Question

Gibson Song Duration. A preliminary behavioral study of the Jingdong black gibbon, a primate endemic to the Wuliang Mountains in China, found that the mean song bout duration in the wet season is 12.59 minutes with a standard deviation of 5.31 minutes. [SOURCE: L. Shecran et al.. "Preliminary Report on the Behavior of the Jingdong Black Gibbon. Tropical Biodiversity, Vol, 5(2). Pp. 113-125] Assuming that song bout is normally distributed, determine the percentage of song bouts that have durations within

a. one standard deviation to either side of the mean.

b. two standard deviations to either side of the mean.

C. three standard deviations to either side of the mean.

Short answer

a. $68.26\%$area will be within one standard deviation from the mean.

b. $95.44\%$area will be within one standard deviation from the mean.

c. $99.74\%$area will be within one standard deviation from the mean.

Step-by-step solution

Given Information

To determine the area of each region for the given normal distribution plot.

Explanation

Given: the normal distribution curve is given.

Calculation:

Calculate the standard deviation and mean using the relation

$\sigma =\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}}$

$\overline{x}=\frac{\sum x_i}{n}$

After solving we will get

$\sigma =1.78$

$\overline{x}=0$

Then calculate the area according to the figure

The area of each section will be

a. $68.26\%$area will be within one standard deviation from the mean.

b.$95.44\%$area will be within one standard deviation from the mean.

c.$99.74\%$area will be within one standard deviation from the mean.

Program:

clc

clear

close all

$x=(-3:0.1:3);$

$y=normpdf(x,0,1)$

$m=mean\left(x\right)$

$s=std\left(x\right)$

$\text{area}=m+x^{*}s;$

$\text{plot}(x,y)\text{;}$

grid on

axis square

Query:

- First, we have defined the sample.

- Then create a figure of normal distribution.