
Monthly expenditure of milk of 100 families in a housing society are given in the following frequency distribution
Monthly expenditure(in) 0-175 175-350 350-525 525-700 700-875 875-1050 1050-1225 Number of families 10 14 15 21 28 7 5
Monthly expenditure(in) | 0-175 | 175-350 | 350-525 | 525-700 | 700-875 | 875-1050 | 1050-1225 |
Number of families | 10 | 14 | 15 | 21 | 28 | 7 | 5 |
Answer
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Hint: The mode and median are the set of data, we can define the mode as the number that often occurs in the group of data, and the median is defined as the number of data that is being at the middle of all.
Complete step-by-step answer:
The median (M) of the distribution:
The number of families monthly expenditure is \[n = 100\]
The lower limit of the median class from the data is \[l = 525\]
The cumulative frequency of class preceding the medium is \[C.F = 39\]
The class size of the given data is \[h = 175\]
The frequency of median class is \[f = 21\]
The equation to find the median of the data is
Let us take median as M
\[M = l + \left( {\dfrac{{\dfrac{n}{2} - C.F}}{f}} \right)h\]
Substituting the values in the above equation, then
\[\begin{array}
M = 525 + \left( {\dfrac{{\dfrac{{100}}{2} - 39}}{{21}}} \right)175\\
= 525 + \left( {\dfrac{{50 - 39}}{{21}}} \right)175\\
= 616.67
\end{array}\]
Therefore, the median of the given data is\[616.67\].
The mode (m) of the given data:
The equation to find the mode of the given data is,
\[M = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right)\]
The frequency corresponding to modal class \[{f_m} = 21\]
The frequency preceding to the modal class \[{f_1} = 28\]
The frequency proceeding to the modal class \[{f_2} = 7\]
The lower limit of the modal class from the data is \[l = 700\]
Substituting the values in the above equation, then
\[\begin{array}
= 700 + \left( {\dfrac{{28 - 21}}{{2 \times 28 - 21 - 7}}} \right)\\
= 700 + \dfrac{{7 \times 175}}{{28}}\\
= 700 + 43.75\\
= 743.75
\end{array}\]
Therefore, the mode of the given data is 743.75.
Note: Here, in the solution, we have to be very clear about the frequencies of the data. There are different frequencies like preceding, proceeding of modal class, and cumulative frequency, so we need to be careful while selecting those things from the data. While solving the problem, we have to be very particular about the formulae because even a symbol will lead to the wrong answer.
Complete step-by-step answer:
The median (M) of the distribution:
The number of families monthly expenditure is \[n = 100\]
The lower limit of the median class from the data is \[l = 525\]
The cumulative frequency of class preceding the medium is \[C.F = 39\]
The class size of the given data is \[h = 175\]
The frequency of median class is \[f = 21\]
The equation to find the median of the data is
Let us take median as M
\[M = l + \left( {\dfrac{{\dfrac{n}{2} - C.F}}{f}} \right)h\]
Substituting the values in the above equation, then
\[\begin{array}
M = 525 + \left( {\dfrac{{\dfrac{{100}}{2} - 39}}{{21}}} \right)175\\
= 525 + \left( {\dfrac{{50 - 39}}{{21}}} \right)175\\
= 616.67
\end{array}\]
Therefore, the median of the given data is\[616.67\].
The mode (m) of the given data:
The equation to find the mode of the given data is,
\[M = l + \left( {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right)\]
The frequency corresponding to modal class \[{f_m} = 21\]
The frequency preceding to the modal class \[{f_1} = 28\]
The frequency proceeding to the modal class \[{f_2} = 7\]
The lower limit of the modal class from the data is \[l = 700\]
Substituting the values in the above equation, then
\[\begin{array}
= 700 + \left( {\dfrac{{28 - 21}}{{2 \times 28 - 21 - 7}}} \right)\\
= 700 + \dfrac{{7 \times 175}}{{28}}\\
= 700 + 43.75\\
= 743.75
\end{array}\]
Therefore, the mode of the given data is 743.75.
Note: Here, in the solution, we have to be very clear about the frequencies of the data. There are different frequencies like preceding, proceeding of modal class, and cumulative frequency, so we need to be careful while selecting those things from the data. While solving the problem, we have to be very particular about the formulae because even a symbol will lead to the wrong answer.
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