Question

it is questions of statics class 10

Answer 1

We have the table

Class Interval ⠀⠀⠀ Frequency⠀⠀⠀⠀⠀⠀⠀

0-10⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀8

10-20⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀10

20-30⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀x

30-40⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀16

40-50⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀12

50-60⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀6

60-70⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀8

_______________________________________

Let f1 = Frequency of modal class

f0= Frequency of previous (f1)

f2 = Frequency after ( f1)

l = lower limit of modal class

h = height of class interval

Thus from the table we conclude that

➠Modal Class = 30-40 ( as it has highest Frequency)

➠l = 30

➠h = 10

➠f0 = x

➠f1 = 16

➠f2 = 12

We know the formula

$\boxed{ \mapsto \sf \: mode = l +h \bigg(\dfrac{f _{1} -f _{0} }{2f _{0} - f _{0} - f _{2}} \bigg)}$

Also given that mode = 36 , Now,

${ \mapsto \sf \:36= 30 +10 \bigg(\dfrac{16 - 0 }{2 \times 16- 0- 12} \bigg)}$

${ \mapsto \sf \:36 - 30= 10 \bigg(\dfrac{16 - x }{32- x- 12} \bigg)}$

${ \mapsto \sf \:6= 10 \bigg(\dfrac{16 - x }{20- x} \bigg)}$

${ \mapsto \sf \: \dfrac{6}{10} = \bigg(\dfrac{16 - x }{20- x} \bigg)}$

${ \mapsto \sf \: \dfrac{3}{5} = \bigg(\dfrac{16 - x }{20- x} \bigg)}$

$\mapsto \sf 80 - 5x = 60 - 3x$

$\mapsto \sf 80 - 60 = 5x - 3x$

$\mapsto \sf 2x = 20$

$\mapsto \sf \: x = 10 \red \bigstar$