Question

The mean of the following distribution is 50. Find the value of a.

X	10	30	50	70	90
Y	17	$5a + 3$	32	$7a - 11$	18

Answer 1

Hint: Consider x as the observations and y as the frequency and use the formula of finding mean with the help of direct mean method given by : $\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}$. Here, $\overline{x}$ is the notation for mean, $\sum{{{x}_{i}}{{f}_{i}}}$ is the summation of product of observation and its corresponding frequency and $\sum{{{f}_{i}}}$ is the summation of all the frequencies. Substitute all the given values to find the value of a.

Complete step-by-step solution:
Here, we have been provided with the following distribution table:

X	10	30	50	70	90
Y	17	$5a + 3$	32	$7a - 11$	18

So, we can see that x are the observations and y are their corresponding frequencies. Now, we know that in such a type of distribution we have to apply a direct method to calculate the mean. So, applying the formula for mean, we get,
$\overline{x}=\dfrac{\sum{{{x}_{i}}{{f}_{i}}}}{\sum{{{f}_{i}}}}$
Here, $\overline{x}$ is the mean od the data, $\sum{{{x}_{i}}{{f}_{i}}}$ is the product of observations and frequencies and $\sum{{{f}_{i}}}$ is the sum of all the frequencies.
Now, here frequency is denoted with y, so we have ${{f}_{i}}={{y}_{i}}$.
$\begin{align}
  & \Rightarrow \overline{x}=\dfrac{\sum{{{x}_{i}}{{y}_{i}}}}{\sum{{{y}_{i}}}} \\
&\Rightarrow\overline{x}=\dfrac{{{x}_{1}}{{y}_{1}}+{{x}_{2}}{{y}_{2}}+{{x}_{3}}{{y}_{3}}+{{x}_{4}}{{y}_{4}}+{{x}_{5}}{{y}_{5}}}{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}+{{y}_{5}}} \\
\end{align}$
Now, substituting the value of $\overline{x}=50$ and the values of x and y from the table, we get,
$\begin{align}
& \Rightarrow 50=\dfrac{10\times 17+30\times \left( 5a+3 \right)+50\times 32+70\times \left( 7a-11 \right)+90\times 18}{17+5a+3+32+7a-11+18} \\
 & \Rightarrow 50=\dfrac{170+150a+90+1600+490a-770+1620}{59+12a} \\
 & \Rightarrow 50=\dfrac{2710+640a}{59+12a} \\
\end{align}$
Dividing both sides by 10, we get,
$\Rightarrow 5=\dfrac{271+64a}{59+12a}$
By cross-multiplication, we get,
$\begin{align}
  & \Rightarrow 5\times \left( 59+12a \right)=271+64a \\
 & \Rightarrow 295+60a=271+64a \\
 & \Rightarrow 295-271=64a-60a \\
 & \Rightarrow 24=4a \\
 & \Rightarrow a=6 \\
\end{align}$
Hence, the value of a is 6.

Note: One may note that in the above question, we cannot consider x as the frequency and y as the observation. Note that the given table is non-grouped data, that’s why the assumed mean method is not applied and the direct mean method is applied. The assumed mean method is generally used for grouped data. In grouped data observations (x) are given in intervals.