Chapter 2.2, Problem 76E

To determine

Whether the distribution of shark length approximately normal.

Answer to Problem 76E

The distribution of shark length is approximately normal.

Explanation of Solution

Given information:

  

According to 68 − 95 − 99.7 rule:

68% of the data of a normal distribution lies with 1 standard deviation from the mean.

95% of the data of a normal distribution lies with 2 standard deviation from the mean.

99.7% of the data of a normal distribution lies with 1 standard deviation from the mean.

Then

The general Normal density graph is represented as:

  

Note that

The sample has 30 out of 44 observations within 1 standard deviation of the mean.

That means

The sample has 68.18% observations 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

The sample has 42 out of 44 observations within 2 standard deviation of the mean.

That means

The sample has 95.45% observations within 2 standard deviation.

Between

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

And

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

Also note that

The sample has 44 out of 44 observations within 3 standard deviation of the mean.

That means

The sample has 100% observations within 3 standard deviation.

Between

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

And

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

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.