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 Arachnology,Vol.$30$,No.$3$,pp.$571-587$), we find that the carapace length of the adult male G.mollicoma is normallydistributed with a mean of $18.14$mm and a standard deviation of $1.76$mm. Let $x$denote carapace length for the adult male G.mollicoma.

a. Sketch the distribution of the variable $x$.

b. Obtain the standardized version,$z$of $x$.

c. Identify and sketch the distribution of $z$.

d. The percentage of adult male G. mollicoma that have carapace length between $16$mm and $17$mm is equal to the area under the standard normal curve between _______ and _______ .

e. The percentage of adult male G. mollicoma that have carapace length exceeding $19$mm is equal to the area under the standard normal curve that lies to the _______ of _______ .

Short answer

Part$\left(a\right):$The distribution of$x$is as follows:

Part $\left(b\right):$The standardized version of$x$is,$z=\frac{x-18.14}{1.76}$.

Part $\left(c\right):$The distribution of $z$is as follows:

Part $\left(d\right)$: The percentage of adult male G. mollicoma that have carapace length between $16$mm and $17$mm is equal to the area under the standard normal curve between $-1.22$and $-0.65$.

Part$\left(e\right)$: The percentage of adult male G. mollicoma that have carapace length exceeding$19$mm is equal to the area under the standard normal curve that lies to therightof$0.49$.

Step-by-step solution

Part a Step 1. Given information

Let $x$denote the carapace length for the adulty male G. mollicoma.

Part a Step 2. The distribution of x is as follows as:

Part b Step 1. The standardized version of x in terms of z is, 

$z=\frac{x-18.14}{1.76}$.

Part c Step 1. The distribution of z is as follows:

Part d Step 1. Substitute x=16 in the standardized version of x.

$$\begin{gathered} z=\frac{16-18.14}{1.76} \\ =-1.22 \end{gathered}$$

Substitute $x=17$in the standardized version of $x$.

$$\begin{gathered} z=\frac{17-18.14}{1.76} \\ =-0.65 \end{gathered}$$

Therefore, the percentage of adult male G. mollicoma that have carapace length between $16$mm and $17$mm is equal to the area under the standard normal curve between $-1.22$and$-0.65$.

Part e Step 1. Substitute x=19 in the standardized version of x.

$$\begin{gathered} z=\frac{19-18.14}{1.76} \\ =0.49 \end{gathered}$$

Therefore, the percentage of adult male G. mollicoma that have carapace length exceeding$19$mmis equal to the area under the standard normal curve that lies to the right of$0.49$.