Given that z is a standard normal random variable, compute the following probabilities. (Round your answers to four decimal places.)
(a)
P(z ≤ −3.0)
(b)
P(z ≥ −3)
(c)
P(z ≥ −1.3)
(d)
P(−2.4 ≤ z)
(e)
P(−1 < z ≤ 0)
Solution :
(a)
=> P(z <= −3.0) = 1 - P(z > -3.0)
= 1 - P(Z < 3)
= 1 - 0.9987
= 0.0013
(b)
=> P(z >= −3) = 1 - P(Z < -3)
= 1 - [1 − P(Z < 3)]
= P(Z < 3)
= 0.9987
(c)
=> P(z >= −1.3) = 1 - P(Z < -1.3)
= 1 - [1 − P(Z < 1.3)]
= P(Z < 1.3)
= 0.9032
(d)
=> P(−2.4 <= z) = P(Z >= -2.4)
= 1 - P(Z < -2.4)
= 1 - [1 − P(Z < 2.4)]
= P(Z < 2.4)
= 0.9918
(e)
=> P(−1 < z <= 0) = P(Z < 0) - P(Z < -1)
= 0.5 - [1 - P(Z < 1)]
= 0.5 - [1 - 0.8413]
= 0.3413
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