Question

find the missing frequencies if N=100 and median is 32.

marks obtained-0-10, 10-20, 20-30, 30-40, 40-50, 50-60, total

no. of students-10 ,? ,25 ,30 ,? ,10 ,100.

Answer 1

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Answer 2

Missing frequencies are x=9 and y= 16.

Given:

• Marks obtained:0-10, 10-20, 20-30, 30-40, 40-50, 50-60
• No. of students-10 ,x ,25 ,30 ,y ,10 ,100.
• Median is 32, total frequency N:100.

To find:

• Find the missing frequencies.

Solution:

Formula to be used:

$\bf Median = l + \left( \frac{ \frac{n}{2} - cf }{f} \right)h$

here,

l: Lower limit of median class.

cf: Cumulative frequency of preceding class.

f: Frequency of median class

h: height of class

Step 1:

Construct cumulative frequency table.

Let the missing frequencies are x and y respectively.

$\begin{tabular}{|c|c|c|}\cline{1-3}Marks\: Obtained&Freq&CF\\\cline{1-3}0-10&10&10\\\cline{1-3}10-20&x&10+x\\\cline{1-3}20-30&25&\bf 35+x\\\cline{1-3}\bf 30-40&\bf 30&65+x\\\cline{1-3}40-50&y&65+x+y\\\cline{1-3}50-60&10&75+x+y\\\cline{1-3}Total&100&\\\cline{1-3}\end{tabular}$

Step 2:

Find the values used in formula.

$\frac{n}{2} = 50$

Median class is 30-40( Upper nearest to 50)

CF=35+x

f=30

h=10

Median=32

Put all values in the formula.

$32 = 30 + \left( \frac{50- 35 - x }{30} \right)10$

or

$32 - 30 = \frac{ 15 - x}{3}$

or

$6 - 15 = - x$

or

$- x = - 9$

or

$\bf x = 9$

Step 3:

Find value of y.

As total frequency is 100.

$75 + x + y = 100$

or

$y = 100 - 75 - 9$

or

$\bf y = 16$

Thus,

Unknown frequencies are x=9 and y= 16.

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