Question

According to a recent​ study, the carapace length for adult males of a certain species of tarantula are normally distributed with a mean of  18.01 mm and a standard deviation of σ=1.69 mm. Complete parts​ (a) through​ (d) below.

a) . Find the percentage of the tarantulas that have a carapace length between 16 mm and 17 mm.

b) Find the percentage of the tarantulas that have a carapace length exceeding 19 mm.

c) Determine and interpret the quartiles for the carapace length of these tarantulas.

the first quartile is _

Interpret the first quartile. Select the correct choice below and fill in the answer​ box(es) to complete your choice.

the second quartile is _

Interpret the second quartile. Select the correct choice below and fill in the answer​ box(es) to complete your choice.

the third quartile is _

Interpret the third quartile. Select the correct choice below and fill in the answer​ box(es) to complete your choice.

d)

Obtain the 95th percentile for the carapace length of these tarantulas.

The 95th percentile is

I do not know how to determine the quartiles, or the 95th percentile.

Answer 1

Let X:carapace length, X~N(18.01,1.69²)

Then , $Z=\frac{X-\mu}{\sigma } \sim N(0,1); \mu =18.01,\sigma =1.69-------(*)$ is a Standard Normal Variable

a)

b)

$P(X>19)=P(Z>0.59)=1-P(Z<0.59)=1-0.7224=0.2776=27.76\%$

c) $Q1\Rightarrow P(X<x')=0.25\Rightarrow P(Z<z')=0.25$

$z'=-0.67\Rightarrow \frac{x'-18.01}{1.69}=-0.67\Rightarrow x'=16.8777$

which means that 25% of the Population has X<16.8777mm

$Q2\Rightarrow P(X<x')=0.5\Rightarrow P(Z<z')=0.5$

$z'=0\Rightarrow \frac{x'-18.01}{1.69}=0\Rightarrow x'=18.01$ which means 50% of the observations are under 18.01mm

$Q3\Rightarrow P(X<x')=0.75\Rightarrow P(Z<z')=0.75$

$z'=0.68\Rightarrow \frac{x'-18.01}{1.69}=0.68\Rightarrow x'=19.1592$ which means 75% of observations are under 19.1592mm

Note that Q represents the quartiles and the X's are transformed to Z using *. Also, Z critical values are obtained from standard normal table.

d) 95th Percentile is found as $P_{95}\Rightarrow P(X<x^*)=0.95 \Rightarrow P(Z<z^*)=0.95$

Using standard normal table, $z^*=1.645 \Rightarrow \frac{x^*-18.01}{1.69}=1.645\Rightarrow x^*=20.79$