Question

Big sharks Here are the lengths (in feet) of 44 great white sharks:

A dot plot of the data and summary statistics are shown below. Is this distribution of shark length approximately Normal? Justify your answer based on the graph and the 68–95–99.7 rule.

n	Mean	SD	Min	Q1	Med	Q3	Max
44	15586	2.55	9.4	13.55	15.75	17.2	22.8

Short answer

The distribution of shark length is approximately normal.

Step-by-step solution

Given information 

Concept

Use of 68 − 95 − 99.7 rule:

Calculation

According to $68–95–99.7$the rule:

In a normal distribution, $68$ percent of the data lies within $1$ standard deviation of the mean.

A normal distribution has $95$ percent of its data within two standard deviations of the mean.

A normal distribution has $99.7\%$ of its data inside $1$ standard deviation of the mean.

Then

The general Normal density graph is represented as:

Calculation

Note that

$30$ of the $44$ observations in the sample are within $1$ standard deviation of the mean.

That means

Within one standard deviation, $68.18$ percent of the observations in the sample are within $1$ standard deviation.

Between

$\overline{x}-s_x=15.586-2.55=13.036$

And

$\overline{x}+s_x=15.586+2.55=18.136$

Also, note that

$42$ of the $44$ observations in the sample are within two standard deviations of the mean.

That means

The sample has $95.45\%$ observations within $2$ standard deviations.

Between

$\overline{x}-2s_x=15.586-2(2.55)=10.486$

And

$\overline{x}+2s_x=15.586+2(2.55)=20.686$

Also, note that

$44$ of the $44$ observations in the sample are within three standard deviations of the mean.

That means

The sample has $100\%$ observations within $3$ standard deviations.

Between

$\overline{x}-3s_x=15.586-3(2.55)=7.936$

And

$\overline{x}+3s_x=15.586+3(2.55)=23.236$

We have that

$68.18\%$ is very close to $68\%$

$95.45\%$ is very close to $95\%$

$100\%$ is very close to $99.7\%$

This implies

The data reasonably follows $68-95-99.7$ rule.

Thus,

The distribution is approximately normal.