question_answer

In a class test, marks obtained by 120 students are given in the following frequency distribution. If it is given that mean is 59, find the missing frequencies x and y.
Marks	No. of students
0 ? 10	1
10 ? 20	3
20 ? 30	7
30 ? 40	10
40 ? 50	15
50 ? 60	x
60 ? 70	9
70 ? 80	27
80 ? 90	18
90 ? 100	y

Answer:

           

Marks	No. of Students \[{{f}_{i}}\]	\[{{X}_{i}}\]	\[{{d}_{i}}=\frac{{{X}_{i}}-55}{10}\]	\[{{f}_{i}}{{d}_{i}}\]
0 ? 10	1	5	\[5\]	\[5\]
10 ? 20	3	15	\[4\]	\[-12\]
20 ? 30	7	25	\[3\]	\[-21\]
30 ? 40	10	35	\[2\]	\[-20\]
40 ? 50	15	45	\[1\]	\[-15\]
50 ? 60	x	A = 55	0	0
60 ? 70	9	65	1	9
70 ? 80	27	75	2	54
80 ? 90	18	85	3	54
90 ? 100	y	95	4	4y
	\[\sum{{{f}_{i}}=90+x+y}\]			\[\sum{{{f}_{i}}{{d}_{i}}=-73+117+4y=44+4y}\]
\[\sum{{{f}_{i}}=90+x+y}\]
But       \[\sum{{{f}_{i}}=120}\]                         [Given]
\[\therefore 90+x+y=120\]
\[x=120-90-y=30-y\]                                  ...(i)
Mean \[=A+\frac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}\times h\]
\[\Rightarrow 59=55+\left( \frac{44+4y}{120}\times 10 \right)\]                   \[\left[ A=55,h=10,\sum{{{f}_{i}}=120} \right]\]
\[\Rightarrow 59-55=\frac{4(11+y)}{12}\]
\[\Rightarrow 4\times 3=11+y\]
\[\Rightarrow y=12-11=1\]
From eq. (i), \[x=30-1=29\]
\[\therefore x=29,y=1\]