Question

given that x is a normal variable with a mean μ = 110 and standard deviation σ = 12, find the following probabilities. round your answers to 4 decimal places.
a) ( p ( x leq 120 ) )
enter answer
b) ( p ( x geq 80 ) )
enter answer
c) ( p ( 80 leq x leq 120 ) )
enter answer

Explanation:

Step1: Calculate z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
For part a), when \(x = 120\), \(z=\frac{120 - 110}{12}=\frac{10}{12}\approx0.83\).
Using the standard normal table or a calculator, \(P(X\leq120)=\Phi(0.83)\approx0.7967\).

Step2: Calculate z - score for part b)

When \(x = 80\), \(z=\frac{80 - 110}{12}=\frac{- 30}{12}=-2.5\).
\(P(X\geq80)=1 - P(X < 80)\). Since \(P(X < 80)=\Phi(-2.5)\), and using the property \(\Phi(-z)=1-\Phi(z)\), \(\Phi(-2.5)=1 - \Phi(2.5)\). From the standard normal table, \(\Phi(2.5)=0.9938\), so \(P(X\geq80)=1-(1 - 0.9938)=0.9938\).

Step3: Calculate z - scores for part c)

For \(x_1 = 80\), \(z_1=\frac{80 - 110}{12}=-2.5\) and for \(x_2 = 120\), \(z_2=\frac{120 - 110}{12}\approx0.83\).
\(P(80\leq X\leq120)=\Phi(0.83)-\Phi(-2.5)\). Since \(\Phi(-2.5)=1 - \Phi(2.5)\), \(P(80\leq X\leq120)=\Phi(0.83)-(1 - \Phi(2.5))\). Substituting \(\Phi(0.83)=0.7967\) and \(\Phi(2.5)=0.9938\), we get \(P(80\leq X\leq120)=0.7967-(1 - 0.9938)=0.7905\).

Answer:

a) \(0.7967\)
b) \(0.9938\)
c) \(0.7905\)