Probability

[newbiestats]

I am having trouble remembering probability, could someone help me with this please

4. One of the larger species of tarantulas is the Grammostola mol licoma, whose common name is the Brazilian
Giant Tawny Red. A Tarantula has two body parts. The anterior part of the body is covered above 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. 571-87), we ?nd the
carapace length of adult male Grammostola mol licoma is normally distributed with mean 18.14mm and
standard deviation is 1.76mm.

(a) Find the percentage of the adult male Grammostola mol licoma that have carapace lengths between
16mm and 17mm.

(b) Find the percentage of the adult male Grammostola mol licoma that have carapace lengths exceeding
19mm.

 

[chrisr]

\(\displaystyle Z=\frac{x-\mu}{\sigma}\)

\(\displaystyle For\ \mu=18.14,\ \sigma=1.76,\ x_1=16,\ x_2=17\)

You must calculate the probability that x is between 16 and 17mm.
You do this by converting the 2 x-values to 2 Z-values.

Since the x values are below the mean, the Z-values are negative.
Your Z-chart may or may not have negative values.
If it doesn't, then make use of the fact that the graph is symmetrical about the mean.
Then, in that case, add 1.14 to 18.14 and add 2.14 to 18.14 to get 2 Z-values you can then work with.

Calculate \(\displaystyle Z_1\) and \(\displaystyle Z_2\) for \(\displaystyle x_1\) and \(\displaystyle x_2\).

Then calculate the probability that Z lies between these 2 values.
This probability is \(\displaystyle P(Z_1<Z<Z_2)\) for x=16 and 17, or vice versa for x=19.28 and 20.28 depending on whether you have negative Z-values or not.

This probabilty is the probability read off for \(\displaystyle Z_1\) subtracted from the reading for \(\displaystyle Z_2\)
if you have negative Z,
or the other way around if you only have positive Z.

Wish I had a pet tarantula!