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Question: If\({L_1} = 44.5\),\(C = 10\),\({F_2} = 10\), \({F_1} = 28\), \(f = 30\), then find mode. A. \(45....

IfL1=44.5{L_1} = 44.5,C=10C = 10,F2=10{F_2} = 10, F1=28{F_1} = 28, f=30f = 30, then find mode.
A. 45.4145.41
B. 47.3147.31
C. 49.6149.61
D. 5353

Explanation

Solution

Hint : First, we shall analyze the given data so that we are able to solve the given problem. Here, we are asked to calculate the mode. We all know that the mode of the data is the most frequently occurring value. In this question, we are directly given the values that are usually found using the frequency distribution. We need to substitute the given values in the formula directly.

Formula used:
The formula to calculate the mode for the grouped data is as follows.
Mode=l+f1−f02f1−f0−f2×hMode = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times h
Where ll is the lower limit, hh is the size of the class interval,f1{f_1} denotes the frequency of the modal class, f0{f_0} is the frequency of class preceding (before) the modal class, andf2{f_2} denotes the frequency of the class succeeding (after) the modal class.

Complete step-by-step solution:
We are givenL1=44.5{L_1} = 44.5,C=10C = 10,F2=10{F_2} = 10, F1=28{F_1} = 28, f=30f = 30
Here, we can note thatL1{L_1}denotes the lower limit of the modal class, CC is the size of the class interval l, ff denotes the frequency of the modal class,F1{F_1} is the frequency of class preceding (before) the modal class andF2{F_2} is the frequency of class succeeding (after) the modal class
Now, we shall apply the given values in the formula.
Thus, we get
Mode=L1+f−F12f−F1−F2×CMode = {L_1} + \dfrac{{f - {F_1}}}{{2f - {F_1} - {F_2}}} \times C
⇒Mode=44.5+30−282×30−28−10×10\Rightarrow Mode = 44.5 + \dfrac{{30 - 28}}{{2 \times 30 - 28 - 10}} \times 10
⇒Mode=44.5+260−38×10\Rightarrow Mode = 44.5 + \dfrac{2}{{60 - 38}} \times 10
⇒Mode=44.5+2022\Rightarrow Mode = 44.5 + \dfrac{{20}}{{22}}
⇒Mode=44.5+0.9090\Rightarrow Mode = 44.5 + 0.9090
⇒Mode=45.409090\Rightarrow Mode = 45.409090
⇒Mode=45.41\Rightarrow Mode = 45.41 (approximately)
Hence, option A is correct.

Note: Here in this problem, we are directly given the values that are needed for the formula. The modal class is the class interval that contains the maximum frequency. Suppose if we are given a frequency distribution table, we need to first find the modal class. The given formula is applicable for finding the mode for grouped data. For ungrouped data, we need to just pick the most frequently occurred value for mode.