STAT 400 hw3 ans

# STAT 400 hw3 ans - STAT 400 Fall 2011 Homework#3(10...

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STAT 400 Fall 2011 Homework #3 (10 points) (due Friday, September 16, by 3:00 p.m.) No credit will be given without supporting work. 1. Homer Simpson is going to Moe’s Bar for some Flaming Moe ’s. Let X denote the number of Flaming Moe ’s that Homer Simpson will drink. Suppose X has the following probability distribution: x f ( x ) x f ( x ) x 2 f ( x ) 0 1 2 3 4 0.1 0.2 0.3 0.3 0.1 0.0 0.2 0.6 0.9 0.4 0.0 0.2 1.2 2.7 1.6 1.0 2.1 5.7 a) Find the probability f ( 4 ) = P(X = 4). f ( 4 ) = 1 [ 0.1 +0.2 + 0.3 + 0.3 ] = 0.10 . b) Find the probability P(X 1). P(X 1) = 0.90 . c) Find the probability P(X 1 | X < 3). P(X 1 | X < 3) = 6 0 5 0 3) P(X 3) X 1 P(X . . = 0.8333 . d) Compute the expected value of X, E(X).

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E(X) = all x (x) x f = 2.1 . e) Compute the standard deviation of X, SD(X). Var (X) = 2 all x 2 E(X) (x) x f = 5.7 (2.1) 2 = 1.29. SD(X) = 29 . 1 = 1.1358 . 1. (continued) Suppose each Flaming Moe costs \$1.50, and there is a cover charge of \$1.00 at the door. Let Y denote the amount of money Homer Simpson spends at the bar. Then Y = 1.50 X + 1.00 . f) Find the probability that Homer would spend over \$5.00. x y f ( x ) = f ( y ) 0 1 2 3 4 \$1.00 \$2.50 \$4.00 \$5.50 \$7.00 0.10 0.20 0.30 0.30 0.10 1.00 P(Y > \$5.00) = P(X 3) = 0.4 . g) Find the expected amount of money that Homer Simpson would spend, E(Y). Y = E(Y) = 1.50 E(X) + 1.00 = \$4.15 . ( On average, Homer drinks 2.1 bottles per day, his expected payment for the drink is \$3.15. His expected total payment is \$4.15 since he has to pay \$1.00 for the over charge. ) OR
x y f ( x ) = f ( y ) y f ( y ) 0 1 2 3 4 \$1.00 \$2.50 \$4.00 \$5.50 \$7.00 0.10 0.20 0.30 0.30 0.10 0.10 0.50 1.20 1.65 0.70 1.00 4.15 Y = E(Y) = y y f y all ) ( = \$ 4.15 . h) Find the standard deviation for the amount of money that Homer Simpson would spend, SD(Y). Y = SD(Y) = | 1.50 | SD(X) = \$1.7037 . OR x y f ( x ) = f ( y ) ( y Y ) 2 f ( y ) y 2 f ( y ) 0 1 2 3 4 \$1.00 \$2.50 \$4.00 \$5.50 \$7.00 0.10 0.20 0.30 0.30 0.10 0.99225 0.54450 0.00675 0.54675 0.81225 0.100 1.250 4.800 9.075 4.900 1.00 2.9025 20.125 y y f y all 2 Y 2 Y ) ( Var(Y) μ σ = 2.9025 . OR 2 all 2 2 Y E(Y) ) ( Var(Y) σ y y f y = 20.125 (4.15) 2 = 20.125 17.2225= 2.9025 . Y = SD(Y) = 2.9025 = \$ 6.6144 .

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2. Suppose that the probability that a duck hunter will successfully hit a duck is 0.40 on any given shot. Suppose also that the outcome of each shot is independent from the others. a) What is the probability that the first successful hit will be on the fourth shot? Miss Miss Miss Hit 0.60 0.60 0.60 0.40 = 0.0864 . Geometric distribution, p = 0.40. b) What is the probability that the third successful hit will be on the ninth shot? [ 8 shots: 2 S’s & 6 F’s ] S 60 0 40 0 2 8 6 2 . . 0.40 0.0836 . OR S S F F F F F F S F S S F F F F F S F F S F S F F F S F F F S F F F S S S F S F F F F F S F S F S F F F F S F F S F F S F F S F F F F S S F F S S F F S F F F F S F S F F S F F F S F F S F F F S F S F F F F S F S F S S F F F S F F F S F S F F F S F F S F F S F F F F S S F F F F S F F S S S F F F F S F F S F S F F F F S F S F F F S S F F F S F F F F F S S F S S F F F F F S F S F S F F F F F S S F F F S F S F F S F F F F F S F S S S F F F F F F S S F F S S F F F F S F F F S F F S F S F F F F F F S S S 28 (0.40) 3 (0.60) 6 0.0836 .
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