Q.6.97 - Elementary Statistics

Question

Giant Tarantulas. One of the larger species of tarantulas is the Grammostola mollicoma, whose common name is the Brazilian giant tawny red. A tarantula has two body parts. The anterior part of the body is covered above by a shell, or carapace. From a recent article by F. Costa and F. Perez-Miles titled "Reproductive Biology of Uruguayan Theraphosids" (The Journal of A rachutology, Vol. 30 , No. 3, pp. 571-587), we find that the carapace length of the adult male $G$. mollicoma is normally distributed with mean $18.14 \mathrm{mm}$ and standard deviation $1.76 \mathrm{mm}$.

a. Find the percentage of adult male $G$. mollicoma that have carapace length between $16 \mathrm{mm}$ and $17 \mathrm{mm}$.

b. Find the percentage of adult male $G$. mollicoma that have carapace length cxcceding $19 \mathrm{mm}$.

c. Determine and interpret the quartiles for carapace length of the adult male $G$. mollicoma.

d. Obtain and interpret the $95th$ percentile for carapace length of the adult male $G$. mollicoma.

Step-by-Step Solution

Answer

a). The required percentage is $14.66\%$.

b). The proportion of $z-$greater than $0.49$ is $31.21\%$.

Part (a) Step 1: Given Information

G.mallicoma's carapace length follows a normal distribution with $\mu =18.14mm$ and $\sigma =1.76mm$.

Part (a) Step 2: Explanation

Calculating the $z-$ scores:

For $16 \mathrm{mm}$ :

$z=\frac{16-18.14}{1.76}$

$\approx -1.22$

For $17 \mathrm{mm}$ :

$z=\frac{17-18.14}{1.76}$

$\approx -0.65$

The difference between the numbers in Table II is then used to calculate the percentage of all observations:

$0.2578-0.1112=0.1466$

$=14.66\%$

Part (b) Step 1: Given Information

Given data:

Mean: $18.14$.

Standard deviation: $1.76$.

Part (b) Step 2: Explanation

Calculating the $z-$ score:

$z=\frac{19-18.14}{1.76}$

 $\approx 0.49$

As a result, the $z-$ scores bigger than $0.49$ are equal to the number of observations greater than $19mm$.

From Table II in Appendix A, the proportion of $z-$ scores greater than $0.49$ is

$1-0.6879 =0.3121$

                    $=31.21\%$