The aim of this question is to find the number of observations in the set which are less than its median value of $38$.
The concept behind this question is the Locator/ Percentile method. We are going to use the Locator/ Percentile method for finding the number of observations in the given five-number summary.
The five-number summary consists of these $5$ values: the minimum value, lower quartile $Q_1$, median $Q_2$, upper quartile $Q_3$, and the maximum value. These $5$ values divide the set of data into four groups with about $25%$ or $1/4$ of the data value in each group. These values are also used to create a box plot/ box and whisker plot. To determine lower quartile $Q_1$ and upper quartile $Q_3$, we will use the Locator/ Percentile method.
Expert Answer
The five-number summary of the total $33$ whole number observations set is given as:
\[[12,24,38,51,64]\]
The given data is in the ascending order, so we can determine the minimum value and the maximum value.
Here, the minimum value is $=12$.
The lower quartile $=Q_1=24$.
Now for the median, we know that for a data set having an odd total number, the position of the median value is found by dividing the total number of elements by $2$ and then rounding off to the next value. When the total value is even, then there is no median value. Instead, there is a mean value which is found by dividing the total number of values by two or by dividing the total number of values by two and adding one to it.
In our case as the total number of values is odd, which in the five-number summary is the middle value:
Median $=Q_2=38$
The upper quartile $=Q_3=51$
The maximum value is $=64$
As the data is divided into $4$ groups:
\[\dfrac{\left( 31-4\right)}{4}=8\]
\[=2\times 8\]
\[=16\]
Therefore, we have two groups less than the median and two groups more than the median.
Numerical Results
For the $33$ unique whole number set, we have two groups of observations which are less than the median of $38$ and two groups more than the median.
Example
Find the $5$ number summary for the given data:
\[[5,8.5,11.1,14.6,14.7,17.7,20.1,23.2,27.8]\]
The given data is in the ascending order, so we can determine the minimum value and the maximum value.
Here, the minimum value is $=5$.
For lower quartile, we know that:
\[L=0.25(N)=2.25\]
Rounding off, the $3rd$ value is our first quartile.
The lower quartile $=Q_1=11.1$.
In this case, as the total number of value is odd, so median value is total number of values divided by $2$.
\[Median=\frac {N}{2}\]
\[Median=\frac {9}{2}\]
\[Median=4.5\]
Rounding the value, we get $5^{th}$ value to be median.
Median $=Q_2=14.7$
For the upper quartile, we have:
\[L=0.75(N)=6.75\]
Rounding off, the $7^{th}$ value is our third quartile.
The upper quartile $=Q_3=20.1$.
The maximum value is $=27.8$.
Our five-number summary is given below:
\[[5,11.1,14.7,20.1,27.8]\]