Find the mode of the following frequency dist | vedclass

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(B) The modal class is the class with the highest frequency. Here,the highest frequency is $37$,which corresponds to the class interval $200-300$.Thus

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$240$

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(B) The modal class is the class with the highest frequency. Here,the highest frequency is $37$,which corresponds to the class interval $200-300$.Thus,the modal class is $200-300$.We use the formula for mode: $Z = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$Where:$l$ (lower limit of modal class) $= 200$$h$ (class size) $= 100$$f_1$ (frequency of modal class) $= 37$$f_0$ (frequency of the class preceding the modal class) $= 21$$f_2$ (frequency of the class succeeding the modal class) $= 13$Substituting these values into the formula:$Z = 200 + \left( \frac{37 - 21}{2(37) - 21 - 13} ight) 	imes 100$$Z = 200 + \left( \frac{16}{74 - 34} ight) 	imes 100$$Z = 200 + \left( \frac{16}{40} ight) 	imes 100$$Z = 200 + 0.4 	imes 100$$Z = 200 + 40 = 240$Therefore,the mode of the given frequency distribution is $240$.

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$3$	$x$	$7$	$7$	$y$

Age (in years)	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Number of patients	$60$	$42$	$55$	$70$	$53$	$20$

Find the mode of the following frequency distribution:

Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$

Frequency $7$ $21$ $37$ $13$ $12$ $10$

Similar Questions

The mean of the following data is $55.5$ and the total frequency is $20$. Find the missing frequencies $x$ and $y$.
Class $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $3$ $x$ $7$ $7$ $y$

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For a given frequency distribution,$n=100, A=20$ and $\bar{x}=20$. Then,$\Sigma f_{i} d_{i}=\ldots \ldots \ldots . .$

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Find the mode of the following frequency distribution: Class $0-100$ $100-200$ $200-300$ $300-400$ $400-500$ $500-600$ Frequency $7$ $21$ $37$ $13$ $12$ $10$