Q. 6.97 - Elementary Statistics

Question

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", we find that the carapace length of the adult male G. mollicoma is normally distributed with mean $18.14mm$ and standard deviation $1.76mm$.

Part (a): Find the percentage of adult male G. mollicoma that have carapace length between $16mm$ and $17mm$.

Part (b): Find the percentage of adult male G. mollicoma that have carapace length exceeding $19mm$.

Part (c): Determine and interpret the quartiles for carapace length of the adult male G. mollicoma.

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

Step-by-Step Solution

Answer

Part (a): $14.66\%$of all adult male G. mollicoma has carapace lengths between $16mm$ and $17mm$.

Part (b): The percentage of adult male G. mollicoma that have carapace length exceeding is$31.21\%$.

Part (c): The first quartile is $16.96mm$. $25\%$ of the adult male G. mollicom has carapace lengths less than $16.96mm$.

The second quartile is $18.14mm$. $50\%$of the adult male G. mollicom has carapace lengths less than $18.14mm$.

The third quartile is $19.32mm$. $75\%$ of the adult male G. mollicom has carapace lengths less than $19.32mm$.

Part (d): The $95th$percentile for length is $21.04$. $95\%$of the adult male G. mollicom have carapace lengths less than $21.04mm$.

Part (a) Step 1. Given information.

The given mean is $18.14mm$ and standard deviation is $1.76mm$.

Part (a) Step 2. Draw the figure showing the required shaded region.

Draw the figure showing the required shaded region and its delimiting x-values, which are $16$ and $17$.

Now, we need to compute the z-scores for the x-values $16$ and $17$.

$$\begin{gathered} x=16\to \\ =\frac{16-18.14}{1.76} \\ =-1.22 \\ x=17\to \\ =\frac{17-18.14}{1.76} \\ =-0.65 \end{gathered}$$

We need to find the area under the standard normal curve that lies between $-1.22$ and $-0.65$.

The area to the left to $-1.22$ is $0.1112$ and the area to the left of $-0.65$ is  $-0.65$. The required area shaded in the figure is between $0.2578-0.1112=0.1466$.

$14.66\%$ of all adult male G. mollicoma has carapace lengths between $16mm$ and $17mm$.

Part (b) Step 1. Find the percentage of adult male that have carapace length exceeding 19 m m .

The figure shows the required shaded region,

Now, we need to compute the z-scores for the x-values $19$.

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

We need to find the area under the standard normal curve that lies above $0.49$. The area to the left of $0.49$ is $0.6879$ and the area to the right of $0.49$ is $1-0.6879=0.3121$.

The required shaded area is $0.3121$. Therefore, the percentage of adult male G. mollicoma that have carapace length exceeding $19mm$ is $31.21\%$.

Part (c) Step 1. Determine the first quartile for carapace length of adult male.

The z-score corresponding to $P_{25}$ is the one having an area $0.25$ to its left under the standard normal curve. From standard normal table, that z-score is $-0.67$ approximately.

Now, we must find the x-value having the z-score $-0.67$, the length that is $0.67$ standard deviations below the mean. It is $16.96$.

The first or $25$th percentile for length is $16.96mm$.

$25\%$ of the adult male has carapace lengths less than $16.96mm$.

Part (c) Step 2. Determine the second quartile for carapace length of adult male.

The z-score corresponding to $P_{50}$ is the one having an area $0.5$ to its left under the standard normal curve. From standard normal table, that z-score is $0$.

Now, we must find the x-value having the z-score $0$, the length that is $0$ standard deviations below the mean. It is $18.14$.

The second or $50$th percentile for length is $18.14mm$.

$50\%$ of the adult male has carapace lengths less than $18.14mm$.

Part (c) Step 3. Determine the third quartile for carapace length of adult male.

The z-score corresponding to $P_{75}$ is the one having an area $0.67$ to its left under the standard normal curve. From standard normal table, that z-score is $0.67$.

Now, we must find the x-value having the z-score $0.67$, the length that is $0.67$ standard deviations above the mean. It is $19.32$.

The third or $75$th percentile for length is $19.32mm$.

$75\%$of the adult male has carapace lengths less than $19.32mm$.

Part (d) Step 1. Determine the 95 t h percentile for carapace length of the adult male.

The z-score corresponding to $P_{75}$ is the one having an area $0.95$ to its left under the standard normal curve. From standard normal table, that z-score is $1.645$.

Now, we must find the x-value having the z-score $1.645$, the length that is $1.645$ standard deviations below the mean. It is $18.14+1.645+1.76=21.04$.

The $95$th percentile for length is $21.04$. Thus, $95\%$ of the adult male have carapace lengths less than $21.04mm$.