Q. 6.37 - 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 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 _______ .

Step-by-Step Solution

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 the right of $0.49$.

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 $19$mm is equal to the area under the standard normal curve that lies to the right of $0.49$.