Question

The marks scored by \[6\] students in a preparatory exam are given here:
\[56,71,49,90,75,88.\]
Find the median score and semi-quartile range.

Answer 1

Hint: Arrange the given data in ascending order and find the median using the formula of statistics. Then separate the upper half and lower half of the data and find the upper quartile and lower quartile which are the medians of upper half of the data and the lower half of the data respectively. Then find the interquartile range subtracting the upper quartile and lower quartile, then divide it by two to find the semi-interquartile range.

Complete step by step solution:
Given data,
\[56,71,49,90,75,88.\]
Arranging the data in ascending order to find the median, we get;
\[49,56,71,75,88,90\]
Number of terms in the list \[ = 6\]
Since the number of terms in the list \[ = n = 6\], which is even, we have the median in the given formula:
Median \[ = \dfrac{{{{\left( {\dfrac{n}{2}} \right)}^{th}}{\text{term}} + {{\left( {\dfrac{n}{2} + 1} \right)}^{th}}{\text{term}}}}{2}\]
Substituting the values, we get;
\[ \Rightarrow \] Median \[ = \dfrac{{{{\left( {\dfrac{6}{2}} \right)}^{th}}{\text{term}} + {{\left( {\dfrac{6}{2} + 1} \right)}^{th}}{\text{term}}}}{2}\]
Simplifying the above terms, we get;
\[ \Rightarrow \] Median \[ = \dfrac{{{3^{rd}}{\text{term}} + {4^{th}}{\text{term}}}}{2}\]
Substituting the third term and the fourth term, we get;
\[ \Rightarrow \] Median \[ = \dfrac{{71 + 75}}{2}\]
Simplifying the above terms, we get;
\[ \Rightarrow \] Median \[ = \dfrac{{146}}{2}\]
Dividing with the denominator, we get;
\[ \Rightarrow \] Median \[ = 73\]
Therefore, the median \[ = 73\].
Let us divide the given data into lower half data and upper half data.
Lower half data \[ = 49,56,71\]
Upper half data \[ = 75,88,90\]
Now, to find the lower quartile, we need to find the median of the lower half of the data.
Number of elements in the lower half of the data \[ = 3\] that is odd.
For odd number of elements, the median is given as follows:
Median \[ = {\dfrac{{n + 1}}{2}^{th}}\] term
Substituting the value of \[n\], we get;
Median \[ = {\dfrac{{3 + 1}}{2}^{th}}\] term
Simplifying the above equation, we get;
Median \[ = {\dfrac{4}{2}^{th}}\] term
Dividing the numerator with the denominator, we get;
Median \[ = {2^{nd}}\] term
Substituting the term in the above equation, we get;
Median \[ = 56\]
The Median of the lower half of the terms \[ = 56\]
That implies, lower quartile \[ = 56\]
Similarly, we find the upper quartile as given below:
Now, to find the upper quartile, we need to find the median of the upper half of the data.
Number of elements in the lower half of the data \[ = 3\] that is odd.
For odd number of elements, the median is given as follows:
Median \[ = {\dfrac{{n + 1}}{2}^{th}}\] term
Substituting the value of \[n\], we get;
Median \[ = {\dfrac{{3 + 1}}{2}^{th}}\] term
Simplifying the above equation, we get;
Median \[ = {\dfrac{4}{2}^{th}}\] term
Dividing the numerator with the denominator, we get;
Median \[ = {2^{nd}}\] term
Substituting the term in the above equation, we get;
Median \[ = 88\]
The Median of the upper half of the terms \[ = 88\]
That implies, upper quartile \[ = 88\]
The interquartile range is the difference between the upper quartile and the lower quartile. We get the interquartile range as given below:
Interquartile range \[ = \] upper quartile \[ - \] lower quartile
Substituting the values, we get;
\[ \Rightarrow \] Interquartile range \[ = 88 - 56\]
\[ \Rightarrow \] Interquartile range \[ = 32\]
The semi-interquartile range is half of the interquartile range, i.e.,
Semi-interquartile range \[ = \dfrac{{32}}{2}\]
\[ \Rightarrow \] Semi-interquartile range \[ = 16\]

$\therefore $ The semi-interquartile range \[ = 16\]

Note: We can observe that quartiles in statistics means the quantity which divides the data sets into four parts of more or less equal sizes when the data is computed in ascending order. They are used to summarize a group of numbers and work with data that is not symmetrically distributed.
Quartile deviation is also known as semi-interquartile range. The difference between the third and first quartiles is called interquartile range. The interquartile range may be taken as a measure of dispersion (that is the extent to which the values are spread out from the average).