The frequency distribution of the age of stud | vedclass

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(B) Given total number of students $N = 40$,so $5 + 8 + 5 + 12 + X + Y = 40 \Rightarrow X + Y = 10 \dots (1)$.The cumulative frequencies are: $5, 13,

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$44$

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(B) Given total number of students $N = 40$,so $5 + 8 + 5 + 12 + X + Y = 40 \Rightarrow X + Y = 10 \dots (1)$.The cumulative frequencies are: $5, 13, 18, 30, 30+X, 30+X+Y$.Since $N=40$,the median is the average of the $20^{th}$ and $21^{st}$ observations. Looking at the cumulative frequencies,the $20^{th}$ and $21^{st}$ observations fall in the age group $18$. Thus,$	ext{Median} (M) = 18$.Mean deviation about median is given by $	ext{M.D.} = \frac{\sum f_i |x_i - M|}{N}$.Given $	ext{M.D.} = 1.25$,so $1.25 = \frac{5|15-18| + 8|16-18| + 5|17-18| + 12|18-18| + X|19-18| + Y|20-18|}{40}$.$1.25 = \frac{5(3) + 8(2) + 5(1) + 12(0) + X(1) + Y(2)}{40}$.$50 = 15 + 16 + 5 + 0 + X + 2Y$.$50 = 36 + X + 2Y \Rightarrow X + 2Y = 14 \dots (2)$.Subtracting $(1)$ from $(2)$:$(X + 2Y) - (X + Y) = 14 - 10 \Rightarrow Y = 4$.Substituting $Y=4$ in $(1)$,$X + 4 = 10 \Rightarrow X = 6$.Therefore,$4X + 5Y = 4(6) + 5(4) = 24 + 20 = 44$.

Age	$15$	$16$	$17$	$18$	$19$	$20$
No. of students	$5$	$8$	$5$	$12$	$X$	$Y$

Class interval	Frequency
$0-2$	$1$
$2-4$	$3$
$4-6$	$5$
$6-8$	$3$
$8-10$	$1$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq	$6$	$7$	$15$	$16$	$4$	$2$

Class Interval	$0$–$2$	$2$–$4$	$4$–$6$	$6$–$8$	$8$–$10$
Frequency	$1$	$3$	$4$	$1$	$2$

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
No of Girls	$6$	$8$	$14$	$16$	$4$	$2$

The frequency distribution of the age of students in a class of $40$ students is given below.

Age $15$ $16$ $17$ $18$ $19$ $20$

No. of students $5$ $8$ $5$ $12$ $X$ $Y$

If the mean deviation about the median is $1.25$,then $4X + 5Y$ is equal to:

Similar Questions

The mean deviation from the median for the following data is
$x_i$ $2$ $3$ $5$ $7$ $9$
$f_i$ $8$ $6$ $4$ $2$ $1$

The mean deviation about the mean for the following data is
Class interval Frequency
$0-2$ $1$
$2-4$ $3$
$4-6$ $5$
$6-8$ $3$
$8-10$ $1$

Calculate the mean deviation about the median for the following data:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Freq $6$ $7$ $15$ $16$ $4$ $2$

The mean deviation about the mean for the following data is
Class Interval $0$–$2$ $2$–$4$ $4$–$6$ $6$–$8$ $8$–$10$
Frequency $1$ $3$ $4$ $1$ $2$

Find the mean deviation about the median for the following data:

Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$

No of Girls $6$ $8$ $14$ $16$ $4$ $2$

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The frequency distribution of the age of students in a class of $40$ students is given below. Age $15$ $16$ $17$ $18$ $19$ $20$ No. of students $5$ $8$ $5$ $12$ $X$ $Y$ If the mean deviation about the median is $1.25$,then $4X + 5Y$ is equal to:

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The mean deviation from the median for the following data is $x_i$$2$$3$$5$$7$$9$$f_i$$8$$6$$4$$2$$1$

The mean deviation about the mean for the following data isClass intervalFrequency$0-2$$1$$2-4$$3$$4-6$$5$$6-8$$3$$8-10$$1$

Calculate the mean deviation about the median for the following data:Class$0-10$$10-20$$20-30$$30-40$$40-50$$50-60$Freq$6$$7$$15$$16$$4$$2$

The mean deviation about the mean for the following data is Class Interval$0$–$2$$2$–$4$$4$–$6$$6$–$8$$8$–$10$Frequency$1$$3$$4$$1$$2$

Find the mean deviation about the median for the following data: Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ No of Girls $6$ $8$ $14$ $16$ $4$ $2$