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	$5$	$8$	$5$	$12$	$X$	$Y$

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

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 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: