Q. 3.140 - Elementary Statistics

Question

Giant Tarantulas. One of the larger species of tarantulas is the Grammostola nollicoma, 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, F. Costa and F. Perez-Miles discussed the carapace length of the adult male G. mollicoma in the article "Reproductive Biology of Uruguayan Theraphosids" . The carapace lengths of a random sample of $50$ adult male G. mollicoma have a mean of $18.14 mm$ and a standard deviation of $1.76 mm$

a. Construct a graph
b. Apply Property $1$ of the empirical rule to make pertinent statements about the observations in the sample.
c. Repeat part (b) for Property $2$ of the empirical rule.
d. Repeat part (b) for Property $3$ of the empirical rule.

Step-by-Step Solution

Answer

Part(a) Required graph is given below.

Part(b) By property $1$ of empirical rule $68\%$ of sample to make statement about observation in sample.

Part(c)  By property $2$ of empirical rule $95\%$ of sample to make statement about observation in sample.

Part(d)  By property $3$ of empirical rule $99.7\%$ of sample to make statement about observation in sample.

Part(a) Step 1 : Given information

We are given that the carapace lengths of a random sample of $50$ adult male G. mollicoma have a mean of $18.14 mm$ and a standard deviation of $1.76 mm$.

Part(a) Step 2 : Simplify

For drawing the graph we need to find standard deviation on either side of mean with the help of property of empirical rule. which is discussed in further parts.

Required graph is 

Part(b) Step 1 : Given information

.The carapace lengths of a random sample of $50$ adult male G. mollicoma have a mean of $18.14 mm$ and a standard deviation of $1.76 mm$

Part(b) Step 2 : Simplify

As we already know by Property I of the empirical rule, We can say that nearly $68\%$ of the sample have carapace lengths within one standard deviation to either side of the mean.

Now, 

$50\times \frac{68}{100}=34$

One standard deviation to either side of the mean is from
$$\begin{gathered} \overline{x}-s=18.14-1.76=16.38 mm \\ \overline{x}+s=18.14+1.76=19.90 mm \end{gathered}$$

From above calculations, we can say that,

nearly $34$ of the $50$ adult in the sample have carapace lengths between $16.38mm$ and $19.90 mm$.

Part(c) Step 1 : Given information

The carapace lengths of a random sample of $50$ adult male G. mollicoma have a mean of $18.14 mm$ and a standard deviation of $1.76 mm$

Part(c) Step 2 : Simplify

As we already know that with the help of  Property 2 of the empirical rule, nearly $95\%$ of the men in the sample have carapace  lengths within two standard deviations to either side of the mean.

 Now,

 $50\times \frac{98}{100}=49$

Two standard deviations to either side of the mean is from
$$\begin{gathered} \overline{x}-2s=18.14-2\left(1.76\right)=14.62 \\ \overline{x}+2s=18.14+2\left(1.76\right)=21.66 \end{gathered}$$
From above calculations, we can say that,
nearly $49$ of the $50$  in the sample have carapace lengths between $14.62$ and $21.66 mm$

Part(d) Step 1 : Given information

The carapace lengths of a random sample of $50$ adult male G. mollicoma have a mean of $18.14 mm$ and a standard deviation of $1.76 mm$

Part(d) Step 2 : Simplify

As we know , by Property 3 of the empirical rule, nearly $99.7\%$ of the men in the sample have forearm lengths within three standard deviations to either side of the mean.

Now,

$50\times \frac{99.7}{100}=49.85\approx 50$

Three standard deviations to either side of the mean is from
$$\begin{gathered} \overline{x}-3s=18.14-3\left(1.76\right)=12.86 \\ \overline{x}+3s=18.14+3\left(1.76\right)=23.42 \end{gathered}$$

From above calculations, we can say that,

Nearly $50$, that is, all of the $50$ men in the sample have carapace lengths between $12.86$ and $23.42 mm$