Lifetimes (in hours) | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
|---|---|---|---|---|---|---|
Frequency | 10 | 35 | 52 | 61 | 28 | 29 |
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 − 80.
Therefore, modal class = 60 − 80
Lower limit (\(l\)) of modal class = 60
Frequency (\(f_1\)) of modal class = 61
Frequency (\(f_0\)) of class preceding the modal class = 52
Frequency (\(f_2\)) of class succeeding the modal class = 38
Class size (\(h\)) = 20
Mode = \(l\) + \((\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h\)
Mode = 60\(+ (\frac{61 - 52 }{ 2(61) - 52 - 38}) \times(20)\)
Mode =\(60+ [\frac{9}{122 - 90}] \times 20\)
Mode = \(60 +( \frac{9 \times20}{ 32})\)
Mode = \(60 + \frac{90}{16}\)
Mode = 60 + 5.625
Mode = 65.625
Therefore, modal lifetime of electrical components is 65.625 hours.
The modal class of the following table will be:
\[ \begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 0-5 & 5 \\ \hline 5-10 & 8 \\ \hline 10-15 & 12 \\ \hline 15-20 & 10 \\ \hline 20-25 & 7 \\ \hline \end{array} \]
The modal class of the following table will be:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0\text{--}5 & 5\text{--}10 & 10\text{--}15 & 15\text{--}20 & 20\text{--}25 \\ \hline \text{Frequency} & 2 & 7 & 11 & 8 & 6 \\ \hline \end{array} \]
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 :
| Number of cars | 0 - 10 | 10 - 20 | 20 - 30 | 30 - 40 | 40 - 50 | 50 - 60 | 60 -70 | 70 - 80 |
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
The following table shows the ages of tha year:
Age (in years) | 5 - 15 | 15 - 25 | 25 - 35 | 35 - 45 | 45 - 55 | 55 - 65 |
|---|---|---|---|---|---|---|
Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
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