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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

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ANSWER:

a. 0.2981

b. 0.0918

c. 0.9525

d. 0.4972

Explanation:

Given:

μ = 100

σ = 15

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


z=(x-\mu)/(\sigma)

Then determine the probability with the normal table.

We calculate in each case:

a. p(X > 108)


\begin{gathered} z=(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=(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=(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

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