Sloution to problem set 6

# Probability and Statistics for Engineering and the Sciences (with CD-ROM and InfoTrac )

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Sol6 1. 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 2. 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 | Z | ) = P( | Z | < c ) = P(-c < Z < c) = Φ (c) - Φ (-c) = 2 Φ (c) – 1 ⇒Φ (c) = .9920 c = 2.41 3. a. Φ (c) = .9100 c 1.34 (.9099 is the entry in the 1.3 row, .04 column) b. 9 th percentile = -91 st percentile = -1.34 c. Φ (c) = .7500 c .675 since .7486 and .7517 are in the .67 and .68 entries, respectively.

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