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,
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,
Now we will find the summation of frequencies and deviation of the given mean data, thus we get,
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.
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,
Marks | Frequency $\left( {{f}_{i}} \right)$ | Mid-value $\left( {{x}_{i}} \right)$ |
0 - 5 | 3 | 2.5 |
5 - 10 | 5 | 7.5 |
10 - 15 | 7 | 12.5 |
15 - 20 | 8 | 17.5 |
20 - 25 | 10 | 22.5 |
25 - 30 | 11 | 27.5 |
30 - 35 | 14 | 32.5 |
35 - 40 | 19 | 37.5 |
40 - 45 | 15 | 42.5 |
45 - 50 | 13 | 47.5 |
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,
Marks | Frequency $\left( {{f}_{i}} \right)$ | Mid-value $\left( {{x}_{i}} \right)$ | ${{d}_{i}}=\dfrac{{{x}_{i}}-22.5}{5}$ |
0 - 5 | 3 | 2.5 | -4 |
5 - 10 | 5 | 7.5 | -3 |
10 - 15 | 7 | 12.5 | -2 |
15 - 20 | 8 | 17.5 | -1 |
20 - 25 | 10 | 22.5 | 0 |
25 - 30 | 11 | 27.5 | 1 |
30 - 35 | 14 | 32.5 | 2 |
35 - 40 | 19 | 37.5 | 3 |
40 - 45 | 15 | 42.5 | 4 |
45 - 50 | 13 | 47.5 | 5 |
Now we will find the summation of frequencies and deviation of the given mean data, thus we get,
Marks | Frequency $\left( {{f}_{i}} \right)$ | Mid-value $\left( {{x}_{i}} \right)$ | ${{d}_{i}}=\dfrac{{{x}_{i}}-22.5}{5}$ | ${{f}_{i}}{{d}_{i}}$ |
0 - 5 | 3 | 2.5 | -4 | -12 |
5 - 10 | 5 | 7.5 | -3 | -15 |
10 - 15 | 7 | 12.5 | -2 | -14 |
15 - 20 | 8 | 17.5 | -1 | -8 |
20 - 25 | 10 | 22.5 | 0 | 0 |
25 - 30 | 11 | 27.5 | 1 | 11 |
30 - 35 | 14 | 32.5 | 2 | 28 |
35 - 40 | 19 | 37.5 | 3 | 57 |
40 - 45 | 15 | 42.5 | 4 | 60 |
45 - 50 | 13 | 47.5 | 5 | 65 |
105 | 172 |
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.
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