The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them
Monthly consumption | Number of consumers |
---|---|
65 - 85 | 4 |
85 - 105 | 5 |
105 - 125 | 13 |
125 - 145 | 20 |
145 - 165 | 14 |
165 - 185 | 8 |
185 - 205 | 4 |
The cumulative frequencies with their respective class intervals are as follows.
Monthly consumption | Number of consumers | Cumulative frequency |
---|---|---|
60 - 85 | 4 | 4 |
85 - 105 | 5 | 4 + 5 = 9 |
105 - 125 | 13 | 9 + 13 = 22 |
125 - 145 | 20 | 22 + 20 = 42 |
145 - 165 | 14 | 42 + 14 = 56 |
165 - 185 | 8 | 56 + 8 = 64 |
185 - 205 | 4 | 64 + 4 = 68 |
Total(n) | 68 |
From the table, we obtain
n = 68
Cumulative frequency just greater \(\frac{n}2 ( i.e., \frac{68}2 = 34)\) than is 42, belonging to class interval 125 - 145.
Median class = 125 - 145
Lower limit (\(l\)) of median class = 125
Frequency (\(f\)) of median class = 20
Cumulative frequency (\(cf\)) of median class = 22
Class size (\(h\)) = 20
Median = \(l + (\frac{\frac{n}2 - cf}f \times h)\)
Median = \(125 + (\frac{34 - 22}{20} \times 20)\)
Median = 125 +12
Median = 137
To find the class mark (xi) for each interval, the following relation is used.
Class mark \((x_i)\) = \(\frac {\text{Upper \,limit + Lower \,limit}}{2}\)
Taking 11.5 as assumed mean (a), \(d_i\), \(u_i\), and \(f_iu_i\) are calculated according to step deviation method as follows.
Monthly consumption | Number of consumers | \(\bf{x_i}\) | \(\bf{d_i = x_i -11.5}\) | \(\bf{u_i = \frac{d_i}{3}}\) | \(\bf{f_iu_i}\) |
---|---|---|---|---|---|
60 - 85 | 4 | 75 | -60 | -3 | -12 |
85 - 105 | 5 | 95 | -40 | -2 | -10 |
105 - 125 | 13 | 115 | -20 | -1 | -13 |
125 - 145 | 20 | 135 | 0 | 0 | 0 |
145 - 165 | 14 | 155 | 20 | 1 | 14 |
165 - 185 | 8 | 175 | 40 | 2 | 16 |
185 - 205 | 4 | 195 | 60 | 3 | 12 |
Total | 68 | 7 |
From the table, it can be observed that
\(\sum f_i = 68\)
\(\sum f_iu_i = 7\)
Mean, \(\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i})\times h\)
\(\overset{-}{x}\) = \(135 + (\frac{7 }{68})\times 20\)
\(\overset{-}{x}\) = 135 + \(\frac{140}{68}\)
Mean, \(\overset{-}{x}\) = 137.058
The data in the given table can be written as
Monthly consumption | Number of consumers |
---|---|
65 - 85 | 4 |
85 - 105 | 5 |
105 - 125 | 13 |
125 - 145 | 20 |
145 - 165 | 14 |
165 - 185 | 8 |
185 - 205 | 4 |
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.
Therefore, Modal class = 125 − 145
Lower limit (\(l\)) of modal class = 125
Frequency (\(f_1\)) of modal class = 40
Frequency (\(f_0\)) of class preceding the modal class = 13
Frequency (\(f_2\)) of class succeeding the modal class = 14
Class size (\(h\)) = 20
Mode = \(l\) + \((\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h\)
Mode = \(125 + (\frac{20 - 13 }{ 2(20) - 13 - 14}) \times 20\)
Mode =\(125+ [\frac{7}{13}] \times 20\)
Mode = \(125 +( \frac{ 140}{ 13})\)
Mode = 135.76
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively. The three measures are approximately the same in this case.
Class Interval | 50-70 | 70-90 | 90-110 | 110-130 | 130-150 | 150-170 |
---|---|---|---|---|---|---|
Number of Students | 15 | 21 | 32 | 19 | 8 | 5 |
Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 |
---|---|---|---|---|---|
No. of students | 10 | 18 | 42 | 13 | 7 |
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm) | Number of leaves |
---|---|
118 - 126 | 3 |
127 - 135 | 5 |
136 - 144 | 9 |
145 - 153 | 12 |
154 - 162 | 5 |
163 - 171 | 4 |
172 - 180 | 2 |
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)
Class interval | Frequency |
---|---|
0 - 10 | 5 |
10 - 20 | x |
20 - 30 | 20 |
30 - 40 | 15 |
40 - 50 | y |
50 - 60 | 5 |
Total | 60 |
The following table gives the distribution of the life time of 400 neon lamps :
Life time (in hours) | Number of lamps |
---|---|
1500 - 2000 | 14 |
2000 - 2500 | 56 |
2500 - 3000 | 60 |
3000 - 3500 | 86 |
3500 - 4000 | 74 |
4000 - 4500 | 62 |
4500 - 5000 | 48 |
Find the median life time of a lamp.