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Marks0 - 55 -1010 - 1515 - 2020 - 2525 - 3030 - 3535 - 4040 - 4545 - 50
Frequency3578101114191513

For the following distribution, find the mean using the step deviation method. (Round your answer to the nearest whole number)
(A) 29
(B) 31
(C) 35
(D) 37

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Answer
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Hint: In the given question, we have to find the mean. Thus, we will use the step deviation method to get the answer. To find the mean, we will use the formula, $a+\dfrac{\Sigma {{f}_{i}}{{d}_{i}}}{\Sigma f}\times h$, where a is the assumed mean, ${{d}_{i}}=\dfrac{\left( {{x}_{i}}-a \right)}{h}$, h is the height and f is the frequency. Thus, we will find the mid-value of the given marks and then substitute all these values in the given formula to get the required solution for the problem.

Complete step by step answer:
According to the problem, we have to find the mean using the step-deviation method. So, we will first find the mid-value of the given interval, which is equal to, $\dfrac{\text{upper limit + lower limit}}{2}$, thus we get,
MarksFrequency $\left( {{f}_{i}} \right)$Mid-value $\left( {{x}_{i}} \right)$
0 - 532.5
5 - 1057.5
10 - 15712.5
15 - 20817.5
20 - 251022.5
25 - 301127.5
30 - 351432.5
35 - 401937.5
40 - 451542.5
45 - 501347.5


So, let us assume that the mid-value $=a=22.5$. Therefore, we will now calculate the value of ${{d}_{i}}$, which is equal to $\dfrac{{{x}_{i}}-a}{h}$, where h is the height.
Thus, h = 5, therefore, we get,
MarksFrequency $\left( {{f}_{i}} \right)$Mid-value $\left( {{x}_{i}} \right)$${{d}_{i}}=\dfrac{{{x}_{i}}-22.5}{5}$
0 - 532.5-4
5 - 1057.5-3
10 - 15712.5-2
15 - 20817.5-1
20 - 251022.50
25 - 301127.51
30 - 351432.52
35 - 401937.53
40 - 451542.54
45 - 501347.55


Now we will find the summation of frequencies and deviation of the given mean data, thus we get,
MarksFrequency $\left( {{f}_{i}} \right)$Mid-value $\left( {{x}_{i}} \right)$${{d}_{i}}=\dfrac{{{x}_{i}}-22.5}{5}$${{f}_{i}}{{d}_{i}}$
0 - 532.5-4-12
5 - 1057.5-3-15
10 - 15712.5-2-14
15 - 20817.5-1-8
20 - 251022.500
25 - 301127.5111
30 - 351432.5228
35 - 401937.5357
40 - 451542.5460
45 - 501347.5565
105172


Now we have,
$\begin{align}
  & \Sigma {{f}_{i}}{{d}_{i}}=172\ldots \ldots \ldots \left( 1 \right) \\
 & \Sigma {{f}_{i}}=105\ldots \ldots \ldots \left( 2 \right) \\
 & h=5\ldots \ldots \ldots \left( 3 \right) \\
 & a=22.5\ldots \ldots \ldots \left( 4 \right) \\
\end{align}$
Now, mean for step=deviation formula is equal to, mean = $a+\dfrac{\Sigma {{f}_{i}}{{d}_{i}}}{\Sigma f}\times h$. So, we will put equation (1), (2), (3) and (4) in the formula and we get mean as,
 $\begin{align}
  & =22.5+\dfrac{172}{105}\times 5 \\
 & \Rightarrow 22.5+\dfrac{172}{21} \\
\end{align}$
On further simplification, we get,
Mean $=22.5+8.19$
$\Rightarrow $ Mean $=30.69\approx 31$
Therefore, for the given distribution, the mean using step-deviation method is equal to 31 approximately.
 So, the correct answer is “Option B”.

Note: While solving this problem, do mention all the steps properly to avoid error and confusion. Do mention the formulas accurately. Do not forget that ${{d}_{i}}$ is calculated by subtracting ${{x}_{i}}$ and assumed mean, divided by the width of the class interval.