Question

Find the missing frequencies and the median for the following distribution if the mean is 1.46

Number of accidents	Frequencies (No. of days)
0	46
1	?
2	?
3	25
4	10
5	5
	$\sum {{f_i}} $

Answer 1

Hint: According to the question given in the question we have to find the missing frequencies and the median for the following distribution if the mean is 1.46. So, first of all we have to let the two missing frequencies.
Now, we have to find the value of ${f_i}{x_i}$ by multiplying number of accidents $({x_i})$with the given frequencies $({f_i})$
Now, we have to determine the summation of $({f_i})$and ${f_i}{x_i}$.
So, to find the both of the missing frequencies we have to use the formula to find the maen for the given data as given below:

Formula used:
Mean$\overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}.....................(a)$
So, with the help of formula (a) above we will obtain a linear expression in form of both the frequencies we let.
Now, as given the we have to find the median for the given data so first of all we have to find the cumulative frequency for the data and we can understand the cumulative frequency as explained below:
Cumulative frequency: The cumulative frequency can be calculated or obtained by adding each given frequency from a frequency distribution table to the sum of its predecessors and its last value will be always equal to the total for all observations, since all the frequencies will already have been added to the previous total.
Now, to find the median we have to use the formula as given below:
Formula used:
Median$ = \dfrac{{{{\left( {\dfrac{N}{2}} \right)}^{th}} + {{\left( {\dfrac{N}{2} + 1} \right)}^{th}}}}{2}................(b)$
Hence, with the help of the formula (b) and we can also obtain the median as asked in the question.


Complete step by step solution:
Step 1: First of all we have to let the missing two frequencies as ${f_1}$and ${f_2}$hence, now the new table will be:

Number of accidents	Frequencies (No. of days)
0	46
1	${f_1}$
2	${f_2}$
3	25
4	10
5	5

Step 2: Now, we have to calculate the value of ${f_i}{x_i}$by multiplying number of accidents $({x_i})$with the given frequencies $({f_i})$hence,

Number of accidents$({x_i})$	Frequencies (No. of days) $({f_i})$	${f_i}{x_i}$
0	46	$0 \times 46 = 0$
1	${f_1}$	$1 \times {f_1} = {f_1}$
2	${f_2}$	$2 \times {f_2} = 2{f_2}$
3	25	$3 \times 25 = 75$
4	10	$4 \times 10 = 40$
5	5	$5 \times 5 = 25$

Step 3: Now, from the above table as obtained in the step 2 we have to obtain the summation of ${f_i}{x_i}$.
Hence,
\[
   \Rightarrow \sum {{f_i}{x_i} = 0 + {f_1} + 2{f_2} + 75 + 40 + 25} \\
   \Rightarrow \sum {{f_i}{x_i} = 140 + {f_1} + 2{f_2}} \\
 \]
Step 4: Same as the step 3 we have to find the summation of ${f_i}$
Hence,
$
  \sum {{f_i} = 46 + {f_1} + {f_2} + 25 + 10 + 5} \\
  \sum {{f_i} = 86 + {f_1} + {f_2}} \\
 $
Step 5: Now, to find the values of ${f_1}$and ${f_2}$we have to use the formula (a) as mentioned in the solution hint and given that $\sum {{f_i}} = 200$
$ \Rightarrow 1.96 = \dfrac{{140 + {f_1} + {f_2}}}{{200}}$
After cross-multiplication,
$
   \Rightarrow 1.96 \times 200 = 140 + {f_1} + 2{f_2} \\
   \Rightarrow {f_1} + 2{f_2} = 292 - 140 \\
   \Rightarrow {f_1} + 2{f_2} = 152............(1) \\
 $
And as know that,
$
   \Rightarrow 86 + {f_1} + {f_2} = 200 \\
   \Rightarrow {f_1} + {f_2} = 114.............(2) \\
 $
Step 6: Now, to find the values of ${f_1}$and ${f_2}$we have to subtract equation (2) by equation (1) hence,
\[
   \Rightarrow ({f_1} + 2{f_2}) - ({f_1} + {f_2}) = 152 - 114 \\
   \Rightarrow {f_1} + 2{f_2} - {f_1} - {f_2} = 38 \\
   \Rightarrow {f_2} = 38 \\
 \]
Now, to obtain the value of ${f_1}$ we have to substitute the value of ${f_2}$in equation (2) hence,
$
   \Rightarrow {f_1} + 38 = 114 \\
   \Rightarrow {f_1} = 114 - 38 \\
   \Rightarrow {f_1} = 76 \\
 $
Step 7: Now, to find the median we have to find the cumulative frequencies as mentioned in the solution hint.

Number of accidents$({x_i})$	Frequencies (No. of days) $({f_i})$	Cumulative frequency
0	46	$46 + 0 = 46$
1	76	$46 + 76 = 122$
2	38	$122 + 38 = 160$
3	25	$160 + 25 = 185$
4	10	$185 + 10 = 195$
5	5	$195 + 5 = 200$

Hence, N = 200 which is equal to $\sum {{f_i}} $
Step 8: Now, to find the value of median we have to apply the formula (b) as mentioned in the solution hint. Hence,
\[
   = \dfrac{{{{\dfrac{{200}}{2}}^{th}} + {{\left( {\dfrac{{200}}{2} + 1} \right)}^{th}}}}{2} \\
   = \dfrac{{{{100}^{th}} + {{101}^{th}}}}{2} \\
 \]
Hence, as we know that ${100^{th}}$and ${101^{th}}$observations are 1 from the given table hence,
$
   = \dfrac{{1 + 1}}{2} \\
   = 1 \\
 $
Hence, with the help of formula (a) and formula (b) we have obtained the frequencies which are 76, and 3 and the median which is = 1

Note: 1. The cumulative frequency of a set of a data or class interval of a frequency table is the sum of frequencies of the data up to a required level and it can also be used to determine the number of items that have values below a particular level.
2. Calculating cumulative frequency gives us the sum of all the frequencies up to a certain point in a data set.