Question

For a normal distribution with a mean of μ = 100 and a standard deviation of o= 15, find each of thefollowing probabilities:a. p(X>108)p=b. p(X<80)p=c. p(X<125)p=d. p(90

Answer 1

ANSWER:

a. 0.2981

b. 0.0918

c. 0.9525

d. 0.4972

STEP-BY-STEP EXPLANATION:

Given:

μ = 100

σ = 15

We must calculate the z-score using the following formula:

$z=\frac{x-\mu}{\sigma}$

Then determine the probability with the normal table.

We calculate in each case:

a. p(X > 108)

$\begin{gathered} z=\frac{108-100}{15}=0.53 \\ \\ p(z>0.53)=1-p(z<0.53) \end{gathered}$

We look for the value of the normal table:

Therefore:

$\begin{gathered} p(z\gt0.53)=1-p(z\lt0.53) \\ \\ p\left(X>108\right)=1-0.7019=0.2981 \end{gathered}$

b. p(X < 80)

$\begin{gathered} z=\frac{80-100}{15}=-1.33 \\ \\ p(z<-1.33) \end{gathered}$

We look for the value of the normal table:

Therefore:

$p\left(X<80\right)=0.0918$

c. p (X < 125)

$\begin{gathered} z=\frac{125-100}{15}=1.67 \\ \\ p(z<1.67) \end{gathered}$

We look for the value of the normal table:

Therefore:

$p(X<125)=0.9525$

d. p (90 < X <110)

[tex]\begin{gathered} z=\frac{90-100}{15}=-0.67 \\ \\ z=\frac{110-100}{15}=0.67 \\ \\ p(-0.67We look for the value of the normal table:

Therefore:

[tex]\begin{gathered} p(-0.67