A study finds that the carapace length of an adult spider is normally distributed with a mean of $18\mathrm{.}02mm$ and a standard deviation of $1\mathrm{.}56mm.$ Let $x$ denote carapace length for the adult spider.

A$.z=\frac{(x-18\mathrm{.}02)}{1\mathrm{.}56}$

B$.z=\frac{(x-1\mathrm{.}56)}{18\mathrm{.}02}$

C$.x=\frac{(z-1\mathrm{.}56)}{18\mathrm{.}02}$

D$.x=\frac{(z-18\mathrm{.}02)}{1\mathrm{.}56}$

c$.$ Identify and sketch the distribution of $z.$ Choose the correct graph below.

A$.$

$.$

D

c$.$

d$.$ Find the z$-$scores that correspond to the percentage of adult spiders that have carapace lengths between $15mm$ anc $16mm.$

The percentage of adult spiders that have carapace lengths between $15mm$ and $16mm$ is equal to the area under the standard normal curve between and

$($Round to two decimal places as needed.$)$