Solutions04 - \ 8. a. P(0 Z 2.17) = (2.17) (0) = .4850 b....

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8. a. P(0 Z 2.17) = Φ (2.17) – Φ (0) = .4850 b. Φ (1) – Φ (0) = .3413 c. Φ (0) – Φ (–2.50) = .4938 d. Φ (2.50) – Φ (–2.50) = .9876 e. Φ (1.37) = .9147 f. P( –1.75 < Z) + [1 – P(Z < –1.75)] = 1 – Φ (–1.75) = .9599 g. Φ (2) – Φ (–1.50) = .9104 h. Φ (2.50) – Φ (1.37) = .0791 i. 1 – Φ (1.50) = .0668 j. P( |Z| 2.50 ) = P( –2.50 Z 2.50) = Φ (2.50) – Φ (–2.50) = .9876 9. a. .9838 is found in the 2.1 row and the .04 column of the standard normal table so c = 2.14. b. P(0 Z c) = .291 Φ (c) = .7910 c = .81 c. P(c Z) = .121 1 – P(c Z) = P(Z < c) = Φ (c) = 1 – .121 = .8790 c = 1.17 d. P(–c Z c) = Φ (c) – Φ (–c) = Φ (c) – (1 – Φ (c)) = 2 Φ (c) – 1 Φ (c) = .9920 c = .97 e. P( c | Z | ) = .016 1 – .016 = .9840 = 1 – P(c
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Solutions04 - \ 8. a. P(0 Z 2.17) = (2.17) (0) = .4850 b....

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