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

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) (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.

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