Question

In the following frequency distribution the weekly Median cost of living Index is 1755. If total number of weeks is 52 then find the missing frequencies . cost of living Index 1500-1600 1600-1700 1700- 1800 1800-1900 1900-2000 2000-2100 2100-2200 NO of weeks 4 x y 8 4 3 2

Answer 1

ok, soo



Let first frequency be x. So, second frequency y = 52 - (4+x+8+4+3+2) = 31-x

Cost of living index No. of weeks(f) c.f.
1500-1600 4 4
1600-1700 x 4+x -->c
1700-1800 31-x -->f 35
1800-1900 8 43
1900-2000 4 47
2000-2100 3 50
2100-2200 2 52 
Given, Median=1755 , which lies in the class 1700−1800.So, L=1700, N2=522=26, h=100, f=31−x, and c=4+xThus,Median=L+(N2−cf)h⇒1755=1700+(26−4−x31−x)100⇒55=(22−x31−x)100⇒55(31−x)=(22−x)100⇒1705−55x=2200−100x⇒100x−55x=2200−1705⇒45x=495⇒x=11And,y=31−x=31−11=20⇒y=20
































































tada

Answer 2

Answer:

p=11 and q=20

Step-by-step explanation:

NOTE* I have used p for x and q for y......Hope you don't mind...^-^

Median=1755

median class- 1700-1800

l=1700

n/2= 52/2=26

cf= 4+p

f= y

h= 100

Total frequeny= 52

21+ p+q=52

p+q= 31

q=31-x                                 -(1)

Median= l+ (n/2-cf)/f *h

⇒1755= 1700 +[26-(4+p)]/q *100

  1755= 1700 +[26-(4+p)]/31-q *100

  55= 22-p/31-p *100

  55/100= 22-p/31-p

  11/20=22-p/31-p

Cross multiply,

  11(31-p)= 20(22-p)

  341-11p= 440-20p

  9p= 99

   p=11

   

  putting p=11 in eq(1)

  q=31-11

  q= 20

  There it is!! Hope it helped....