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 :
Expenditure (in Rs) | Number of families |
|---|---|
1000 - 1500 | 24 |
1500 - 2000 | 40 |
2000 - 2500 | 33 |
2500 - 3000 | 28 |
3000 - 3500 | 30 |
3500 - 4000 | 22 |
4000 - 4500 | 16 |
4500 - 5000 | 7 |
It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.
Therefore, modal class = 1500 - 2000
Lower limit (\(l\)) of modal class = 1500
Frequency (\(f_1\)) of modal class = 40
Frequency (\(f_0\)) of class preceding the modal class = 24
Frequency (\(f_2\)) of class succeeding the modal class = 33
Class size (\(h\)) = 500
Mode = \(l\) + \((\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h\)
Mode = \(1500 + (\frac{40 - 24 }{ 2(40) - 24 - 33}) \times 500\)
Mode =\(1500+ [\frac{16}{80 - 57}] \times 500\)
Mode = \(1500 +( \frac{8000}{ 23})\)
Mode = 1500 + 347.826
Mode = 1847.826
Mode = 1847.83
Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark (\(x_i\)) for each interval, the following relation is used.
Class mark \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)
class size (h) of the data = 500
Taking 2750 as assured mean (a), \(d_i\), \(u_i\), and \(f_iu_i\) can be calculated as follows.
| Expenditure (in Rs) | Number of families (fi) | \(\bf{x_i}\) | \(\bf{d_i = x_i -2750}\) | \(\bf{u_i = \frac{d_i}{500}}\) | \(\bf{f_iu_i}\) |
|---|---|---|---|---|---|
1000 - 1500 | 24 | 1250 | -1500 | -3 | -72 |
1500 - 2000 | 40 | 1750 | -1000 | -2 | -80 |
2000 - 2500 | 33 | 2250 | -500 | -1 | -33 |
2500 - 3000 | 28 | 2750 | 0 | 0 | 0 |
3000 - 3500 | 30 | 3250 | 500 | 1 | 30 |
3500 - 4000 | 22 | 3750 | 1000 | 2 | 44 |
4000 - 4500 | 16 | 4250 | 1500 | 3 | 48 |
4500 - 5000 | 7 | 4750 | 2000 | 4 | 28 |
Total | 200 |
| -35 |
From the table, it can be observed that
\(\sum f_i = 200\)
\(\sum f_iu_i = -35\)
Mean, \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i})h\)
x = \(2750 + (\frac{-352 }{200})\times 500\)
x = 2750 - 87.5
x = 2662.5
Therefore, mean monthly expenditure was Rs 2662.50.
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:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval} & 0\text{--}5 & 5\text{--}10 & 10\text{--}15 & 15\text{--}20 & 20\text{--}25 \\ \hline \text{Frequency} & 2 & 7 & 11 & 8 & 6 \\ \hline \end{array} \]
Lifetimes (in hours) | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 | 100 - 120 |
|---|---|---|---|---|---|---|
Frequency | 10 | 35 | 52 | 61 | 28 | 29 |
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 |
On the day of her examination, Riya sharpened her pencil from both ends as shown below. 
The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.