Question

If $\mu $ is the mean of the distribution of $\left( {{Y}_{i}},{{f}_{i}} \right)$ , then $\sum\nolimits_{i}{{{f}_{i}}}\left( {{Y}_{i}}-\mu \right)$ is equal to
a)MD
b)SD
c)0
d)Relative Frequency

Answer 1

Hint: In this question, we are given that $\mu $ is the mean of the distribution$\left( {{Y}_{i}},{{f}_{i}} \right)$ . Thus, we can expand the expression to be evaluated and then use the definition of the mean, $\mu =\sum\limits_{i}{{{Y}_{i}}{{f}_{i}}}$ to obtain the required answer.

Complete step-by-step answer:
In this question, the expression to be evaluated contains the mean of the distribution, $\mu $ . Therefore, we should first note the definition of $\mu $ . The mean of a distribution $\left( {{Y}_{i}},{{f}_{i}} \right)$ where ${{Y}_{i}}$ occurs with a frequency ${{f}_{i}}$ is given by
 $\begin{align}
  & \mu =\dfrac{\sum\limits_{i}{{{Y}_{i}}{{f}_{i}}}}{\sum\limits_{i}{{{f}_{i}}}} \\
 & \Rightarrow \sum\limits_{i}{{{Y}_{i}}{{f}_{i}}}=\mu \sum\limits_{i}{{{f}_{i}}}.....................................(1.1) \\
\end{align}$
The given expression is
$\sum\nolimits_{i}{{{f}_{i}}}\left( {{Y}_{i}}-\mu \right)$
Which can be expanded
\[\sum\nolimits_{i}{{{f}_{i}}}\left( {{Y}_{i}}-\mu \right)=\sum\nolimits_{i}{{{f}_{i}}}{{Y}_{i}}-\sum\nolimits_{i}{{{f}_{i}}}\mu ..............(1.2)\]
However, as $\mu $ is a constant and does not depend on I, it can be taken out in the second term. Also, we can use equation (1.1) in the first term and rewrite equation (1.2) as
\[\sum\nolimits_{i}{{{f}_{i}}}\left( {{Y}_{i}}-\mu \right)=\mu \sum\limits_{i}{{{f}_{i}}}-\mu \sum\nolimits_{i}{{{f}_{i}}}=0..............(1.3)\]
Thus, we obtain the answer as 0 which matches option (c) of the question. Therefore, the answer to this question should be option (c).

Note: We should note that in equation (1.1), we should be careful to divide $\sum\limits_{i}{{{f}_{i}}}$ in the denominator if we are given the distribution in terms of frequencies. However, if the distribution is given as $\left( {{Y}_{i}},{{p}_{i}} \right)$ where ${{p}_{i}}$ is the probability of getting ${{Y}_{i}}$ , then we don't have to divide$\sum\limits_{i}{{{p}_{i}}}$ in (1.1) as the sum of total probabilities is equal to one.