
The table below shows the daily expenditure on food of \[25\] households in a locality.
Daily expenditure $100-150$ $150-200$ $200-250$ $250-300$ $300-350$ No of households $4$ $5$ $12$ $2$ $2$
Find the mean daily expenditure on food by a suitable method.
| Daily expenditure | $100-150$ | $150-200$ | $200-250$ | $250-300$ | $300-350$ |
| No of households | $4$ | $5$ | $12$ | $2$ | $2$ |
Answer
573.3k+ views
Hint: In this question, we have the set of given observations.we have to find out the mean.
The mean (or average) of observations, is the sum of the values of all the observations divided by the total number of observations.
If \[{x_1},{x_2},{x_3},......,{x_n}\] are observations with respective frequencies \[{f_1},{f_2},{f_3},........,{f_n}\] then this means observation \[{x_1}\] occurs \[{f_1}\] times, \[{x_2}\] occurs \[{f_2}\] times, and so on.
Now, the sum of the values of all the observations = \[{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}\], and sum of the number of observations = \[{f_1} + {f_2} + {f_3} + ........ + {f_n}\]
So, the mean x of the data is given by
\[x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}}}{{{f_1} + {f_2} + {f_3} + ........ + {f_n}}}\]
Or
\[x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
Complete step-by-step solution:
We have daily expenditure on food of \[25\] households in a locality.
We need to find out the mean daily expenditure on food.To find the mean first we need to calculate the midpoint \[\left( {{x_i}} \right)\] of the respective expenditures and \[{f_i}{x_i}\] value for each case then we can put it in the mean formula.
Now, the mean x of the data is given by,
\[x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}}}{{{f_1} + {f_2} + {f_3} + ........ + {f_n}}}\]
Or
\[x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
That is, \[x = \dfrac{{500 + 875 + 2700 + 550 + 650}}{{4 + 5 + 12 + 2 + 2}} = \dfrac{{5275}}{{25}} = 211\]
Hence the mean daily expenditure on food is \[211\].
Note: In statistics, the mean is one of the three measures of central tendency, which are single numbers that try to pinpoint the central location within a data set. The mean, or average, is used most commonly, but it is important to differentiate it from the two other measures: median and mode. The median is the middle number when the numbers are listed in ascending order, while the mode is the most frequently occurring number.
The mean (or average) of observations, is the sum of the values of all the observations divided by the total number of observations.
If \[{x_1},{x_2},{x_3},......,{x_n}\] are observations with respective frequencies \[{f_1},{f_2},{f_3},........,{f_n}\] then this means observation \[{x_1}\] occurs \[{f_1}\] times, \[{x_2}\] occurs \[{f_2}\] times, and so on.
Now, the sum of the values of all the observations = \[{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}\], and sum of the number of observations = \[{f_1} + {f_2} + {f_3} + ........ + {f_n}\]
So, the mean x of the data is given by
\[x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}}}{{{f_1} + {f_2} + {f_3} + ........ + {f_n}}}\]
Or
\[x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
Complete step-by-step solution:
We have daily expenditure on food of \[25\] households in a locality.
We need to find out the mean daily expenditure on food.To find the mean first we need to calculate the midpoint \[\left( {{x_i}} \right)\] of the respective expenditures and \[{f_i}{x_i}\] value for each case then we can put it in the mean formula.
| Daily expenditure(in C) | Number of households\[\left( {{f_i}} \right)\] | Mid-point\[\left( {{x_i}} \right)\] | \[{f_i}{x_i}\] |
| \[100 - 150\] | \[4\] | \[\dfrac{{100 + 150}}{2} = \dfrac{{250}}{2} = 125\] | \[125 \times 4 = 500\] |
| \[150 - 200\] | \[5\] | \[\dfrac{{150 + 200}}{2} = \dfrac{{350}}{2} = 175\] | \[175 \times 5 = 875\] |
| \[200 - 250\] | \[12\] | \[\dfrac{{200 + 250}}{2} = \dfrac{{450}}{2} = 225\] | \[225 \times 12 = 2700\] |
| \[250 - 300\] | \[2\] | \[\dfrac{{250 + 300}}{2} = \dfrac{{550}}{2} = 275\] | \[275 \times 2 = 550\] |
| \[300 - 350\] | \[2\] | \[\dfrac{{300 + 350}}{2} = \dfrac{{650}}{2} = 325\] | \[325 \times 2 = 650\] |
Now, the mean x of the data is given by,
\[x = \dfrac{{{f_1}{x_1} + {f_2}{x_2} + ...... + {f_n}{x_n}}}{{{f_1} + {f_2} + {f_3} + ........ + {f_n}}}\]
Or
\[x = \dfrac{{\sum\limits_{i = 1}^n {{f_i}{x_i}} }}{{\sum\limits_{i = 1}^n {{f_i}} }}\]
That is, \[x = \dfrac{{500 + 875 + 2700 + 550 + 650}}{{4 + 5 + 12 + 2 + 2}} = \dfrac{{5275}}{{25}} = 211\]
Hence the mean daily expenditure on food is \[211\].
Note: In statistics, the mean is one of the three measures of central tendency, which are single numbers that try to pinpoint the central location within a data set. The mean, or average, is used most commonly, but it is important to differentiate it from the two other measures: median and mode. The median is the middle number when the numbers are listed in ascending order, while the mode is the most frequently occurring number.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

