# 4 - STAT 350 Fall 2008 Homework#4 Solution Corrected...

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STAT 350 – Fall 2008 Homework #4 – Solution Corrected through Lecture #8 Text Exercises Chapter 5, Problems #18, 20, 46b&c, 50, 52, 54 5.18 Let A 1 denote the event that person 1 has blood type A (etc.) a. P(A 1 and A 2 ) = P(A 1 )P(A 2 ) = (0.42)(0.42) = 0.1764 b. P(B 1 and B 2 ) = 0.10 2 = 0.01 P(AB 1 and AB 2 ) = 0.04 2 = 0.0016 P(O 1 and O 2 ) = 0.44 2 = 0.1936 c. P(two randomly chosen people have the same blood type) = P(A 1 and A 2 ) + P(B 1 and B 2 ) + P(AB 1 and AB 2 ) + P(O 1 and O 2 ) = 0.3816 d. This is the complement of the event in (c): Answer = 1 – 0.3816 = 0.6184 5.20 Let A = the event that the top subsystem works, B = the event the bottom subsystem works P(A) = P(1 works OR 2 works) = P(1 works) + P(2 works) – P(1 works and 2 works) = 0.9 + 0.9 – 0.9 2 = 0.99 P(B) = P(3 works AND 4 works) = 0.9 2 = 0.81 P(system works) = P(A OR B) = P(A) + P(B) – P(A and B) = P(A) + P(B) – P(A)P(B) = 0.99 + 0.81 – (0.99)(0.81) = 0.9981 5.46 Let X = the inside diameter of a randomly selected piston ring in cm. E(X) = 12, SD(X) = 0.04 b. () () 12 EX EX == () 0.04 0.005 64 SD X SD X n = c. The mean of a random sample of size 64 Homework #4 – Solution Page 1 of 5

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5.50 a. () () 50 EX EX == , () ( ) 1 0.1 100 SD X SD X n = 49.75 50 50 50.25 50 49.75 50.25 0.1 0.1 0.1 2.5 2.5 (2.5) ( 2.5) 0.9938 0.0062 0.9876 X PX P PZ ⎛⎞ −− << = < < ⎜⎟ ⎝⎠ =−< < = Φ Φ = = b. 49.8 , ( ) 1 0.1 100 SD X SD X n = 49.75 49.8 49.8 50.25 49.8 49.75 50.25 0.1 0.1 0.1 0.5 4.5 (4.5) ( 0.5) 1 0.3085 0.6915 X P = < < < = Φ Φ = = 5.52 a. 50 , ( ) 1.5 0.5 9 SD X SD X n = 50 52 50 52 4 0 0.5 0.5 X P ≥= = On the standard normal table, the best you can say is the probability is approx. zero, the actual probability is about 0.000032 b. 50 , ( ) 1.5 0.23717 40 SD X SD X n = 50 52 50 52 8.43 0 1.5 40 1.5 40 X P = Even statistical packages will return this probability as 0.
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4 - STAT 350 Fall 2008 Homework#4 Solution Corrected...

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