
The following data shows the age distribution of patients of malaria in a village during a particular month. Find the average age of the patients.
Age (in years) No. of cases \[5 - 14\] 6 \[15 - 24\] 11 \[25 - 34\] 21 \[35 - 44\] 23 \[45 - 54\] 14 \[55 - 64\] 5 \[65 - 74\] 3
A.36.12 years
B.36.13 years
C.13.36 years
D.23.36 years
| Age (in years) | No. of cases |
| \[5 - 14\] | 6 |
| \[15 - 24\] | 11 |
| \[25 - 34\] | 21 |
| \[35 - 44\] | 23 |
| \[45 - 54\] | 14 |
| \[55 - 64\] | 5 |
| \[65 - 74\] | 3 |
Answer
572.7k+ views
Hint: Since, the given frequency distribution has intervals of constant class width, we can directly use the formula to find mean as \[\overline x = \dfrac{{\sum {xf} }}{{\sum f }}\], where \[\overline x \] is the mean of the distribution, \[x\] is the middle value of an interval and \[f\] is the frequency of the interval. First we will find the class mark of each interval, then multiply the class mark with their corresponding frequencies and find the sum of both the frequencies and the product of frequency and class mark. Finally we can substitute these values and get the mean.
Complete step-by-step answer:
Class marks is the average of the upper and lower limit of an interval. For the interval \[5 - 14\], the class mark will be \[\dfrac{{5 + 14}}{2} = \dfrac{{19}}{2} = 9.5\]. Similarly we can find the class mark of other intervals.
Now, as we have to find the class mark of each interval and then multiply it with the frequency we will organize the data in the form of a table.
The table formed will be as follows.
Now, we have the values of \[\sum {xf} \] and \[\sum f \], which we can substitute in the formula to get the mean as.
\[
\Rightarrow \overline x = \dfrac{{\sum {xf} }}{{\sum f }} \\
\Rightarrow \overline x = \dfrac{{2998.5}}{{83}} = 36.1265 \simeq 36.13 \\
\]
Thus, the mean for the given frequency distribution will be 36.13 years.
Hence, option (B) will be the correct option.
Note: As we know that the mean of a data set represents the most likely number in the data set, we can observe that here easily. Maximum number of patients have their ages in the range of \[35 - 44\], and the mean also comes out to be 36.13, which is very close to the class mark of this interval thus verifying its definition.
Complete step-by-step answer:
Class marks is the average of the upper and lower limit of an interval. For the interval \[5 - 14\], the class mark will be \[\dfrac{{5 + 14}}{2} = \dfrac{{19}}{2} = 9.5\]. Similarly we can find the class mark of other intervals.
Now, as we have to find the class mark of each interval and then multiply it with the frequency we will organize the data in the form of a table.
The table formed will be as follows.
| Age (in years) | No. of cases (\[f\]) | Class mark(\[x\]) | \[xf\] |
| \[5 - 14\] | 6 | 9.5 | 57 |
| \[15 - 24\] | 11 | 19.5 | 214.5 |
| \[25 - 34\] | 21 | 29.5 | 619.5 |
| \[35 - 44\] | 23 | 39.5 | 908.5 |
| \[45 - 54\] | 14 | 49.5 | 963 |
| \[55 - 64\] | 5 | 59.5 | 297.5 |
| \[65 - 74\] | 3 | 65.5 | 208.5 |
| \[\sum f = 83\] | \[\sum {xf} = 2998.5\] |
Now, we have the values of \[\sum {xf} \] and \[\sum f \], which we can substitute in the formula to get the mean as.
\[
\Rightarrow \overline x = \dfrac{{\sum {xf} }}{{\sum f }} \\
\Rightarrow \overline x = \dfrac{{2998.5}}{{83}} = 36.1265 \simeq 36.13 \\
\]
Thus, the mean for the given frequency distribution will be 36.13 years.
Hence, option (B) will be the correct option.
Note: As we know that the mean of a data set represents the most likely number in the data set, we can observe that here easily. Maximum number of patients have their ages in the range of \[35 - 44\], and the mean also comes out to be 36.13, which is very close to the class mark of this interval thus verifying its definition.
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