Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1) what is the probability that
Solution,
Using standard normal table,
a) P( -1.23 < Z < 1.64)
= P( Z < 1.64) - P( Z < -1.23)
= 0.9495 - 0.1093
= 0.8402
b) P( -1.27 > z or z > 1.74)
= P( z < -1.27) + P( z > 1.74)
= 0.1020 + { 1 - P( z < 1.74)}
= 0.1020 + { 1 - 0.9591}
= 0.1020 + 0.0409
= 0.1429
c) Using standard normal table,
P( -z < Z < z) = 95%
= P(Z < z) - P(Z <-z ) = 0.95
= 2P(Z < z) - 1 = 0.95
= 2P(Z < z) = 1 + 0.95
= P(Z < z) = 1.95 / 2
= P(Z < z) = 0.975
= P(Z < 1.96) = 0.975
= z ± 1.96
z = -1.96, 1.96
d) P( z > z ) = 2.5%
= 1 - P(Z < z) = 0.025
= P(Z < z) = 1 - 0.025
= P(Z < z ) = 0.975
= P(Z < 1.96 ) = 0.975
z = 1.96
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