
The runs scored in a cricket match by \[11\] players are as follows:\[6,15,120,50,100,80,10,15,8,10,15\] Find the mean, mode and median of this data. Are the three the same?
Answer
475.8k+ views
Hint: Here, we have to arrange the given data in ascending order
Then we use the respective formula and respective mean, mode and median.
Finally we get the required answer.
Formula used: Mean, $\bar x = \dfrac{{\sum x }}{n}$
\[{\text{Median=}}{\left({\dfrac{{{\text{n+1}}}}{{\text{2}}}}\right)^{{\text{th}}}}{\text{term}}\], if \[n\] is odd
Complete step-by-step solution:
It is given that the data \[6,15,120,50,100,80,10,15,8,10,15\]
We have to arrange the given data in ascending order as follows
$6,8,10,10,15,15,15,50,80,100,120$
Now we have to find one by one,
First we have to find the mean value,
So, we have to add all the observations and divide it by the total number of observations.
Now we use the formula for mean, $\bar x = \dfrac{{\sum x }}{n}$
Here, $\sum x $ is the sum of all observations and $n$ is the total number of observations.
So, we can write it as,
$\bar x = \dfrac{{6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120}}{{11}}$
On adding we get,
$\bar x = \dfrac{{429}}{{11}}$
Let us divide the terms and we get,
$ \Rightarrow \bar x = 39$
Hence, the mean value is $39$.
Secondly we have to find the mode value; we have to find the most occurring observation.
Here, \[15\] occurs \[3\] times.
Hence, the value of mode is $15$.
Finally we have to find the median value; we have to find whether the total number of observations is odd or even.
Here, the total number of observations is \[11\] which is odd.
Now we use the formula is
\[{\text{Median=}}{\left({\dfrac{{{\text{n+ 1}}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{term}}\]
So, \[{\text{Median = }}{\left( {\dfrac{{{\text{11 + 1}}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{ = }}{\left( {\dfrac{{{\text{12}}}}{{\text{2}}}} \right)^{{\text{th}}}}\]
\[{\text{ = }}{{\text{6}}^{{\text{th}}}}{\text{ term}}\]
Now, the \[{{\text{6}}^{{\text{th}}}}{\text{ term}}\] of the given data is \[15\].
Now, we have Mean = \[39\]
Mode = \[15\]
Median = \[15\]
Here, the mean value is different from the value of mode and median but mode and median is the same.
\[\therefore \] The three values are not the same.
Note: Mean value is always found to determine the average value of observations.
While finding the values of median, it is essential to see whether the total number of observations is odd or even.
If it is odd, the formula can be written as \[{\text{Median = }}{\left( {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{observation}}\].
If it is even, the formula can be written as \[{\text{Median = }}{\left( {\dfrac{{\text{n}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{observation}}\].
Finally mode is the frequently occurring observation.
Then we use the respective formula and respective mean, mode and median.
Finally we get the required answer.
Formula used: Mean, $\bar x = \dfrac{{\sum x }}{n}$
\[{\text{Median=}}{\left({\dfrac{{{\text{n+1}}}}{{\text{2}}}}\right)^{{\text{th}}}}{\text{term}}\], if \[n\] is odd
Complete step-by-step solution:
It is given that the data \[6,15,120,50,100,80,10,15,8,10,15\]
We have to arrange the given data in ascending order as follows
$6,8,10,10,15,15,15,50,80,100,120$
Now we have to find one by one,
First we have to find the mean value,
So, we have to add all the observations and divide it by the total number of observations.
Now we use the formula for mean, $\bar x = \dfrac{{\sum x }}{n}$
Here, $\sum x $ is the sum of all observations and $n$ is the total number of observations.
So, we can write it as,
$\bar x = \dfrac{{6 + 8 + 10 + 10 + 15 + 15 + 15 + 50 + 80 + 100 + 120}}{{11}}$
On adding we get,
$\bar x = \dfrac{{429}}{{11}}$
Let us divide the terms and we get,
$ \Rightarrow \bar x = 39$
Hence, the mean value is $39$.
Secondly we have to find the mode value; we have to find the most occurring observation.
Here, \[15\] occurs \[3\] times.
Hence, the value of mode is $15$.
Finally we have to find the median value; we have to find whether the total number of observations is odd or even.
Here, the total number of observations is \[11\] which is odd.
Now we use the formula is
\[{\text{Median=}}{\left({\dfrac{{{\text{n+ 1}}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{term}}\]
So, \[{\text{Median = }}{\left( {\dfrac{{{\text{11 + 1}}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{ = }}{\left( {\dfrac{{{\text{12}}}}{{\text{2}}}} \right)^{{\text{th}}}}\]
\[{\text{ = }}{{\text{6}}^{{\text{th}}}}{\text{ term}}\]
Now, the \[{{\text{6}}^{{\text{th}}}}{\text{ term}}\] of the given data is \[15\].
Now, we have Mean = \[39\]
Mode = \[15\]
Median = \[15\]
Here, the mean value is different from the value of mode and median but mode and median is the same.
\[\therefore \] The three values are not the same.
Note: Mean value is always found to determine the average value of observations.
While finding the values of median, it is essential to see whether the total number of observations is odd or even.
If it is odd, the formula can be written as \[{\text{Median = }}{\left( {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{observation}}\].
If it is even, the formula can be written as \[{\text{Median = }}{\left( {\dfrac{{\text{n}}}{{\text{2}}}} \right)^{{\text{th}}}}{\text{observation}}\].
Finally mode is the frequently occurring observation.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

Raindrops are spherical because of A Gravitational class 11 physics CBSE

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE
