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Q2:

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : 

CBSE Class 10 Maths ...

Determine the modal lifetimes of the components.

Solution :

Given: The frequency distribution of the lifetimes (in hours) of 225 electrical components is as follows:

Lifetimes (hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency ($f_i$) 10 35 52 61 38 29

To Find: The modal lifetime of the electrical components.

Step 1: Identifying the Modal Class
The modal class is the class interval with the highest frequency. Looking at the frequency distribution table, the highest frequency is $61$, which corresponds to the class interval $60-80$.
Therefore, the Modal Class is $60-80$.

Step 2: Stating the Formula for Mode
The formula for calculating the mode of a grouped frequency distribution is:
$\text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h$
Where:
$l$ = Lower limit of the modal class
$h$ = Size of the class interval
$f_1$ = Frequency of the modal class
$f_0$ = Frequency of the class preceding the modal class
$f_2$ = Frequency of the class succeeding the modal class

Step 3: Extracting the Values
From the given data:
$l = 60$
$h = 80 - 60 = 20$
$f_1 = 61$
$f_0 = 52$
$f_2 = 38$

Step 4: Substituting Values into the Formula
$\text{Mode} = 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20$
[Substituting the identified values into the formula]

Step 5: Performing the Arithmetic Calculations
$\text{Mode} = 60 + \left( \frac{9}{122 - 90} \right) \times 20$
$\text{Mode} = 60 + \left( \frac{9}{32} \right) \times 20$
$\text{Mode} = 60 + \left( \frac{180}{32} \right)$
$\text{Mode} = 60 + 5.625$
$\text{Mode} = 65.625$

Final Answer: The modal lifetime of the electrical components is 65.625 hours.


More Questions from Class 10 Mathematics Statistics EXERCISE 13.2


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