Using standard normal z-table we, want to find the z-score for which the area to the right is 0.52.
That is to find, z-score such that, P(Z > z) = 0.52
P(Z > z) = 0.52
=> 1 - P(Z < z) = 0.52
=> P(Z < z) = 1 - 0.52
=> P(Z < z) = 0.48
Using z-table we get, z-score corresponding probability of 0.48 is, z = -0.05
=> P(Z > -0.05) = 0.52
Answer : B) -0.05
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Help with the two please!
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