
ASSERTION: The arithmetic mean of the following given frequency distribution table is \[13.81\].
\[x\] \[4\] \[7\] \[10\] \[13\] \[16\] \[19\] \[f\] \[7\] \[10\] \[15\] \[20\] \[25\] \[30\]
REASON: Arithmetic Mean can be found using the formula \[\overline x = \dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\]
\[A.)\] Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
\[B.)\] Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
\[C.)\] Assertion is correct but Reason is incorrect.
\[D.)\] Assertion is incorrect but Reason is correct.
| \[x\] | \[4\] | \[7\] | \[10\] | \[13\] | \[16\] | \[19\] |
| \[f\] | \[7\] | \[10\] | \[15\] | \[20\] | \[25\] | \[30\] |
Answer
569.7k+ views
Hint: First of all, we need to calculate the arithmetic mean of the given frequency distribution using a general formula. And, then we will tally the arithmetic mean with the given statement.
Complete step by step solution:
It is given that the question is stated as, the frequency distribution table.
Also, we can arrange the given table as follow:
Here we have to multiply \[{x_i}\] and \[{f_i}\], we can write it as in the table.
As it is an ungrouped data set, we need to multiply each data value by the frequency and then up to calculate the total.
Then it can be divided by the total of frequencies to get the arithmetic mean of the frequency distribution.
So, we can write it as the sum of the multiplication value of \[{x_i}\] and \[{f_i}\]
\[\sum {f{}_i{x_i}} = 28 + 70 + 150 + 260 + 400 + 570 = 1478.\]
Also, we can add the function value,
\[\sum {{f_i} = } 7 + 10 + 15 + 20 + 25 + 30 = 107.\]
If we divide \[\sum {{f_i}{x_i}} \]by \[\sum {{f_i}} \] then we will get the required arithmetic mean.
So, the required arithmetic mean would be \[ = \] \[\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\]
Putting the values and we get
\[ \Rightarrow \dfrac{{1478}}{{107}}\]
Let us divide the term and we can write it as approximately, we get
\[ \approx 13.81\]
So, we can say that both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
\[\therefore \] Option A is the correct choice.
Note: Always remember that an assertion is a forceful statement or fact. Every fact has its own reason to prove it right. We have to look around for that reason only.
Arithmetic mean of any frequency distribution is depending on the nature of the data.
And, always check if the distribution of data is grouped data or ungrouped data.
Complete step by step solution:
It is given that the question is stated as, the frequency distribution table.
Also, we can arrange the given table as follow:
| \[{x_i}\] | \[{f_i}\] | \[({f_i} \times {x_i})\] \[{f_i} \times {x_i}\] |
| \[4\] | \[7\] | \[28\] |
| \[7\] | \[10\] | \[70\] |
| \[10\] | \[15\] | \[150\] |
| \[13\] | \[20\] | \[260\] |
| \[16\] | \[25\] | \[400\] |
| \[19\] | \[30\] | \[570\] |
Here we have to multiply \[{x_i}\] and \[{f_i}\], we can write it as in the table.
As it is an ungrouped data set, we need to multiply each data value by the frequency and then up to calculate the total.
Then it can be divided by the total of frequencies to get the arithmetic mean of the frequency distribution.
So, we can write it as the sum of the multiplication value of \[{x_i}\] and \[{f_i}\]
\[\sum {f{}_i{x_i}} = 28 + 70 + 150 + 260 + 400 + 570 = 1478.\]
Also, we can add the function value,
\[\sum {{f_i} = } 7 + 10 + 15 + 20 + 25 + 30 = 107.\]
If we divide \[\sum {{f_i}{x_i}} \]by \[\sum {{f_i}} \] then we will get the required arithmetic mean.
So, the required arithmetic mean would be \[ = \] \[\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\]
Putting the values and we get
\[ \Rightarrow \dfrac{{1478}}{{107}}\]
Let us divide the term and we can write it as approximately, we get
\[ \approx 13.81\]
So, we can say that both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
\[\therefore \] Option A is the correct choice.
Note: Always remember that an assertion is a forceful statement or fact. Every fact has its own reason to prove it right. We have to look around for that reason only.
Arithmetic mean of any frequency distribution is depending on the nature of the data.
And, always check if the distribution of data is grouped data or ungrouped data.
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