direct Method

Using the values from our consolidated table, we apply the direct method formula. We substitute the sum of $f_i x_i$, which is $1390$, and the sum of frequencies $\sum f_i$, which is $35$. We perform the division to get the mean.

Step 7: Calculate Mean by Assumed Mean Method

For the assumed mean method, we take the assumed mean $A=50$. We calculate $\sum f_i d_i$. From the table, $\sum f_i d_i = (6 \times -30) + (11 \times -20) + (7 \times -10) + (4 \times 0) + (4 \times 10) + (2 \times 20) + (1 \times 30) = -180 - 220 - 70 + 0 + 40 + 40 + 30 = -470 + 110 = -360$. We substitute the values into the formula along with $\sum f_i = 35$ to find the mean.

Step 8: Calculate Mean by Step Deviation Method

For the step deviation method, we use the assumed mean $A=50$ and class width $h=10$. We substitute the value of $\sum f_i u_i = -36$ and $\sum f_i = 35$ into the formula. We perform the calculation to arrive at the mean.

Step 9: Final Calculations

We complete the arithmetic for each method. In the direct method, $1390 \div 35 = 39.7142...$. For the assumed mean method, we have $50 + \frac{-360}{35} = 50 - 10.2857... = 39.7142...$. Similarly, for the step deviation method, $50 + (\frac{-36}{35}) \times 10 = 50 - \frac{360}{35} = 50 - 10.2857... = 39.7142...$. All three methods yield the same mean value, approximately $39.71\%$. We round the result to two decimal places.

More from this chapter

• **Example 6.** Find mean of the following data. | Class | 0-15 | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 | |:---: |:---: |:---: |:---: |:---: |:---: |:---: | | **
• Example 2. The mean of the following data is 14. Find the value of k. |c|c|c|c|c|c| x & 5 & 10 & 15 & 20 & 25 f & 7 & k & 8 & 4 & 5
• Find the mean of the following frequency distribution: class intervals 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 5, 8, 15, 16, 6.
• Find the mean of 4, 8, 15, 16, 23, 42
• The marks obtained by 50 students in a test are tabulated below: | Marks: | 0 - 15 | 15 - 30 | 30 - 45 | 45 - 60 | 60 - 75 | |---|---|---|---|---|---| | Number
• A survey regarding the heights (in cm) of 50 girls of class X of a school was conducted and the following data was obtained: | Height (in cm) | Number of girls
• In the following table, the median age of 200 spectators of a football match is 32. Find the missing frequencies p and q. | Age (in years) | Number of Spectator
• The government rescued 100 people after a train accident. Their ages were recorded in the following table. Find their mean age. | Age (in years) | Number of peo

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