Chapter 3.1, Problem 33E

To determine

To describe: the shape of the distribution and any outliers.

Answer to Problem 33E

Shape: right skewed

Outliers: no

Center: median is 5.6mg

Spread: interquartile range is 5.5mg

Explanation of Solution

Given:

  

Calculation:

Consider the below table, for weights (in milligrams) of 58 diamonds.

Class Interval	Tally Marks	x	Frequency (f)	Cumulative frequency(c.f.)
0-2	|||| ||	1	7	7
2-4	|||| |||| ||||	3	14	21
4-6	|||| ||||	5	10	31
6-8	|||| ||||	7	10	41
8-10	||||	9	4	45
10-12	|||	11	3	48
12-14	|	13	1	49
14-16	|	15	1	50
16-18		17	0	50
18-20	|||	19	3	53
20-22	|	21	1	54
22-24	|	23	1	55
24-26	|	25	1	56
26-28	|	27	1	57
28-30		29	0	57
30-32		31	0	57
32-34	|	33	1	58
		Total	58

A histogram shows that the data are skewed right, not symmetric. Hence, the distribution of weights of these diamonds is skewed to right. The median would then be a better measure of the center.

It is clear in histogram that class 18-20 is an outlier because it that falls outside the overall pattern. The spread of a data set refers roughly to how wide be the data relative to its center. The range is the difference between the maximum value and the minimum value in the data, or simply those two numbers. The spread is the range of values from the lowest value to the largest value. The spread is a more precise measure than the center.

Range= Highest value- Lowest value

  $\begin{array}{l}=33.8–0.1\hfill \\ =33.7\hfill \end{array}$

  $\begin{array}{l}N=58\\ \frac{N}{2}=\frac{58}{2}\\ =29\end{array}$

Cumulative frequency just greater than 29, is 31 and the class corresponding to 31 is 4-6 is median class. Median is obtained by the formula:

  $Median=l+\frac{h}{f}\left(\frac{N}{2}-c\right)$

Where:

l is the lower limit of the median,

f is the frequency of the median class,

h is the magnitude of the median class,

c is the c.f. of the class preceding the median class,

and $N={\displaystyle \sum f}$

  $\begin{array}{l}Median=4+\frac{2}{10}\left(\frac{58}{2}-21\right)\\ =5.6\end{array}$

Consider the following table:

Class Interval	x	f	Cumulative frequency	fx
0-2	1	7	7	7
2-4	3	14	21	42
4-6	5	10	31	50
6-8	7	10	41	70
8-10	9	4	45	36
10-12	11	3	48	33
12-14	13	1	49	13
14-16	15	1	50	15
16-18	17	0	50	0
18-20	19	3	53	57
20-22	21	1	54	21
22-24	23	1	55	23
24-26	25	1	56	25
26-28	27	1	57	27
28-30	29	0	57	0
30-32	31	1	57	0
32-34	33	1	58	33
	Total	${\displaystyle \sum f}$ = 58		${\displaystyle \sum f_x=452}$

  $\begin{array}{l}Mean=\frac{{\displaystyle \sum f_x}}{{\displaystyle \sum f}}\\ =\frac{452}{58}=7.79\end{array}$

  $\begin{array}{l}{\text{Skewness(S}}_{\text{k}}\text{)=Mean-Median}\\ \begin{array}{l}\text{= 7}\text{.79 – 5}\text{.6}\hfill \\ \text{ =2}\text{.19}\hfill \end{array}\end{array}$

The first quartile is the median of the data values below the median. Since there are 29 data values below the median, the first quartile is the ${\text{15}}^{\text{th}}$ data value$Q_1=3.5$

The third quartile is the median of the data values above the median. Since there are 29 data values above the median, the first quartile is the ${\text{15}}^{\text{th}}$ data value above the median, which corresponds with the $29+15={44}^{th}$ data value in the sorted data set $Q_3=9$

The interquartile range IQR is the difference of the third and first quartile.

  $\begin{array}{l}IQR=Q_3-Q_1\\ =9-3.5\\ =5.5\end{array}$

Shape: right skewed, because the highest bar in the histogram is to the left and the tall to the right.

Outliers: no, because there are no gaps in the histogram between bars.

Center: median is 5.6mg

Spread: interquartile range is 5.5mg

Conclusion:

Hence,

Shape: right skewed

Outliers: no

Center: median is 5.6mg

Spread: interquartile range is 5.5mg