We are tasked with finding the arithmetic mean using the step deviation method. Let's break down the solution step by step.

Step 1: Define midpoints ( $x_i$ )

For each class interval, we find the midpoint ( $x_i$ ) as follows:

$x_i = \frac{\text{lower limit} + \text{upper limit}}{2}$

• Class 0-10: $x_1 = \frac{0 + 10}{2} = 5$
• Class 10-20: $x_2 = \frac{10 + 20}{2} = 15$
• Class 20-30: $x_3 = \frac{20 + 30}{2} = 25$
• Class 30-40: $x_4 = \frac{30 + 40}{2} = 35$
• Class 40-50: $x_5 = \frac{40 + 50}{2} = 45$

Step 2: Choose an assumed mean ( $A$ )

Select one of the midpoints as an assumed mean. Let’s choose $A = 25$ (from the class 20-30).

Step 3: Calculate the deviation ( $d_i$ )

The deviation for each class is calculated as:

$d_i = x_i - A$

So:

• $d_1 = 5 - 25 = -20$
• $d_2 = 15 - 25 = -10$
• $d_3 = 25 - 25 = 0$
• $d_4 = 35 - 25 = 10$
• $d_5 = 45 - 25 = 20$

Step 4: Calculate $u_i$ using step size ( $h$ )

Let the class width or step size $h = 10$ (which is the common difference between the class intervals).

$u_i = \frac{d_i}{h}$

Thus:

• $u_1 = \frac{-20}{10} = -2$
• $u_2 = \frac{-10}{10} = -1$
• $u_3 = \frac{0}{10} = 0$
• $u_4 = \frac{10}{10} = 1$
• $u_5 = \frac{20}{10} = 2$

Step 5: Multiply frequency ( $f_i$ ) by $u_i$

Now, we calculate $f_i u_i$ for each class:

• $f_1 u_1 = 7 \times (-2) = -14$
• $f_2 u_2 = 8 \times (-1) = -8$
• $f_3 u_3 = 20 \times 0 = 0$
• $f_4 u_4 = 10 \times 1 = 10$
• $f_5 u_5 = 5 \times 2 = 10$

Step 6: Calculate the total

Now, sum the frequencies and the $f_i u_i$ values:

$\sum f_i = 7 + 8 + 20 + 10 + 5 = 50$ $\sum f_i u_i = -14 + (-8) + 0 + 10 + 10 = -2$

Step 7: Apply the step deviation formula for the mean

The formula for the arithmetic mean using the step deviation method is:

$\text{Mean} = A + h \times \frac{\sum f_i u_i}{\sum f_i}$

Substitute the known values:

$\text{Mean} = 25 + 10 \times \frac{-2}{50} = 25 - 0.4 = 24.6$

Final Answer:

The arithmetic mean of the data is $24.6$.

Would you like more details or have any questions on this method? Here are 5 related questions to explore further:

1. How is the assumed mean chosen in the step deviation method?
2. Why is it necessary to calculate the midpoint for each class interval?
3. What role does the step size ( $h$ ) play in simplifying calculations?
4. How does this method differ from the direct method for calculating the mean?
5. Can this method be applied to ungrouped data, and if so, how?

Tip: The step deviation method is especially useful when the data involves large numbers or intervals, as it simplifies the calculation of the arithmetic mean.