Let Z be a standard normal random variable and calculate the following probabili
ID: 3219070 • 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 2.23) (b) P(0 lessthanorequalto Z lessthanorequalto 2) (c) P(-2.30 lessthanorequalto Z lessthanorequalto 0) (d) P(-2.30 lessthanorequalto Z lessthanorequalto 2.30) (e) P(Z lessthanorequalto 1.07) (f)P(-1.95 lessthanorequalto Z) (g) P(- 1.30 lessthanorequalto Z lessthanorequalto 2.00) (h) P(1.07 lessthanorequalto Z lessthanorequalto 2.50) (i) P(1.30 lessthanorequalto Z) (j) P(|Z| lessthanorequalto 2.50) You may need to use the appropriate table in the Appendix of Tables to answer this question.Explanation / Answer
a) P(0 < Z < 2.23) = P(Z < 2.23) - P(Z < 0)
= 0.9871 - 0.5
= 0.4871
b) P(0 < Z < 2) = P(Z < 2) - P(Z < 0)
= 0.9772 - 0.5
= 0.4772
c) P(-2.3 < Z < 0) = P(Z < 0) - P(Z < -2.3)
= 0.5 - 0.0107
= 0.4893
d)P(-2.3 < Z < 2.3) = P(Z < 2.3) - P(Z < -2.3)
= 0.9893 - 0.0107
= 0.9786
e) P(Z < 1.07) = 0.8577
f) P(Z > -1.95) = 1 - P(Z < -1.95)
= 1 - 0.0256
= 0.9744
g) P(-1.3 < Z < 2) = P(Z < 2) - P(Z < -1.3)
= 0.9772 - 0.0968
= 0.8804
h) P(1.07 < Z < 2.5) = P(Z < 2.5) - P(Z < 1.07)
= 0.9938 - 0.8577
= 0.1361
i) P(Z > 1.3) = 1 - P(Z < 1.3)
= 1 - 0.9032
= 0.0968
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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