Using the Standard Normal Table found in your textbook, find the z-scores such that: (a) The area under the standard normal curve to its left is 0.5 z = (b) The area under the standard normal curve to its left is 0.9826 z = (c) The area under the standard normal curve to its right is 0.1423 z = (d) The area under the standard normal curve to its right is 0.9394
a)
We have to find a such that
P( Z < a) = 0.5
In Z table find z score such that probabiity is 0.5
We get Z- score = 0
[ That is P( Z < 0) = 0.5 ]
b)
P( Z < a) = 0.9826
In Z table, find z-score for the probability 0.9896
z-score = 2.111
c)
P( Z > a) = 0.1423
P( Z < a ) = 1 - 0.1423
P( Z < a) = 0.8577
In Z table , find z-score such that probability is 0.8577,
z-score = 1.07
d)
P( Z > a) = 0.9394
P( Z < (-a) ) = 0.9394
In z table, z-score for the probability 0.9394 is
z-score = 1.55
So z-score = -1.55 .
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