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Stat HW4

# Stat HW4 - IE 111 Spring Semester 2009 Homework#4 Solutions...

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IE 111 Spring Semester 2009 Homework #4 Solutions Question 1. Suppose we have a biased coin such that P(head) P(tail). Suppose I flip the coin three times. Let the random variable X be the number of heads in the 3 flips. You are given the following information: P(X=0) = 2/25 P(X=1) = 6/25 P(X=2) = ? P(X=3) = 9/25 a) Find P(X=2) 2/25 + 6/25 + P(X=2) + 9/25 = 1 there fore P(X=2) = 8/25 b) Find P(X 1) P(X 1) = P(X=0) + P(X=1) = 8/25 c) Find P( 1 X < 2) P(1 X<2) = P(X=1) = 6/25 d) Find P( X=1 | X 1) P(X=1 | X 1) = P(X=1and X 1) / P(X 1) = P(X=1) / P(X 1) = (6/25) / (8/25) = 6/8 Question 2. A die has 4 sides (not 6 like a regular die). The four sides are labeled 1, 3, 5, and 10 respectively. It is equally likely that you will get a 1 or a 3. It is equally likely that you will get a 5 or a 10. It is three times more likely that you will get a 5 or10 than a 1 or 3. a) Let X = the outcome of a roll of the die. Find the probability mass function of X. x 1 3 5 10 P X (x) p p 3p 3p 8p = 1 thus p=1/8 and the PMF is x 1 3 5 10 P X (x) 1/8 1/8 3/8 3/8

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b) If Y=2X 2 +1, find P Y (y) x 1 3 5 10 y 3 19 51 201 P Y (y) 1/8 1/8 3/8 3/8 c) If I roll the die 10 times, what is the probability I get exactly 3 tens? Let Z = number of tens in 10 rolls. Z is Binomial( n=10, p=1/3) P(Z=3) = 10 C 3 (3/8) 3 (5/8) 7 =0.2357 Question 3. Let the random variable X be the value of the up face of a certain unfair die. For this die, the probability of getting a specific number (1 through 6) is proportional to twice that number. a) Find the Probability Mass Function P X (x). x 1 2 3 4 5 6 P X (x) 2p 4p 6p 8p 10p 12p 2p+4p+6p+8p+10p+12p = 1 42p=1 thus p=1/42 so the PMF is x 1 2 3 4 5 6 P X (x) 2/42 4/42 6/42 8/42 10/42 12/42 b) Find P(X 2) P( X 2 ) = 1 - P(X=1) = 1 - 2/42 = 40/42 c) P(X 4 | 3 < X 5) P(X 4 | 3<X 5) = P(X 4 and 3<X 5) / P(3<X 5) = P(X=5) / P(X=4 or X=5) = (10/42) / (18/42) = 5/9 d) Find the probability of rolling a number greater than or equal to “3” four times in a row. P(X 3) = P(X=3) + P(X=4) + P(X=5) + P(X=6) = (6+8+10+12)/42 = 36/42 P(three times in a row) = (36/42) 4 e) The Cumulative Distribution Function F X (x) is Defined as F X (x) = P(X x) Find the C.D.F. for the random variable X.
F X (x) = 0 x<1 F X (x) = 2/42 1 x<2 F X (x) = 5/42 2 x<3 F X (x) = 12/42 3 x<4 F X (x) = 20/42 4 x<5 F X (x) = 30/42 5 x<6 F X (x) = 1 x 6 Question 4.

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Stat HW4 - IE 111 Spring Semester 2009 Homework#4 Solutions...

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