Question

The runs scored in a cricket match by 11 players is as follows:
9,15,121,51,101,81,50,16,82,11,11
Find the mean, mode and median respectively of this data.\[\]
A.48, 11, 51\[\]
B.48, 50, 11\[\]
C.48, 11, 50\[\]
D.49, 81, 11\[\]

Answer 1

Hint: We recall the definitions mean median and mode. We find the mean of the data by dividing the sum of the data values by number data values. We find the median finding the data values in descending order of data values. We find the mode as the data values which appear most number of times in the data sample. \[\]

Complete step by step answer:
We know that if ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ is a data sample with $n$ data values then sample mean is given by
\[\overline{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}=\dfrac{S}{n}\]
Here $S$ is the sum of data values. We find the median of the data value by the term in middle position. If the data values ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ are arranged in ascending order, the median is
\[m=\left\{ \begin{matrix}
   {{x}_{\dfrac{n+1}{2}}} & \text{if }n\text{ is odd} \\
   \dfrac{1}{2}\left( {{x}_{\dfrac{n}{2}}}+{{x}_{\dfrac{n}{2}+1}} \right) & \text{if }n\text{ is even} \\
\end{matrix} \right.\]
The mode of the data values is the highest data value in the data sample if all the data values appear same number of times. If data values of do not occur same number of time then the mode is the data value that occur highest number of times. If the ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ occur ${{f}_{1}},{{f}_{2}},...,{{f}_{n}}$ number of times then the mode is
\[M={{x}_{M}},{{f}_{M}} > {{f}_{i}},\forall i=1,2,3,..n\]
We are given the data sample of runs scored in a cricket match by 11 players as follows 9,15,121,51,101,81,50,16,82,11,11. Let us arrange them in ascending order and have the data the sample as 9, 11, 11, 15, 16, 50, 51, 81, 82, 101 and 121. \[\]
Let us find the mean first. We have the sum of data values as
\[S=9+11+11+15+16+50+51+81+82+101+121=548\]
We divide it by number of data values $n=11$ and find the mean as
\[\overline{x}=\dfrac{S}{n}=\dfrac{548}{11}=48\]
We now find the mode in accordance with the question. We see that all data values except 11 occur once while 11 occur 2 times in the data sample. So the data values that occur most number of times is 11. So the mode is,
\[M=11\]
We see that the number of data values $n=11$ is an odd number. So the median will be the data the value at ${{\left( \dfrac{n+1}{2} \right)}^{\text{th}}}={{\left( \dfrac{11+1}{2} \right)}^{\text{th}}}={{6}^{\text{th}}}$ position which is 50. So we have
\[m=50\]

So, the correct answer is “Option C”.

Note: We need to be careful that we are asked to find the order mean, mode and median not in mean, median, mode. If the 11 would have occurred once, we would have 121 as the mode. The number times a data value occurs is called its frequency. If all the data values are frequented we find mean, median, mode from frequency table.