Question

In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows 60, 70, 90, 95, 50, 65, 70, 80, 85, 95. Find the median of the weights of tomatoes. Prepare a frequency distribution table for the data.

Answer 1

Hint: We arrange the given data values tomato weights in the question in ascending order and where the number of data values is $n=10$. If $n$ is odd we find the median as the data a value at ${{\left( \dfrac{n+1}{2} \right)}^{\text{th}}}$ position in ascending order and if $n$ is even then we take the average of data values at ${{\left( \dfrac{n}{2} \right)}^{\text{th}}}$ and ${{\left( \dfrac{n}{2}+1 \right)}^{\text{th}}}$ position. We recall the definition of frequency as the number of times a data value appears in the data sample and draw the frequency distribution table.

Complete step-by-step answer:
We know that a median of a data set is any value such that at most half of the data set is less than the proposed median and at most half is greater than the proposed median. If there are $n$ data values say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ arranged in ascending order and $n$ is odd then median is the data a value at ${{\left( \dfrac{n+1}{2} \right)}^{\text{th}}}$ position
\[m={{x}_{\dfrac{n+1}{2}}}\]
If $n$ is even , then the median is average of two middle values at ${{\left( \dfrac{n}{2} \right)}^{\text{th}}}$ and ${{\left( \dfrac{n}{2}+1 \right)}^{\text{th}}}$ position.
\[m=\dfrac{1}{2}\left( {{x}_{\dfrac{n}{2}}}+{{x}_{\dfrac{n}{2}+1}} \right)\]
Let us observe the data set given in the question as the weights of 10 tomatoes 60, 70, 90, 95, 50, 65, 70, 80, 85, 95. So the number of data values is odd and we have
\[n=10\]
So we need to arrange the data in ascending order. We have 50, 60, 65, 70, 70, 80, 85, 90, 95, 95. We can denote them as
\[{{x}_{1}}=50,{{x}_{2}}=60,...,{{x}_{10}}=95\]
We have to find the two middle values as the number of data values is even. The position of the middle values in the arranged data in ascending order is
\[\begin{align}
  & {{\left( \dfrac{n}{2} \right)}^{\text{th}}}={{\left( \dfrac{10}{2} \right)}^{\text{th}}}={{5}^{\text{th}}} \\
 & {{\left( \dfrac{n}{2}+1 \right)}^{\text{th}}}={{\left( \dfrac{10}{2}+1 \right)}^{\text{th}}}={{\left( 5+1 \right)}^{\text{th}}}={{6}^{\text{th}}} \\
\end{align}\]
The data values at the ${{5}^{\text{th}}}$ and ${{6}^{\text{th}}}$positions are
\[{{x}_{5}}=70,{{x}_{6}}=80\]
The median is the average of the middle values. We take the average of data values at the ${{5}^{\text{th}}}$ and ${{6}^{\text{th}}}$positions and have the median as
\[m=\dfrac{{{x}_{5}}+{{x}_{6}}}{2}=\dfrac{70+80}{2}=\dfrac{150}{2}=75\]
So the median weight of the tomatoes is 75. We know that the number of times a data value appears in the data sample is called the frequency of that data value. We see that in the ascending ordered data sample only 70 appears twice and the rest of data values appear once. So we draw the table with data values and their corresponding frequencies row by row. \[\]

Weight of the tomatoes (in grams)	Frequency
50	1
60	1
65	1
70	2
80	1
85	1
90	1
95	2

Note: We need to be careful of the confusion between median, mean and mode. Mean is the sum values divided by number of values and mode is the highest value. If we divide the whole data into four parts having equal numbers of data values, the median is called the second quartile.