Question

The following distribution shows the weekly pocket allowance of 64 children of a locality.If the mean pocket allowance is ₹ 180, find the values of x and y.

Pocket allowance (in ₹):	110-130	130-150	150-170	170-190	190-210	210-230	230-250
Number of children:	7	6	9	13	x	5	y

Answer 1

Given

\[ N = 7 + 6 + 9 + 13 + x + 5 + y = 40 + x + y = 64 \Rightarrow x + y = 24 \] ...(i)

Class	`(x_i)`	`(f_i)`	`(u_i = \frac{x_i - 180}{20})`	`(f_i u_i)`
110-130	120	7	-3	-21
130-150	140	6	-2	-12
150-170	160	9	-1	-9
170-190	180	13	0	0
190-210	200	(x)	1	(x)
210-230	220	5	2	10
230-250	240	(y)	3	(3y)
		`N = 64`		`\sum f_i u_i = x + 3y - 32`

Formula & Substitution

\[ 180 = 180 + 20 \times \frac{x + 3y - 32}{64} \]

\[ 0 = x + 3y - 32 \Rightarrow x + 3y = 32 \] ...(ii)

Solving (i) and (ii):

\[ x + y = 24,\quad x + 3y = 32 \]

\[ 2y = 8 \Rightarrow y = 4,\quad x = 20 \]

Final answer: (x = 20), (y = 4)