SOLUTION: Let be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(-c≤Z≤c)=0.9412 Carry your intermediate computations to at

Algebra ->  Probability-and-statistics -> SOLUTION: Let be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(-c≤Z≤c)=0.9412 Carry your intermediate computations to at      Log On


   

Question 1175071: Let be a standard normal random variable. Use the calculator provided, or this table, to determine the value of c. P(-c≤Z≤c)=0.9412
Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi  
(1 - .9412)/2 = .0294
invNorm(.0294) = 1.8897
P(-1.8897 ≤ z ≤1.8897) =  0.9412

Wish You the Best in your Studies.