Statistics | Exercise 13.2

Question 2

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

Determine the modal lifetimes of the components.

Question diagram 1
Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • The modal class is the class interval having the highest frequency in a grouped frequency distribution.
  • The mode for grouped data is calculated using the formula:

Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h

where:

  • l=l = lower limit of the modal class
  • h=h = size of the class interval
  • f1=f_1 = frequency of the modal class
  • f0=f_0 = frequency of the class preceding the modal class
  • f2=f_2 = frequency of the class succeeding the modal class

Step 1 · Identify Modal Class and Parameters

The highest frequency is 6161, which corresponds to the class interval 60−8060 - 80.Diagram 1

Therefore, the modal class is 60−8060 - 80.

From the modal class, we have:

  • Lower limit (ll) =60= 60
  • Frequency of modal class (f1f_1) =61= 61
  • Frequency of preceding class (f0f_0) =52= 52
  • Frequency of succeeding class (f2f_2) =38= 38
  • Class size (hh) =80−60=20= 80 - 60 = 20

Step 2 · Calculate Mode

Using the mode formula Mode=l+[f1−f02f1−f0−f2]×h\text{Mode} = l + \left[ \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \right] \times h

Substitute the values:

Mode=60+[61−522×61−52−38]×20=60+[9122−52−38]×20=60+[970−38]×20=60+[932]×20=60+18032=60+5.625=65.625\begin{aligned} \text{Mode} &= 60 + \left[ \dfrac{61 - 52}{2 \times 61 - 52 - 38} \right] \times 20 \\[0.6em] &= 60 + \left[ \dfrac{9}{122 - 52 - 38} \right] \times 20 \\[0.6em] &= 60 + \left[ \dfrac{9}{70 - 38} \right] \times 20 \\[0.6em] &= 60 + \left[ \dfrac{9}{32} \right] \times 20 \\[0.6em] &= 60 + \dfrac{180}{32} \\[0.6em] &= 60 + 5.625 \\[0.6em] &= 65.625 \end{aligned}
Answer

65.625 hours65.625\text{ hours}

Common Mistakes
  • Swapping f0f_0 and f2f_2: f0f_0 is the frequency before the modal class (5252), while f2f_2 is the frequency after the modal class (3838). Swapping them gives an incorrect denominator.
  • Denominator Sign Errors: In 2f1−f0−f22f_1 - f_0 - f_2, ensure both f0f_0 and f2f_2 are subtracted from 2f12f_1: 2(61)−52−38=122−90=322(61) - 52 - 38 = 122 - 90 = 32.
  • Forgetting to Multiply by Class Size (hh): Ensure you multiply f1−f02f1−f0−f2\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} by h=20h = 20 before adding to ll.

More questions in Exercise 13.2

Q1
  1. The following table shows the ages of the patients admitted in a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Q2
  1. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Q3
  1. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :
Q4

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Q5

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Q6

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

← Back to Statistics