Question

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes(in hours): 0-20 20-40 40-60 60-80 80-100 100-120
No.of components: 10 35 52 61 38 29
Determine the modal lifetimes of the components.

Answer 1

The modal lifetimes of the components is 65.625

Lifetimes:                        No.of components:

(in hours)                               (frequency)

0-20                                              10

20-40                                            35

40-60                                            52

60-80                                             61

80-100                                            38

100-120                                          29

We have a formula to calculate the mode and is given by

$Mode = l + \dfrac{f_1-f_0}{2f_1-f_0-f_2} * h$

Modal class is a class with highest frequency

Modal class interval is 60 - 80 with a highest frequency of 61.

f1 = 61

Frequency of class interval before the modal class = f0 = 52

Frequency of class interval before the modal class = f2 = 38

h = class interval = 80 - 60 = 20

l = lower class limit of modal class = 60

Mode = 60 + [ (61-52) / (2*61 - 52 - 38) ] * 20

= 60 + 9/32 * 20

= 60 + 9/8 * 5

= 60 + 5.625

= 65.625

Answer 2

$\Huge\sf\pink{Solution \:!!}$

$\rule{200}{1}$

$\large\begin{tabular}{c | l}</p><p></p><p>Class Interval & Frequencies \\</p><p></p><p>\cline{1-2}</p><p></p><p>0-20 & 10 \\</p><p></p><p>20-40 & 35 \\</p><p></p><p>40-60 & 52 \\</p><p></p><p>60-80 & 61 \\</p><p></p><p>80-100 & 38 \\</p><p></p><p>100-120 & 29</p><p>\end{tabular}$

$\rule{200}{1}$

$\sf{Since\: the \:maximum\: frequency \:61 \:is\: in\:the}$

$\sf{class \:60-80,\: this\: is\: the \:required \:modal\: class.}$

$\sf\boxed{mode = l + \frac{(f_1 \: - \: f_0)}{2f_1 \: - \: (f_0 \: + \: f_2)} \times h}$

• Mode class frequency, f1 = 61
• Frequency of the class preceding the modal class f0 = 52
• Frequency of the class succeeding the modal class f2 = 38
• Lower boundary of the modal class, l = 60
• Height of the class, h = 20

$\tt{=\:60 \: + \: (\frac{61 - 52}{2 \times 61 - (52 + 38)} ) \times 20}$

$\tt{= \:60 \: + \: \frac{9}{122-90} \times 20}$

$\tt{=\: 60 \: + \frac{9}{32} \: \times \: 20 \:}$

$\tt{= \: 60 \: + \: 5.625}$

$\tt=\red{ \: 65.625 \: hours}$

$\rule{200}{2}$