Q. 3.183 - Elementary Statistics

Question

The Great White Shark. In an article titled "Great White. Deep Trouble" (National Geographic, Vol. $197\left(4\right)$, pp. $2-29$). Peter Benchley-the author of JAWS-discussed various aspects of the Great White Shark Carcharodon carcharias). Data on the number of pups borne in a lifetime by each of $80$ Great White Shark females are provided on the WeissStats site.

a. obtain and interpret the quartiles. 

b. determine and interpret the interquartile range.

c. find and interpret the five-number summary

d. identify potential outliers, if any.

e. obtain and interpret boxplot.

Step-by-Step Solution

Answer

(a) The quartiles are $6$, $7$, $8$

(b) The interquartile range is, $2$

(c) Five-number summary is, $3$, $6$, $7$, $8$, $12$

(d) Potential outliers is, $12$

(e) The required boxplot is given below.

Part (a) Step 1: Given information

We are given that,

Sorted data is given in the Weiss stats which are as follows,

$$\begin{gathered} 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, \\ 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, \\ 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12 \end{gathered}$$

Part (a) Step 2: Simplify

As we know that median is the middle value of a sorted data set. Since the number of data values is even, the median is the average of two middle values:-

    $Q_2=\frac{7+7}{2}=\frac{14}{2}=7$

So, roughly $50\%$ the number of pups born below or equal to $7$ pups.

Now, the first quartile is the median of values below $Q_2$ i.e.

       $Q_1=6$

So, roughly $25\%$ the number of pups born below or equal to $6$ pups

Now,  the third quartile is the median of values below  $Q_2$

        $Q_3=8$

So, roughly $75\%$ the number of pups born below or equal to $8$ pups

Part (b) Step 1: Given information

We need to find out the interquartile range

Part (b) Step 2: Simplify

The interquartile range is the difference between $Q_1$ and $Q_3$

     $IQR=Q_3-Q_1=8-6=2$

The middle $50\%$ of the no. of pups born vary about $2$ pups.

Part (c) Step 1: Given information

We need to find out the five-number summary

Part (c) Step 2: Explanation

The five-number summary is minimum$=3$, first quartile$Q_1=6$, second quartile $Q_2=7$, third quartile $Q_3=8$ and maximum$=12$.

Part (d) Step 1: Given information

We need to find out the potential outliers.

Part (d) Step 2: Simplify

An outlier is more than $1.5 IQR$ or greater than $Q_3$ or less than $Q_1$

Therefore,

        $$\begin{gathered} Q_3+1.5 IQR=8+\left(1.5\times 2\right)=11 \\ {Q}_1-1.5 IQR=6-\left(1.5\times 2\right)=3 \end{gathered}$$

Hence, there is one outlier i.e. $12$ because it does not lie between $3$ and $11$

Part (e) Step 1: Given information

We need to find the boxplot which is given below

Part(e) Step 2: Simplify

The whiskers of the boxplot are at a low and high value. The box starts at $Q_1$ and ends at $Q_3$ and has a straight line at the median.