91Ó°ÊÓ

Given that,

$\begin{array}{|ccccccc|}\hline 4& 4& 12& 18& 9& 6& 12\\ 3& 6& 15& 7& 3& 55& 1\\ 10& 13& 5& 7& 1& 23& 9\\ \hline\end{array}$

First, sort the information in ascending order.

$\begin{array}{|lllllllllllllllllllll|}\hline 1& 1& 3& 3& 4& 4& 5& 6& 6& 7& 7& 9& 9& 10& 12& 12& 13& 15& 18& 23& 55\\ \hline\end{array}$

The total number of observations is21.We have the median is the middle term of the data. So the median is 7

The data set's second quartile is$Q_2=7$

Consider the first part of the entire data set that is at or below the median of the entire data set is:

$\begin{array}{|lllllllllll|}\hline 1& 1& 3& 3& 4& 4& 5& 6& 6& 7& 7\\ \hline\end{array}$

The median is the middle term of the supplied data because there are 11 observations.

The data set's first quartile is$Q_1=4$

Consider the second part of the complete data set, which is at or below the data set's median.

$\begin{array}{|lllllllllll|}\hline 7& 9& 9& 10& 12& 12& 13& 15& 18& 23& 55\\ \hline\end{array}$

The Number of observations is 11 . So the median is the middle term of the given data.

The third quartile of the data set is$Q_3=12$

Therefore, the third quartile of the data is 12

It can be seen from the above result that 25% of the length of stay by patients is fewer than 12 days.

Given that,

$\begin{array}{|ccccccc|}\hline 4& 4& 12& 18& 9& 6& 12\\ 3& 6& 15& 7& 3& 55& 1\\ 10& 13& 5& 7& 1& 23& 9\\ \hline\end{array}$

Calculation of the inter quartile range

The difference between the first and third quartiles is the inter quartile range$(IQR).$

$IQR=Q_3-Q_1$

= 12 - 4

= 8

Patients spend the middle 50% of their time in the hospital for 12 days.

As a result, the interquartile range is eight.

Given that,

$\begin{array}{|ccccccc|}\hline 4& 4& 12& 18& 9& 6& 12\\ 3& 6& 15& 7& 3& 55& 1\\ 10& 13& 5& 7& 1& 23& 9\\ \hline\end{array}$

Calculation of the five numbers summery

The data set's minimum value is 1.

The data set's lower quartile is$Q1=4.$

$Q2=7$is the median of the data set.

The data set's top quartile is $Q3=12$

The provided data set's maximum value is 55.

The measure of variation of the middle quarter is

$=Q_2-Q_1$

= 7 - 4

= 3

The measure of variation of the third quarter is

$=Q_3-Q_2$

= 12 - 7

= 5

The Variation of the first quarter is

$=Q_1-\min$

= 4 - 1

= 3

The Variation of the fourth quarter is

$=\max -Q_3$

= 55 - 1 2

= 43

It can be seen from the given results that the first, second, and third quartiles have less fluctuation. However, there is a lot of variety in the fourth quarter.

Given in the question that, A random sample of 21 patients yielded the following data on length of stay, in days.

$\begin{array}{|ccccccc|}\hline 4& 4& 12& 18& 9& 6& 12\\ 3& 6& 15& 7& 3& 55& 1\\ 10& 13& 5& 7& 1& 23& 9\\ \hline\end{array}$

Let's compute the lower and upper limit of data set.

Lower limit $=Q_1-1.5\left(IQR\right)$

= 14 - 1.5(8)

= -8

Upper limit $=Q_3+1.5\left(IQR\right)$

= 12 + 1.5 (8)

= 24

Potential outliers are observations that fall below or over the lower or higher limits.

According to the statistics, the highest limits are set at 55.

As a result, the possible outlier is 55.

Interpretation: The length of 55 days spent by patients falls outside the data sheet's trend.

Consider the given table,

$\begin{array}{|ccccccc|}\hline 4& 4& 12& 18& 9& 6& 12\\ 3& 6& 15& 7& 3& 55& 1\\ 10& 13& 5& 7& 1& 23& 9\\ \hline\end{array}$

We use MINITAB, for the box plot

The right whisker and asterisks represent the fourth quarter spread, as seen in the above box plot.