Annual sales, in millions of dollars, for 21 pharmaceutical companies follow. 8408 1374 1872 8879 2459 11413 608 14138 6452 1850 2818 1356 10498 7478 4019 4341 739 2127 3653 5794 8305 a. Provide a five-number summary. b. Compute the lower and upper limits. c. Do the data contain any outliers?

The lowest value, median, Quartile-3, and maximum value in any data collection are all represented by the five-number summary.
When the data are sorted in ascending order, the median is the middle value. The median and middle value of the data distribution is known as the Quartile-1. The midway value of the highest value and median of the data distribution is known as the quartile-3.
Calculation: 
The lowest value of the data distribution is 608 and highest value is 14138. 
The median is the middle value of the data when arranged in ascending order which is ${\displaystyle {11}^{th}}$ value 
Median=4019 
First quartile is average of ${\displaystyle 5^{th}}$ and ${\displaystyle 6^{th}}$ values 
Quartile−1=1850+18722=1861 
Third Quartile is average of ${\displaystyle {16}^{th}}$ and ${\displaystyle {17}^{th}}$ values 
Quartile−3=8305+84042=8354.5 
Conclusion: 
The five number summary for annual sales of 21 pharmaceutical companies is 608, 1861, 4019, 8354.5, 14138. 
(b)Concept Used: 
The formula for boundaries of outliers are, 
Lower Boundary=Q1−1.5*IQR 
Upper Boundary=Q3+1.5*IQR 
The difference between third quartile and first quartile is Inter Quartile range. 
Calculation: 
Substituting the values in Inter quartile range formula, 
Inter Quartile​​​ Range=Q3−Q1=8354.5−1861=6493.5 
Substituting in the boundaries of outliers formula, 
Lower Boundary=Q1−1.5*IQR=1861−1.5*6493.5=−7880.25 
Upper Boundary=Q3+1.5*IQR 
=8354.5+1.5*6493.5=18094.75 
Conclusion: 
For 21 pharmaceutical businesses, the lowest and upper bounds for outliers of annual sales are -7880.25 and 18094.75, respectively. (b) For 21 pharmaceutical firms, the lowest and upper ranges for outliers of annual sales are -7880.25 and 18094.75, respectively. The bounds of outliers encompass all of the data points in the provided data. As a result, we may conclude that the data contains no outliers.

a)

Minimum: The minimum value is 608.

First Quartile: The first quartile is middle value of the data below the median, in the other words , value at the 25% of the sorted data array.

$Q_1=0.25\cdot 21=5.25\approx 6\Rightarrow Q_1=1872$

The percentile is not integer, in that case round up and read the value at the given place in the sorted array.

Median:Median is middle value of the sorted data array. In this case, there is odd number of data given, so the median is the middle value.

$Median=Q_2=4019$

Third Quartile: The third quartile is middle value of the data above the median, in the other words , value at the 75% of the sorted data array.

$Q_3=0.75\cdot 21=15.75\approx 16\Rightarrow Q_3=8305$

The percentile is not integer, in that case round up and read the value at the given place in the sorted array.

Maximum: The maximum value is 14138.

b)

To determine lower and upper limits, first find the interquartile range. IQR is the difference between third and first quartile.

$IQR=Q_3-Q_1=8305-1872=6433$

Lower limit is the first quartile value decreased by 1.5 times the IQR.

$LowerLimit=1872-1.5\cdot 6433=-7777.5$

Upper limit is the third quartile value increased by 1.5 times the IQR.

$UpperLimit=8305+1.5\cdot 6433=17954.5$

c)

The data does not cointain any outliers because there is no value that is below the lower limit nor above the upper limit.