The relation between mean, median and mode for a moderately asymmetrical distribution is:
Show Hint
- \(\text{Mode}=3\text{Median}-2\text{Mean}\)
- \(\text{Mode}=3\text{Mean}-2\text{Median}\)
- \(\text{Median}=3\text{Mode}-2\text{Mean}\)
- \(\text{Mean}=3\text{Median}-2\text{Mode}\)
The Correct Option is A
Solution and Explanation
Step 1: Expand the right-hand side. \[ \text{Mean}-\text{Mode} = 3\text{Mean}-3\text{Median}. \]
Step 2: Rearrange the equation. \[ -\text{Mode} = 2\text{Mean}-3\text{Median}. \] Multiplying by \(-1\), \[ \text{Mode} = 3\text{Median}-2\text{Mean}. \] Conclusion: \[ \boxed{\text{Mode}=3\text{Median}-2\text{Mean}} \] Hence, the correct answer is Option (A).
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\[ \begin{array}{|c|l|c|l|} \hline \text{List-I} & & \text{List-II} & \\ \hline (a) & \text{Mean} & (i) & \text{List of individual values} \\ \hline (b) & \text{Median} & (ii) & \text{Value appears most often} \\ \hline (c) & \text{Mode} & (iii) & \text{Average for complete data} \\ \hline (d) & \text{Frequency Distribution} & (iv) & \text{Location average} \\ \hline \end{array} \]
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Statements:
(i) Average number of working hours of employees is 9.42 hours.
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\[ \begin{array}{|c|c|c|c|c|c|} \hline x_i & 2 & 4 & 6 & P & 10 \\ \hline f_i & 3 & 2 & 1 & 4 & 2 \\ \hline \end{array} \]
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