Q. 6.102 - Elementary Statistics

Question

Elephant Pregnancies. G. Wintemeyer et al. studied demographic data on African elephants living in Kenya in the article "Comparative Demography of an At-risk African Elephant Population (PLOS ONE, Vol. 8. No. 1). Based on this study, we will assume that the time between pregnancies of the African elephant in Kenya, for elephants that have more than one calf, is normally distributed with mean of 4.01 years and a standard deviation of 0.94 years. Determine the percentage of such times that are

a. less than 2 years.

b. between 3 and 5 years.

Step-by-Step Solution

Answer

a) The percentage of time of pregnancies for elephants $=1.62\%$

b) The percentage of time of pregnancies for elephants $=71.08\%.$

Step 1: Given Information (Part b)

Mean $\mu =4.01$

Standard deviation $\sigma =0.94$

Step 2: Explanation (Part b)

Find the z score,

For $x=3$

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

$$\begin{gathered} =\frac{3-4.01}{0.94}\text{ for } \\ x=5 \end{gathered}$$

For $x=5$

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

$=\frac{5-4.01}{0.94}$

$=1.05$

The area between z scores is 

$=(\text{ area to the left of }1.05)\text{-(area to the left of }-1.07\text{ ) }$

$=0.8531-0.1423$

$=0.7108$

Step 1: Given Information(Part a)

To find the percentage of time of pregnancies for elephants less than  2  years.

Step 2: Explanation (Part a)

Mean $\mu =4.01$

Standard deviation $\sigma =0.94$

Find the z score,

For $x=2$

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

$=\frac{2-4.01}{0.94}$

$=-2.14$

The area to the left of the z score of 2.14  is  0.0162.