The mode is 37.5 and the total frequency is 169. Find [tex]\( x \)[/tex] and [tex]\( y \)[/tex].

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
C.I & [tex]$0-10$[/tex] & [tex]$10-20$[/tex] & [tex]$20-30$[/tex] & [tex]$30-40$[/tex] & [tex]$40-50$[/tex] & [tex]$50-60$[/tex] & [tex]$60-70$[/tex] & [tex]$70-80$[/tex] \\
\hline
[tex]$F$[/tex] & 5 & 18 & [tex]$x$[/tex] & 45 & 40 & [tex]$y$[/tex] & 10 & 6 \\
\hline
\end{tabular}



Answer :

Let's go through the problem step-by-step:

1. Identify the modal class:
The mode (37.5) lies in the class interval 30-40.

Therefore, the modal class is 30-40.

2. Determine the necessary elements for the mode formula:
- [tex]\( L \)[/tex]: Lower limit of the modal class = 30
- [tex]\( f_1 \)[/tex]: Frequency of the modal class = 45
- [tex]\( f_0 \)[/tex]: Frequency of the class before the modal class = [tex]\( X \)[/tex]
- [tex]\( f_2 \)[/tex]: Frequency of the class after the modal class = 40
- [tex]\( h \)[/tex]: Class width (40 - 30) = 10

3. Using the mode formula:
The mode formula is given by:

[tex]\[ \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h \][/tex]

Substitute the known values into the formula:

[tex]\[ 37.5 = 30 + \left( \frac{45 - X}{2(45) - X - 40} \right) \times 10 \][/tex]

4. Simplify and solve for [tex]\(X\)[/tex]:
[tex]\[ 37.5 = 30 + \left( \frac{45 - X}{90 - X - 40} \right) \times 10 \][/tex]

[tex]\[ 37.5 = 30 + \left( \frac{45 - X}{50 - X} \right) \times 10 \][/tex]

Subtract 30 from both sides:
[tex]\[ 7.5 = \left( \frac{45 - X}{50 - X} \right) \times 10 \][/tex]

Divide both sides by 10:
[tex]\[ 0.75 = \frac{45 - X}{50 - X} \][/tex]

Cross-multiply to solve for [tex]\( X \)[/tex]:
[tex]\[ 0.75(50 - X) = 45 - X \][/tex]

[tex]\[ 37.5 - 0.75X = 45 - X \][/tex]

Move [tex]\( X \)[/tex] terms to one side and constant terms to the other:
[tex]\[ 45 - 37.5 = 0.25X \][/tex]

[tex]\[ 7.5 = 0.25X \][/tex]

[tex]\[ X = \frac{7.5}{0.25} = 30 \][/tex]

5. Solve for [tex]\( Y \)[/tex] using the total frequency:
The total frequency is given as 169.

The frequencies sum equation is:
[tex]\[ 5 + 18 + X + 45 + 40 + Y + 10 + 6 = 169 \][/tex]

Substitute [tex]\( X = 30 \)[/tex] into the equation:
[tex]\[ 5 + 18 + 30 + 45 + 40 + Y + 10 + 6 = 169 \][/tex]

Sum the known values:
[tex]\[ 154 + Y = 169 \][/tex]

Solve for [tex]\( Y \)[/tex]:
[tex]\[ Y = 169 - 154 = 15 \][/tex]

6. Summarize the results:
[tex]\[ X = 30 \quad \text{and} \quad Y = 15 \][/tex]

Therefore, the values of [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] are 30 and 15, respectively.