Summary statistic for ungrouped data
Summary statistic summarize and provide information about the sample data. It includes the minimum value of the data, first quartile ($Q_1$), median (i.e., $Q_2$), mean ($\overline{x}$), third quartile ($Q_3$) and the minimum value of the data.
Summary statistic includes
- minimum value ($\min$),
- first quartile ($Q_1$),
- $\text{median }$ ($Q_2$),
- sample mean ($\overline{x}$),
- third quartile ($Q_3$),
- maximum value ($\max$).
Formula
$\min$, $Q_1$, $\text{median}$, $\overline{x}$, $Q_3$ and $\max$
The mean of $X$ is denoted by $\overline{x}$ and is given by
$\overline{x} =\dfrac{1}{n}\sum_{i=1}^{n}x_i$
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$
where $n$ is the total number of observations.
Example 1
The age (in years) of 6 randomly selected students from a class are
22,25,24,23,24,20.
Find the summary statistics of the data.
Solution
Arrange the data in ascending order
20, 22, 23, 24, 24, 25.
Minumum Value
The minimum age of students is $\min = 20$ years.
Maximum Value
The maximum age of students is $\max = 25$ years.
Sample mean
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{138}{6}\\ &=23\text{ years} \end{aligned} $$
The sample mean of age of students is $23$ years.
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$
where $n$ is the total number of observations.
First Quartile $Q_1$
The first quartle $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(6+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(1.75\big)^{th} \text{ observation}\\ &= \text{Value of }\big(1\big)^{th} \text{ obs.}+0.75 \big(\text{Value of } \big(2\big)^{th}\text{ obs.}-\text{Value of }\big(1\big)^{th} \text{ obs.}\big)\\ &=20+0.75\big(22 -20\big)\\ &=21.5 \text{ years} \end{aligned} $$
Thus, $25$ % of the students had age less than or equal to $21.5$ years.
Median (M) (i.e., Second Quartile $Q_2$)
The median ($M$) or second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} M &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(6+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(3.5\big)^{th} \text{ observation}\\ &= \text{Value of }\big(3\big)^{th} \text{ obs.}+0.5 \big(\text{Value of } \big(4\big)^{th}\text{ obs.}-\text{Value of }\big(3\big)^{th} \text{ obs.}\big)\\ &=23+0.5\big(24 -23\big)\\ &=23.5 \text{ years} \end{aligned} $$
Thus, $50$ % of the students had age less than or equal to $23.5$ years.
Third Quartile $Q_3$
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(6+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{ Value of }\big(5.25\big)^{th} \text{ observation}\\ &= \text{Value of }\big(5\big)^{th} \text{ obs.}+0.25 \big(\text{Value of } \big(6\big)^{th}\text{ obs.}-\text{Value of }\big(5\big)^{th} \text{ obs.}\big)\\ &=24+0.25\big(25 -24\big)\\ &=24.25 \text{ years} \end{aligned} $$
Thus, $75$ % of the students had age less than or equal to $24.25$ years.
Thus the summary statistics of given data set is
$\min = 20$ years, $Q_1 = 21.5$ years, $\text{median }=23.5$ years,$\text{mean }=23$ years, $Q_3=24.25$ years and $\max = 25$ years.
Example 2
A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.
5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12.
Find the summary statistics of the data.
Solution
Arrange the data in ascending order
5, 6, 8, 9, 9, 10, 10, 10, 10, 12, 12, 13, 13, 14, 15.
Minumum Value
The minimum age of students is $\min = 5$ days.
Maximum Value
The maximum age of students is $\max = 15$ days.
Sample mean
The sample mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{156}{15}\\ &=10.4\text{ days} \end{aligned} $$
The sample mean of length of stay in the hospital of students is $10.4$ days.
Quartiles
The formula for $i^{th}$ quartile is
$Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$
where $n$ is the total number of observations.
First Quartile $Q_1$
The first quartle $Q_1$ can be computed as follows:
$$ \begin{aligned} Q_{1} &=\text{Value of }\bigg(\dfrac{1(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{1(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(4\big)^{th} \text{ observation}\\ &=9 \text{ days} \end{aligned} $$
Thus, $25$ % of the patients had length of stay in the hospital less than or equal to $9$ days.
Second Quartile $Q_2$
The second quartile $Q_2$ can be computed as follows:
$$ \begin{aligned} Q_{2} &=\text{Value of }\bigg(\dfrac{2(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{2(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(8\big)^{th} \text{ observation}\\ &=10 \text{ days} \end{aligned} $$
Thus, $50$ % of the patients had length of stay in the hospital less than or equal to $10$ days.
Third Quartile $Q_3$
The third quartile $Q_3$ can be computed as follows:
$$ \begin{aligned} Q_{3} &=\text{Value of }\bigg(\dfrac{3(n+1)}{4}\bigg)^{th} \text{ observation}\\ &=\text{Value of }\bigg(\dfrac{3(15+1)}{4}\bigg)^{th} \text{ observation}\\ &= \text{Value of }\big(12\big)^{th} \text{ observation}\\ &=13 \text{ days} \end{aligned} $$
Thus, $75$ % of the patients had length of stay in the hospital less than or equal to $13$ days.
Thus the summary statistics of given data set is
$\min = 5$ days, $Q_1 = 9$ days, $\text{median }=10$ days,$\text{mean }=10.4$ days, $Q_3=13$ days and $\max = 15$ days.
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