The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption(in units)No. of customers
65-854
85-1055
105-12513
125-14520
145-16514
165-1858
185-2054

Solution:

Track down the aggregate recurrence of the given information as follows:

Class Interval      Frequency           Cumulative recurrence

65-85     4             4

85-105  5             9

105-125               13           22

125-145               20           42

145-165               14           56

165-185               8             64

185-205               4             68

N=68

From the table, it is seen that, n = 68 and thus n/2=34

Thus, the middle class is 125-145 with total recurrence = 42

Where, l = 125, n = 68, Cf = 22, f = 20, h = 20

Middle is determined as follows:

Ncert arrangements class 10 part 14-1

=125+((34−22)/20) × 20

=125+12 = 137

Along these lines, middle = 137

To compute the mode:

Modular class = 125-145,

f1=20, f0=13, f2=14 and h = 20

Mode recipe:

    \[\begin{array}{*{35}{l}} <!-- /wp:paragraph --> <!-- wp:paragraph -->    Mode\text{ }=\text{ }l+\text{ }\left[ \left( f1-f0 \right)/\left( 2f1-f0-f2 \right) \right]\times h  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    ~  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    Mode\text{ }=\text{ }125\text{ }+\text{ }\left( \left( 20-13 \right)/\left( 40-13-14 \right) \right)\times 20  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    ~  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    =125+\left( 140/13 \right)  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    ~  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    =125+10.77  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    ~  \\ <!-- /wp:paragraph --> <!-- wp:paragraph -->    =135.77  \\ <!-- /wp:paragraph --> <!-- wp:paragraph --> \end{array}\]

Consequently, mode = 135.77

Compute the Mean:

Class Interval      fi             xi           

    \[di=xi-a~~ui=di/h~fiui\]

 

65-85     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 fi= 68                                                      Sum fiui= 7

    \[x\text{ }=a+h\text{ }\sum fiui/\sum fi\text{ }=135+20\left( 7/68 \right)\]

Mean=137.05 For this situation, mean, middle and mode are more/less equivalent in this conveyance.