Question

if the median of the following distribution is 46, find the missing frequency p and q
CLASS INTERVAL FREQUENCY
10-20 12
20-30 30
30-40 p
40-50 65
50-60 q
60-70 25
70-80 18

total - 230

Answer 1

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Answer 2

Answer:

34 and 46 are the required value of p and q

Step-by-step explanation:

Explanation:

Given , Class interval : 10-20,20-30,30-40,40-50,50-60,60-70,70-80

Frequency : 12 , 30 , p ,65 ,q ,25 ,18

Sum of total frequency = 230

and median = 46

Formula of median = $l + (\frac{\frac{n}{2} -c.f}{f} )h$

Step1:

$\sum{f}$ = 150 +p+q

but sum of total frequency is 230 given

so , 150+p+q = 230

⇒p + q = 80  ............(i)

where n = 230

therefore $\frac{n}{2} = \frac{230}{2} = 115$

Here we have , median is 46 , which clearly lies in the class interval 40-50.Therefore  c.f = 42+p

l = 40 (lower limit of median class )

h = 10 (difference between two consecutive class interval ) and

f = 65 (highest frequency )

Step2:

Median =$l + (\frac{\frac{n}{2} -c.f}{f} )h$  put all the value in the given formula

46=  $40 + (\frac{ 115-42-p}{65} )10$

⇒ 46 = 40 +$(\frac{73-p}{65})10$

⇒598 = 520 +146-2p

⇒2p = 68 ⇒p = 34 .

now put the value of p = 34 in equation(i) we get ,

p + q = 80 ⇒ 34 +q = 80

⇒ q = 80-34 = 46 .

Final answer :

Hence , the value of p and q are 34 and 46 .