Question

use 2nd vars normalcdf to find the following probabilities: (mean =0, s =1)
p(-1.75 < z < 1.25) ncdf(-1.75,1.25,0,1)=?
p(z > 1.45) ncdf(1.45,10,0,1) =?
p(z < -0.25) ncdf(-10,-1.25,0,1) =?
use 2nd vars invnorm to find:
z that represents the 70th percentile
invnorm(.70,0,1) z =?
z that represents the top 5% cutoff
invnorm(.95,0,1) z =?

Explanation:

Step1: Calculate \(P(-1.75 < Z < 1.25)\)

Using a calculator or software with the normalcdf function (for a standard normal distribution with mean \(0\) and standard deviation \(1\)), \(Ncdf(-1.75,1.25,0,1)\approx0.8822\)

Step2: Calculate \(P(Z > 1.45)\)

Since \(P(Z > 1.45)=1 - P(Z\leq1.45)\), and \(Ncdf(1.45,10,0,1)=1 - Ncdf(-\infty,1.45)\approx1 - 0.9265 = 0.0735\)

Step3: Calculate \(P(Z < - 0.25)\)

\(Ncdf(-10,-0.25,0,1)\approx0.4013\)

Step4: Find the \(z\) - value for the 70th percentile

Using the invnorm function for a standard normal distribution (\(\text{Invnorm}(0.70,0,1)\)), the \(z\) - value is approximately \(0.5244\)

Step5: Find the \(z\) - value for the top \(5\%\) cutoff

The top \(5\%\) cutoff means the area to the left is \(1 - 0.05=0.95\). Using \(\text{Invnorm}(0.95,0,1)\), the \(z\) - value is approximately \(1.6449\)

Answer:

\(P(-1.75 < Z < 1.25)\approx0.8822\); \(P(Z > 1.45)\approx0.0735\); \(P(Z < - 0.25)\approx0.4013\); \(z\) - value for 70th percentile\(\approx0.5244\); \(z\) - value for top \(5\%\) cutoff\(\approx1.6449\)