Question

50 - 60
60 – 70
If the mode of the distribution is 43.75, find p
Class
20 – 30 30 - 40 40 - 50
interval
Frequency
25
47
62
р
10​

Answer 1

$Modal \:class : ( 40 - 50 ) ,$

$Boundary \: limit (l) of \:modal \:class (f_{0})=40$

$Class \:size (h) = 10$

$Frequency \:of \:the \: modal \:class (f_{1}) = 62$

$Frequency \:of \:class \: preceding \: the$

$modal \:class (f_{0}) = 47$

$Frequency \:of \:class \: succeeding \: the$

$modal \:class (f_{2}) = p$

Now, let us substitute these values in the formula:

$Mode= l + \Big( \frac{f_{1} - f_{0}}{2f_{1}-f_{0}-f_{2}}\Big) \times h$

$\implies 40+\Big(\frac{62-47}{2\times 62-47-p}\Big)\times 10 = 43.75 \:(given)$

$\implies \frac{15 \times 10}{124-47-p} =43.75-40$

$\implies \frac{150}{77-p} = 3.75$

$\implies 150 = 3.75(77-p)$

$\implies 150 = 288.75 -3.75p$

$\implies 3.75p = 288.75 - 150$

$\implies 3.75p = 138.75$

$\implies p = \frac{138.75}{3.75}$

$\implies p = 37$

Therefore.,

$\red{ Value \: of \:p}\green { = 37 }$

•••♪

Answer 2

Step-by-step explanation:

Mode= l + \Big( \frac{f_{1} - f_{0}}{2f_{1}-f_{0}-f_{2}}\Big) \times hMode=l+(2f1−f0−f2f1−f0)×h

\implies 40+\Big(\frac{62-47}{2\times 62-47-p}\Big)\times 10 = 43.75 \:(given)⟹40+(2×62−47−p62−47)×10=43.75(given)

\implies \frac{15 \times 10}{124-47-p} =43.75-40⟹124−47−p15×10=43.75−40

\implies \frac{150}{77-p} = 3.75⟹77−p150=3.75

\implies 150 = 3.75(77-p)⟹150=3.75(77−p)

\implies 150 = 288.75 -3.75p⟹150=288.75−3.75p

\implies 3.75p = 288.75 - 150⟹3.75p=288.75−150

\implies 3.75p = 138.75⟹3.75p=138.75

\implies p = \frac{138.75}{3.75}⟹p=3.75138.75

\implies p = 37⟹p=37

Therefore.,

\red{ Value \: of \:p}\green { = 37 }Valueofp=37