The mode of the following frequency distribution is $46$ and the total frequency is $150$. Find the missing frequencies $x$ and $y$.
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $6$ $8$ $17$ $x$ $42$ $30$ $y$ $8$
| Class | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ |
| Frequency | $6$ | $8$ | $17$ | $x$ | $42$ | $30$ | $y$ | $8$ |
(A) Given total frequency $N = 150$.
Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$
$111 + x + y = 150 \implies x + y = 39$ --- $(1)$
Since the mode is $46$,the modal class is $40-50$.
Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.
Mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} \right) \times 10$
$6 = \left( \frac{42 - x}{84 - 30 - x} \right) \times 10$
$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$
$162 - 3x = 210 - 5x$
$2x = 48 \implies x = 24$
Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.
Thus,$x = 24$ and $y = 15$.
Sum of frequencies: $6 + 8 + 17 + x + 42 + 30 + y + 8 = 150$
$111 + x + y = 150 \implies x + y = 39$ --- $(1)$
Since the mode is $46$,the modal class is $40-50$.
Here,$l = 40, f_1 = 42, f_0 = x, f_2 = 30, h = 10$.
Mode formula: $\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
$46 = 40 + \left( \frac{42 - x}{2(42) - x - 30} \right) \times 10$
$6 = \left( \frac{42 - x}{84 - 30 - x} \right) \times 10$
$6 = \frac{10(42 - x)}{54 - x} \implies 3(54 - x) = 5(42 - x)$
$162 - 3x = 210 - 5x$
$2x = 48 \implies x = 24$
Substituting $x = 24$ in $(1)$: $24 + y = 39 \implies y = 15$.
Thus,$x = 24$ and $y = 15$.
Explore More
Similar Questions
For the calculation of mode of the following frequency distribution,$f_{1} = \dots \dots \dots \dots \dots \dots$
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ Frequency $12$ $18$ $27$ $20$ $17$ $6$
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ |
| Frequency | $12$ | $18$ | $27$ | $20$ | $17$ | $6$ |
Easy
View SolutionIn a frequency distribution,the modal class is $70-85$ with frequency $25$. The frequencies of the classes succeeding and preceding the modal class are $20$ and $8$ respectively. Then,the values of $f_{0}, f_{1}$ and $f_{2}$ are respectively.
Medium
View SolutionFind the mode of the following frequency distribution:
Class $0-15$ $15-30$ $30-45$ $45-60$ $60-75$ $75-90$ $90-105$ Frequency $4$ $5$ $23$ $45$ $66$ $42$ $15$
| Class | $0-15$ | $15-30$ | $30-45$ | $45-60$ | $60-75$ | $75-90$ | $90-105$ |
| Frequency | $4$ | $5$ | $23$ | $45$ | $66$ | $42$ | $15$ |
Medium
View SolutionThe modal class of a frequency distribution is $20-24$. Then,in the calculations of mode,$l=$ ..........
Easy
View SolutionFind the mode of the following frequency distribution:
Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ Frequency $7$ $21$ $37$ $13$ $12$ $10$
| Class | $0-100$ | $100-200$ | $200-300$ | $300-400$ | $400-500$ | $500-600$ |
|---|---|---|---|---|---|---|
| Frequency | $7$ | $21$ | $37$ | $13$ | $12$ | $10$ |
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo