The daily income of a sample of 50 employees | vedclass

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(D) Since the given data is not continuous,we first convert it into continuous classes by subtracting $0.5$ from the lower limit and adding $0.5$ to t

Vedclass · Accepted Answer

$356$

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(D) Since the given data is not continuous,we first convert it into continuous classes by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class.Now,we calculate the class mark $(x_i)$ for each class and use the step-deviation method:						Income (in $Rs.$)			Class mark $(x_i)$			Frequency $(f_i)$			$u_i = \frac{x_i - a}{h}$			$f_i u_i$							$0.5-200.5$			$100.5$			$14$			$-1$			$-14$							$200.5-400.5$			$300.5 (a)$			$15$			$0$			$0$							$400.5-600.5$			$500.5$			$14$			$1$			$14$							$600.5-800.5$			$700.5$			$7$			$2$			$14$							Total			-			$\Sigma f_i = 50$			-			$\Sigma f_i u_i = 14$			Here,assumed mean $a = 300.5$,class width $h = 200$,and total observations $N = 50$.Using the step-deviation formula:Mean $(\bar{x}) = a + h 	imes \left( \frac{\Sigma f_i u_i}{\Sigma f_i} ight)$$\bar{x} = 300.5 + 200 	imes \left( \frac{14}{50} ight)$$\bar{x} = 300.5 + 4 	imes 14$$\bar{x} = 300.5 + 56 = 356.5$Thus,the mean daily income is $Rs. 356.5$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$10$	$12$	$13$	$16$	$9$

The daily income of a sample of $50$ employees is tabulated as follows:

Income (in $Rs.$) $1-200$ $201-400$ $401-600$ $601-800$

Number of employees $14$ $15$ $14$ $7$

Find the mean daily income of the employees. (in $.5$)

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Income (in $Rs.$)	$1-200$	$201-400$	$401-600$	$601-800$
Number of employees	$14$	$15$	$14$	$7$

The daily income of a sample of $50$ employees is tabulated as follows: Income (in $Rs.$) $1-200$ $201-400$ $401-600$ $601-800$ Number of employees $14$ $15$ $14$ $7$ Find the mean daily income of the employees. (in $.5$)