Question

(B) The following frequency distribution shows the marks of 50 students in a test. Find the mode of
the given data.
Marks:
20 - 30
30 - 40
40 - 50
50 - 60
70 - 80
60 - 70
No. of students:
2
8
20
10
3
7​

Answer 1

Question:-

The following frequency distribution shows the marks of 50 students in a test. Find the mode of the given data.

$\boxed{\begin{array}{c|c} \bf{Marks} & \bf{No.\:of\:Students} \\ \sf{20-30} & \sf{2} \\ \sf{30-40} & \sf{8} \\ \sf{40-50} & \sf{20} \\ \sf{50-60} & \sf{10} \\ \sf{60-70} & \sf{7} \\ \sf{70-80} & \sf{3}\end{array}}$

Solution:-

Here the class with highest frequency will be the modal class.

The class with highest frequency = 40 - 50

We know,

The formula of mode is as follows:-

$\dag\boxed{\underline{\rm{\orange{Mode = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h}}}}$

Here:-

f₁ = Frequency of the modal class

f₀ = Frequency of the previous class of the modal class

f₂ = Frequency of the next class of the modal class

l = Lower limit of the modal class

h = Height of the class

Height of the class = Upper limit - Lower limit

=> h = 50 - 40 = 10

Putting all the values in the formula:-

= $\sf{l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h}$

= $\sf{40 + \dfrac{20 - 8}{2\times 20 - 8 - 10} \times 10}$

$=\sf{40 + \dfrac{12}{40 - 18}\times 10}$

$=\sf{40 + \dfrac{12}{22}\times 10}$

$=\sf{40 + \dfrac{120}{22}}$

$=\sf{40 + \dfrac{60}{11}}$

$=\sf{40 + 5.45}$

$= \sf{45.45}$

$\boxed{\underline{\pink{\rm{\therefore\:The\:mode\:of\:the\:data\:is\:45.4\:[Approx]}}}}$

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