Chapter 7.CR, Problem 7CR

To determine

To find:

The probability that the mean length of an adult blue whale from a random sample.

Answer to Problem 7CR

Solution:

The probability that the mean length of an adult blue whale from a random sample of 31 whales differs from the population mean by less than 2 meters is 0.9494.

Explanation of Solution

Given:

$n=31$

$\mu =30.0$

$\sigma =5.7$

$\text{difference}\le 2$

Formula:

In statistics, the normal distribution probability density function is given by:

$f\left(x|\mu ,{\sigma }^2\right)=\frac{1}{\sqrt{2\pi {\sigma }^2}}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Where, $x$ denotes random variable value while $\mu$ is the mean and the variance is ${\sigma }^2$.

The normal Gaussian distribution with the following characteristics:

$\mu =0$, and $\sigma =1$

is called standard normal distribution with probability density function:

$f\left(x|0,1\right)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

When this function is plotted on the $x$-axis, the area under between the curve and $x$-axis is equal to one, because the sum of probabilities is unity.

Area on the left of given $z$-value (say $z=z_a$) is the probability $P\left(X\le a\right)$ given by the integration:

${\displaystyle \underset{-\infty }{\overset{a}{\int }}f\left(x\right)dx}$.

A normal distribution can be converted to standard normal distribution by defining the $z$-score that is represented by using the following transformation:

$z=\frac{x-\mu }{\sigma }$

Where $x$ is the value of random variable of $\mu$ and $\sigma$ are the mean and standard deviation of original normal distribution respectively.

The area beyond two $z$-values, the smaller being $z_1$ and larger being $z_2$ is:

$P\left(z_1\le z\le z_2\right)=P\left(z\le z_2\right)-P\left(z\le z_1\right)$

It is noted that by Central Limit Theorem, that sampling distribution is approximately normal with:

a. mean ${\mu }_{\overline{x}}=\mu$, and

b. standard deviation ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Initially the range of values within which the sample mean lies is found by subtracting and adding the difference from and to population mean.

Thereafter, the corresponding $z$-values are obtained and then the area between them is obtained by subtraction using the relation above.

Calculation:

Since the sample size follows the following condition:

$n\ge 30$,

The sample mean is given by:

${\mu }_{\overline{x}}=30$,

And sample standard deviation is:

${\sigma }_{\overline{x}}=\frac{5.7}{\sqrt{31}}$.

The range of values where the mean lies is:

$\left(30-2,30+2\right)=\left(28,30\right)$.

Corresponding $z$-values are:

$\begin{array}{rll}{z}_1& =& \frac{28-30}{\left(\frac{5.7}{\sqrt{31}}\right)}\\ & =& \frac{\left(-2\right)}{1.0238}\\ & =& -1.9536\end{array}$

And

$\begin{array}{rll}{z}_2& =& \frac{32-30}{\left(\frac{5.7}{\sqrt{31}}\right)}\\ & =& \frac{2}{1.0238}\\ & =& 1.9536\end{array}$

Then, the area between the $z$-values is:

$\begin{array}{rll}P\left(-1.9536\le z\le 1.953\right)& =& P\left(z\le 1.9536\right)-P\left(z\le -1.9536\right)\\ & =& \frac{0.97441+0.97500}{2}-\frac{0.02559+0.02500}{2}\\ & =& 0.974705-0.025295\\ & =& 0.94941\end{array}$

Rounding off the number to $4^{\text{th}}$ decimal place gives 0.9494 which is the required probability.

Conclusion:

The probability that the mean length of an adult blue whale from a random sample of 31 whales differs from the population mean by less than 2 meters is 0.9494.