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.
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]
Median class of the following frequency distribution will be:
\[ \begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 0-10 & 7 \\ \hline 10-20 & 12 \\ \hline 20-30 & 18 \\ \hline 30-40 & 15 \\ \hline 40-50 & 10 \\ \hline 50-60 & 3 \\ \hline \end{array} \]
The median class of the following frequency distribution will be:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-Interval} & \text{$0$--$10$} & \text{$10$--$20$} & \text{$20$--$30$} & \text{$30$--$40$} & \text{$40$--$50$} \\ \hline \text{Frequency} & \text{$7$} & \text{$8$} & \text{$15$} & \text{$10$} & \text{$5$} \\ \hline \end{array}\]
The following data shows the number of family members living in different bungalows of a locality:
Number of Members | 0−2 | 2−4 | 4−6 | 6−8 | 8−10 | Total |
---|---|---|---|---|---|---|
Number of Bungalows | 10 | p | 60 | q | 5 | 120 |
If the median number of members is found to be 5, find the values of p and q.
‘दीवार खड़ी करना’ मुहावरे का वाक्य में इस प्रकार प्रयोग करें कि अर्थ स्पष्ट हो जाए।
Select from the following a statement which is not true about the burning of magnesium ribbon in air:
Analyze the significant changes in printing technology during 19th century in the world.
निम्नलिखित विषय पर संकेत बिंदुओं के आधार पर लगभग 120 शब्दों में एक अनुच्छेद लिखिए |
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