Question

The following data gives the information on the observed life times (in hours) of 225 electrical components.

Lifetime(in Hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

Determine the modal life times of components.

Answer 1

Hint: To find the modal class we have to identify the class interval with maximum frequency in this question 61 is the highest frequency so modal class is 60-80
$\begin{align}
  & l\text{ is lower limit of modal class} \\
 & h\text{ is class interval} \\
 & {{f}_{1}}\text{ is frequency of class modal class} \\
 & {{f}_{0}}\text{is frequency of class before modal class} \\
 & {{f}_{2}}\text{ is frequency of class after modal class} \\
\end{align}$
$l=60,\text{ }{{f}_{0}}=52,\text{ }{{f}_{1}}=61,\text{ }{{f}_{2}}=38$

Formula used:
 $Mode=\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times h$
Where
Modal class: Interval with highest frequency
 $\begin{align}
  & l=\text{lower limit of modal class} \\
 & h=\text{class interval} \\
 & {{f}_{1}}=\text{frequency of class modal class} \\
 & {{f}_{0}}=\text{frequency of class before modal class} \\
 & {{f}_{2}}=\text{frequency of class after modal class} \\
\end{align}$

 Complete step-by-step answer:
 Highest frequency of the data is 61
61 belongs to the interval 60-80
Therefore $l=60\text{ }\!\!\And\!\!\text{ }{{f}_{1}}=61$
$\begin{align}
  & \text{frequency of the previous class i}\text{.e}\text{.}\left( 40-60 \right)\text{ is }=52={{f}_{0}} \\
 & \text{frequency of the ne }\!\!~\!\!\text{ xt class i}\text{.e}\text{. }\left( 80-100 \right)\text{ is }=38={{f}_{2}} \\
 & \text{class interval }=80-60 \\
 & \text{ }=20=h \\
\end{align}$
$\begin{align}
  & Mode=l+\dfrac{{{f}_{1}}+{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times h \\
 & \text{ }=60+\dfrac{61-52}{2\left( 61 \right)-52-38}\times 20\text{ }\left( \text{adding 52 }\!\!\And\!\!\text{ 38} \right) \\
 & \text{ }=60+\dfrac{9}{122-90}\times 20 \\
 & \text{ }=60+\dfrac{9}{32}\times 20\text{ }\left( \text{Divide by 4} \right) \\
 & \text{ }=60+\dfrac{45}{8} \\
\end{align}$
  $\begin{align}
  & =60+5\cdot 625 \\
 & =65\cdot 625 \\
\end{align}$
$\text{Modal lifetimes of the components}=65\cdot 625\left( \text{hours} \right)$

Additional information:

\[\] Life times hours	Frequency
0-20	10
20-40	35
40-60	52
60-80	61
80-100	38
100-120	29

Whereas,
$\begin{align}
  & 52={{f}_{0}} \\
 & 61={{f}_{1}} \\
 & 38={{f}_{2}} \\
\end{align}$
Students can solve directly
$Mode=l+\dfrac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}}\times h$
Subletting the values and simplifying.




Note: For this particular question 61 is the highest frequency so modal class is 60-80 so lower limit is 60, frequency of modal class is 61, frequency of previous class is 62, frequency of next class is 38 and class interval is 80. By substituting the values in the formula we can calculate the modal life times Please do mention that Modal lifetimes in have for this particular case $65\cdot 625\text{ hours}$ .