Question 1182236
1.5 hr is 90 minutes mean
sd is 15 minutes
z=(x-mean)/sd
=(115-90)/15 since 1hr55min is 115 min.
25/15=1.67
prob z>1.67is 0.0478
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z(0.85)=1.036.  Use 2ndVARS3invnorm(0.85,0,1) ENTER
1.036=(x-90)/15
15.54=x-90
x=105.54 min
or about 1h45 min
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z=(63-90)/5=-1.8
z=(110-90)/15=1.33
probability z is between those two is 0.8729
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also 2ndVARS2normalcdf(63,110,90,15)ENTER =0.8729 or 87.29%