Question

In a frequency distribution table, the modal value of the wages of 130 workers is Rs. \[97.50\]. Also, \[L = 94.5\], \[{f_m} = x + 15\], \[{f_1} = x\], and \[{f_2} = x + 5\]. Find the upper limit of the modal class.
(a) \[96.5\]
(b) \[97.5\]
(c) \[98.5\]
(d) \[99.5\]

Answer 1

Hint:
Here, we need to find the upper limit of the modal class. We will use the given information and the formula of the mode of a continuous series to form an equation. Then, we will solve the equation to find the class size of the modal class. Finally, using the class size and the lower limit of the modal class, we can calculate the upper limit of the modal class.

Formula Used: We will use the formula of the mode of a continuous series, \[L + \dfrac{{{f_m} - {f_1}}}{{2{f_m} - {f_1} - {f_2}}} \times h\]. Here \[L\] is the lower limit of the modal class, \[{f_m}\] is the frequency of the modal class, \[{f_1}\] is the frequency of the class interval preceding the modal class, \[{f_2}\] is the frequency of the class interval succeeding the modal class, and \[h\] is the class size of the modal class.

Complete step by step solution:
We will use the formula for mode of a continuous series to solve the given problem.
Substituting \[L = 94.5\], \[{f_m} = x + 15\], \[{f_1} = x\], and \[{f_2} = x + 5\] in the formula for mode, \[L + \dfrac{{{f_m} - {f_1}}}{{2{f_m} - {f_1} - {f_2}}} \times h\], we get
\[ \Rightarrow \]Mode \[ = 94.5 + \dfrac{{\left( {x + 15} \right) - x}}{{2\left( {x + 15} \right) - x - \left( {x + 5} \right)}} \times h\]
It is given that the modal value of the wages of 130 workers id Rs. \[97.50\].
Therefore, we get
\[ \Rightarrow 97.5 = 94.5 + \dfrac{{\left( {x + 15} \right) - x}}{{2\left( {x + 15} \right) - x - \left( {x + 5} \right)}} \times h\]
Multiplying the terms 2 and \[x + 15\] using the distributive law of multiplication, we get
\[ \Rightarrow 97.5 = 94.5 + \dfrac{{\left( {x + 15} \right) - x}}{{2x + 30 - x - \left( {x + 5} \right)}} \times h\]
Simplifying the expression, we get
\[ \Rightarrow 97.5 = 94.5 + \dfrac{{x + 15 - x}}{{2x + 30 - x - x - 5}} \times h\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 97.5 = 94.5 + \dfrac{{15}}{{25}} \times h\]
Simplifying the expression, we get
\[ \Rightarrow 97.5 = 94.5 + \dfrac{3}{5} \times h\]
Subtracting \[94.5\] from both sides of the equation, we get
\[\begin{array}{l} \Rightarrow 97.5 - 94.5 = 94.5 + \dfrac{3}{5} \times h - 94.5\\ \Rightarrow 3 = \dfrac{3}{5} \times h\end{array}\]
Multiplying both sides of the equation by \[\dfrac{5}{3}\], we get
\[ \Rightarrow 3 \times \dfrac{5}{3} = \dfrac{3}{5} \times h \times \dfrac{5}{3}\]
Therefore, we get
\[ \Rightarrow h = 5\]
\[\therefore \] The class size of the modal class is 5.
Class size is the difference in the upper limit and the lower limit of a class interval. It is also known as class width.
The formula to calculate class size of a class interval is upper limit of the class interval \[ - \] lower limit of the class interval.
Let the upper limit of the modal class be \[U\].
Therefore, we get
\[ \Rightarrow \]Class size of the class interval \[94.5\] – \[U\] is \[U - 94.5 = 5\].
Adding \[94.5\] to both sides of the equation, we get
\[ \Rightarrow U - 94.5 + 94.5 = 5 + 94.5\]
Therefore, we get
\[ \Rightarrow U = 99.5\]

Thus, the upper limit of the modal class is \[99.5\]. The correct option is option (d).

Note:
We have used the distributive law of multiplication in the solution. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].
We added and subtracted the like terms in a step of the solution. Like terms are the terms whose variables and their exponents are the same. For example, \[100x,150x,240x,600x\] all have the variable \[x\] raised to the exponent 1. Terms which are not like cannot be added together. For example, we cannot add 1 to \[600x\], but we can add \[2x\] to \[600x\].