Question

The median value of the following data is $49.23$ and the following table represents the percentage of marks in the second term examination and number of students acquired the respective percentage. Calculate the value of x and y. The total frequency is $100$

Percentage	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
No.of students	$5$	$12$	x	$26$	y	$18$	$12$

• A. $x=9,y=13$
• B. $x=12,y=15$
• C. $x=9,y=18$
• D. $x=18,y=19$

Answer 1

The correct option is C $x=9,y=18$

b'Finding the cumulative frequency table

Percentage	Number of students	Cumulative frequency
$10-20$	$5$	$5$
$20-30$	$12$	$17$
$30-40$	x	$17+x$
$40-50$	$26$	$43+x$
$50-60$	y	$43+x+y$
$60-70$	$18$	$61+x+y$
$70-80$	$12$	$73+x+y$
	$73+x+y$

$n=100$
$73+x+y=100$
$x+y=100-73$
$x+y=27$
The median is $49.23$ which lies in the class $40-50$
Mode$=l+displaystylefracfracn2-cfftimesh$

Where l is the lower limit of the median
class$=40$

n is the number of observation$=100$

cf is the cumulative frequency of class
preceding the median class$=17+x$

f is the frequency of the median class$=26$

h is the class size (assuming class size to be equal)$=10$

Median$=l+displaystylefracfracn2-cfftimesh$

$49.23=40+left\left(displaystylefrac50-(17+x)26right\right)times10$

$49.23=40+left\left(displaystylefrac50-17-x26right\right)times10$

$49.23=40+displaystylefrac500-170-10x26$

$49.23-40=displaystylefrac500-170-10x26$

$9.23\ast 26=330-10x$

$239.98=330-10x$

$10x=33-239.98$

$10x=90.02$

$X=90.02/10$

$X=9.02$

Hence $x=9.02$, Applying this value in $73+x+y=100$

$73+9.02+y=100$

$82.02+y=100$

$Y=100-82.02$

$Y=17.98$

Rounding the value of x and y, the value
of x is $9$ and the value of y is $18$.

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