SOLUTION: If z represents a standard normal random variable, then the P( -0.16 ≤ z ≤ 2.18) is approximately A .04 B .42 C .48 D .49 E .55

SOLUTION: If z represents a standard normal random variable, then the P( -0.16 ≤ z ≤ 2.18) is approximately A .04 B .42 C .48 D .49 E .55

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Question 929438: If z represents a standard normal random variable, then the P( -0.16 ≤ z ≤ 2.18) is approximately

A .04
B .42
C .48
D .49
E .55
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
P( -0.16 ≤ z ≤ 2.18) = normalcdf(-.16, 2.18) = .5489 0r 54.89
E
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