Question

if the mode of the given data is 45 find x and y

1 class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 total
2 frequency 4 8 x 12 10 4 y 50

Answer 1

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my answer is 100 % right.

x = 10 and y= 2.
mode = l + f1 - fo/ 2f1 - f2 -f0 × h

45-40= 12- x/ 14 - x ×10
14- x = 24-2x
by solving this x= 10
and for y
y =50- total of frequency
y= 50 - 48
y = 2

Answer 2

Given is the mode of a frequency distribution, Find the value of x and y.

Explanation:

• Given is the proper table for classes and their frequency distributions for total $50$ frequencies.  
• $class\ \ \ \ \ \ \ \ \ f_i\\10-20\ \ \ \ \ \ 4 \\20-30\ \ \ \ \ \ 8 \\30-40\ \ \ \ \ \ x \\40-50\ \ \ \ \ \ 12 \\50-60\ \ \ \ \ \ 10 \\60-70\ \ \ \ \ \ 4 \\70-80\ \ \ \ \ \ y\ \ \ \ \ ->\sum\ f_i=N=50$  
• Now given that mode is $M_o=45$ therefore clearly the modal class is $40-50$
• Therefore we have, the lower bound, frequency of modal class, frequency of classes  preceding and succeeding the modal class, and the class size:
• $L=40,\ \ \ f_m=12,\ \ \ f_1=x,\ \ \ f_2=10,\ \ \ C=10\\->M_o=L+(\frac{f_m-f_1}{2f_m-f_1-f_2} )C\\\\->45=40+(\frac{12-x}{24-x-10} )10\\->2(12-x)=14-x\\->x=10$  
• Now we are given $N=50$ hence we get, $N=\sum f\\->4+8+x+12+10+4+y=50\\->48+y=50\\->y=2$  
• The values of x and y are $10\ and\ 2$ respectively.