Chapter 2.2, Problem 63E

(a)

To determine

To describe: the shape, center, and spread of the distribution of shark lengths

Answer to Problem 63E

Shape: symmetric and roughly bell-shaped.

Center: Mean is 15.5884 and median is 15.75

Spread: standard deviation is 2.5499

Explanation of Solution

Given:

18.7	12.3	18.6	16.4	15.7	18.3	14.6	15.8	14.9	17.6	12.1
16.4	16.7	17.8	16.2	12.6	17.8	13.8	12.2	15.2	14.7	12.4
13.2	15.8	14.3	16.6	9.4	18.2	13.2	13.6	15.3	16.1	13.5
19.1	16.2	22.8	16.8	13.6	131	15.7	19.7	18.7	13.2	16.8

Lengths of sharks are in feet of 44

Calculation:

Descriptive statistics and a histogram are provided below:

Variable	N	Mean	St.Dev.	Min	$Q_1$	M	$Q_3$	Max
Shark length	44	15.586	2.550	9.40	13.525	15.75	17.40	22.80

  

Shark lengths are roughly symmetric with a peak at 16 and vary from minimum value of 9.40 feet to maximum value 22.8 feet.

Shape: symmetric and roughly bell-shaped.

Center: Mean is 15.5884 and median is 15.75

Spread: standard deviation is 2.5499

Conclusion:

Shape: symmetric and roughly bell-shaped.

Center: Mean is 15.5884 and median is 15.75

Spread: standard deviation is 2.5499

(b)

To determine

To describe: the percent of observations that fall within one, two, and three standard deviations of the mean.

Answer to Problem 63E

One standard deviations = 68.2%

Two standard deviations = 95.5%

Three standard deviations = 100%

Explanation of Solution

Calculation:

From the above output,

Mean, $\overline{x}=15.586$

Sample standard deviation, $s=2.550$

  $\begin{array}{l}\overline{x}-s=13.036\\ \overline{x}+s=18.136\end{array}$

Percent of interval

  $\left(68.26\%\right) 68.2\%$

  $\begin{array}{l}\overline{x}-2s=10.487\\ \overline{x}+2s=20.686\end{array}$

Percent of interval

  $\left(95.44\%\right) 95.5\%$

  $\begin{array}{l}\overline{x}-3s=7.937\\ \overline{x}+3s=23.236\end{array}$

Percent of interval

  $\left(99.73\%\right) 100.0\%$

According to the 68-95-99.7 rule, approximately 68.2% of the data values fall within one standard deviation. 95.5% of the data values fall within two standard deviations, and 100% of the data values fall within three standard deviations of the mean.

Conclusion:

Therefore,

One standard deviations = 68.2%

Two standard deviations = 95.5%

Threestandard deviations = 100%

(c)

To determine

To construct: a Normal probability plot and interpret the plot.

Answer to Problem 63E

The pattern in the Normal probability plot is fairly linear and it is roughly normally distributed.

  

Explanation of Solution

Calculation:

A Normal probability plot from Minitab is shown below:

  

The plot is fairly linear except from one small shark and one large shark lengths, indicating that the Normal distribution is appropriate.

Conclusion:

Therefore, a Normal probability plot is plotted. The pattern in the Normal probability plot is fairly linear and it is roughly normally distributed.

(c)

To determine

To explain:whether the data are approximately Normal

Answer to Problem 63E

The data are approximately Normal

Explanation of Solution

Calculation:

Results from the parts (a), (b), and (c) indicates that shark lengths are approximately normal.

Approximately normal, because the normal probability plot was roughlynormal and the histogram were roughly bell-shaped.

Conclusion:

Therefore, the data are approximately Normal