Let Z be a standard normal random variable and calculate the following probabili
ID: 3219104 • Letter: L
Question
Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.) (a) P(0 lessthanorequalto Z lessthanorequalto 2.37) (b) P(0 lessthanorequalto Z lessthanorequalto 2) (c) P(-2.60 lessthanorequalto Z lessthanorequalto 0) (d) P(-2.60 lessthanorequalto Z lessthanorequalto 2.60) (e) P(Z lessthanorequalto 1.64) (f) P(-1.95 lessthanorequalto Z) (g) P(-1.60 lessthanorequalto Z lessthanorequalto 2.00) (h) P(1.64 lessthanorequalto Z lessthanorequalto 2.50) (i) P(1.60 lessthanorequalto Z) (j) P(|Z| lessthanorequalto 2.50)Explanation / Answer
a) P(0 < Z < 2.37) = P(Z < 2.37) - P(Z < 0)
= 0.9911 - 0.5
= 0.4911
b) P(0 < Z < 2) = P(Z < 2) - P(Z < 0)
= 0.9772 - 0.5
= 0.4772
c) P(-2.6 < Z < 0) = P(Z < 0) - P(Z < -2.6)
= 0.5 - 0.0047
= 0.4953
d) P(-2.6 < Z < 2.63) = P(Z < 2.63) - P(Z < -2.6)
= 0.9953 - 0.0047
= 0.9906
e) P(Z < 1.64) = 0.9495
f) P(Z > -1.95) = 1 - P(Z < -1.95)
= 1 - 0.0256
= 0.9744
g) P(-1.6 < Z < 2) = P(Z < 2) - P(Z < -1.6)
= 0.9772 - 0.0548
= 0.9224
h) P(1.64 < Z < 2.5) = P(Z < 2.5) - P(Z < 1.64)
= 0.9938 - 0.9495
= 0.0443
i) P(Z > 1.6) = 1 - P(Z < 1.6)
= 1 - 0.9452
= 0.0548
j) P(|Z| < 2.5) = P(-2.5 < Z < 2.5)
= P(Z < 2.5) - P(Z < -2.5)
= 0.9938 - 0.0062
= 0.9876
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