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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

01

Part (a) Step 1: Given Information

G.mallicoma's carapace length follows a normal distribution with $\mathrm{μ}=18.14mm$ and $\mathrm{σ}=1.76mm$.

02

Part (a) Step 2: Explanation

Calculating the $z-$scores:

For $16\mathrm{mm}$:

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

$≈-1.22$

For $17\mathrm{mm}$:

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

$≈-0.65$

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

$0.2578-0.1112=0.1466$

$=14.66\%$

03

Part (b) Step 1: Given Information

Given data:

Mean:$18.14$.

Standard deviation:$1.76$.

04

Part (b) Step 2: Explanation

Calculating the $z-$score:

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

$≈0.49$

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

$1-0.6879=0.3121$

$=31.21\%$

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