Question

Find \[X\] and \[Y\] if the mean is given by 261.

CI	100-150	150-200	200-250	250-300	300-350	Total
\[{f_i}\]	4	\[X\]	12	\[Y\]	2	25

Answer 1

Hint:
We will find \[X\] and \[Y\] by solving 2 linear equations. We will get the first equation by equation sum of all frequencies with 25. We will get the second equation by finding the mean and equating it to 261.
Formulas used: 1. \[{\text{mean}} = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\] where \[{f_i}\] is the frequency of the \[{i^{th}}\] class and \[{x_i}\] id the median of the \[{i^{th}}\] class.
2. Median of a class\[\left( {{x_i}} \right) = \dfrac{{L + U}}{2}\] where \[L\] is the lower limit of the class and \[U\] is the upper limit of the class.

Complete step by step solution:
First, let’s create a table and find the median of each class by using the second formula.

Class Interval	\[{f_i}\]	\[{x_i}\]
100-150	4	\[\dfrac{{100 + 150}}{2} = 125\]
150-200	\[X\]	\[\dfrac{{150 + 200}}{2} = 175\]
200-250	12	\[\dfrac{{200 + 250}}{2} = 225\]
250-300	\[Y\]	\[\dfrac{{250 + 300}}{2} = 275\]
300-350	2	\[\dfrac{{300 + 350}}{2} = 325\]
Total	25

Let’s multiply the $2^{\text{nd}}$ and $3^{\text{rd}}$ column and find \[{f_i}{x_i}\] for each class interval.

Class Interval	\[{f_i}\]	\[{x_i}\]	\[{f_i}{x_i}\]
100-150	4	125	\[4 \cdot 125 = 500\]
150-200	\[X\]	175	\[X \cdot 175 = 175X\]
200-250	12	225	\[12 \cdot 225 = 2700\]
250-300	\[Y\]	275	\[Y \cdot 275 = 275Y\]
300-350	2	325	\[2 \cdot 325 = 650\]
Total	25

Let’s add all the elements of the third column and write the sum in the last row.

Class Interval	\[{f_i}\]	\[{x_i}\]	\[{f_i}{x_i}\]
100-150	4	125	500
150-200	\[X\]	175	\[175X\]
200-250	12	225	2700
250-300	\[Y\]	275	\[275Y\]
300-350	2	325	650
Total	25		\[3850 + 175X + 275Y\]

Let’s find the sum of all frequencies and equate the sum to 25.
\[\begin{array}{l}4 + X + 12 + Y + 2 = 25\\ \Rightarrow 18 + X + Y = 25\end{array}\]
Let’s subtract \[18 + Y\] from both sides of the equation.
\[\begin{array}{l}18 + X + Y - \left( {18 + Y} \right) = 25 - \left( {18 + Y} \right)\\ \Rightarrow 18 + X + Y - 18 - Y = 25 - 18 - Y\\ \Rightarrow X = 7 - Y{\text{ }}\left( 1 \right)\end{array}\]
Let’s substitute 25 for \[\sum {{f_i}} \], \[3850 + 175X + 275Y\]for \[\sum {{f_i}{x_i}} \] and 261 for mean in the 1st formula.
\[\begin{array}{l}261 = \dfrac{{3850 + 175X + 275Y}}{{25}}\\ \Rightarrow 261 = 154 + 7X + 11Y\end{array}\]
Let’s subtract 154 from both sides.
\[\begin{array}{l}261 - 154 = 154 + 7X + 11Y - 154\\ \Rightarrow 107 = 7X + 11Y\end{array}\]
To find \[X\] and \[Y\], let’s substitute \[7 - Y\] for \[X\] in the above equation.
\[\begin{array}{l}107 = 7\left( {7 - Y} \right) + 11Y\\ \Rightarrow 107 = 49 - 7Y + 11Y\end{array}\]
Let’s subtract 49 from both sides of the equation.
\[\begin{array}{l}107 - 49 = 49 - 7Y + 11Y - 49\\ \Rightarrow 58 = 4Y\\ \Rightarrow \dfrac{{58}}{4} = Y\\ \Rightarrow \dfrac{{29}}{2} = Y\end{array}\]
Substitute \[\dfrac{{29}}{2}\] for \[Y\] in the 1st equation.
\[\begin{array}{l}X = 7 - \dfrac{{29}}{2}\\ \Rightarrow X = - 7.5\end{array}\]
We will substitute \[ - 7.5\] for \[X\] in equation (1).
\[\begin{array}{l}Y = 7 - \left( { - 7.5} \right)\\ \Rightarrow Y = 7 + 7.5\\ \Rightarrow Y = 14.5\end{array}\]
The given question is incorrect as the frequency can never be negative.

Note:
Frequency of an entry indicates the number of times that entry has been made. Therefore, the frequency can never be negative. It’s obvious that we should avoid calculation mistakes in such questions because a minor mistake of sign or value can get you a wrong answer.