Stat HW3

# Stat HW3 - Statistics 285 Solutions to HW#3 Problems 4.16 a...

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Unformatted text preview: Statistics 285 Solutions to HW #3 Problems 4.16 a. μ = E ( x ) = ∑ ) ( x xp = 10(.05) + 20(.20) + 30(.30) + 40(.25) + 50(.10) + 60(.10) = .5 + 4 + 9 + 10 + 5 + 6 = 34.5 σ 2 = E ( x- μ ) 2 = ) ( ) ( 2 x p x ∑- μ = (10 - 34.5) 2 (.05) + (20 - 34.5) 2 (.20) + (30 - 34.5) 2 (.30) + (40 - 34.5) 2 (.25) + (50 - 34.5) 2 (.10) + (60 - 34.5) 2 (.10) = 30.0125 + 42.05 + 6.075 + 7.5625 + 24.025 + 65.025 = 174.75 σ = 174.75 = 13.219 b. c. μ ± 2 σ ⇒ 34.5 ± 2(13.219) ⇒ 34.5 ± 26.438 ⇒ (8.062, 60.938) P (8.062 < x < 60.938) = p (10) + p (20) + p (30) + p (40) + p (50) + p (60) = .05 + .20 + .30 + .25 + .10 + .10 = 1.00 4.20 a. Yes. Relative frequencies are observed values from a sample. Relative frequencies are commonly used to estimate unknown probabilities. In addition, relative frequencies have the same properties as the probabilities in a probability distribution, namely 1. all relative frequencies are greater than or equal to zero 2. the sum of all the relative frequencies is 1 b. Using MINITAB, the graph of the probability distribution is: 30 25 20 0.15 0.10 0.05 0.00 age p(age) c. Let x = age of employee. Then P ( x > 30) = .13 + .15 + .12 = .40. P ( x > 40) = 0 P ( x < 30) = .02 + .04 + .05 + .07 + .04 + .02 + .07 + .02 + .11 + .07 = .51 d. P ( x = 25 or x = 26) = .02 + .07 = .09 4.28 a. E ( x ) = ∑ ) ( x xp Firm A: E ( x ) = 0(.01) + 500(.01) + 1000(.01) + 1500(.02) + 2000(.35) + 2500(.30) + 3000(.25) + 3500(.02) + 4000(.01) + 4500(.01) + 5000(.01) = 0 + 5 + 10 + 30 + 700 + 750 + 750 + 70 + 40 + 45 + 50 = 2450 Firm B: E ( x ) = 0(.00) + 200(.01) + 700(.02) + 1200(.02) + 1700(.15) + 2200(.30) + 2700(.30) + 3200(.15) + 3700(.02) + 4200(.02) + 4700(.01) = 0 + 2 + 14 + 24 + 255 + 660 + 810 + 480 + 74 + 84 + 47 = 2450 b. σ = 2 σ σ 2 = ) ( ) ( 2 x p x ∑- μ 2 Firm A: σ 2 = (0 - 2450) 2 (.01) + (500 - 2450) 2 (.01) + ⋅ ⋅ ⋅ + (5000 - 2450) 2 (.01) = 60,025 + 38,025 + 21,025 + 18,050 + 70,875 + 750 + 75,625 + 22,050 + 24,025 + 42,025 + 65,025 = 437,500 σ = 661.44 Firm B: σ 2 = (0 - 2450) 2 (.00) + (200 - 2450) 2 (.01) + ⋅ ⋅ ⋅ + (4700 - 2450) 2 (.01) = 0 + 50,625 + 61,250 + 31,250 + 84,375 + 18,750 + 84,375 + 31,250 + 61,250 + 50,625 = 492,500 σ = 701.78 Firm B faces greater risk of physical damage because it has a higher variance and standard deviation. 4.44 a. We will check the 5 characteristics of a binomial random variable. 1. The experiment consists of n = 5 identical trials. We have to assume that the number of bottled water brands is large. 2. There are only 2 possible outcomes for each trial. Let S = brand of bottled water used tap water and F = brand of bottled water did not use tap water. 3. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) = .25 and q = 1 – p = 1 - .25 = .75....
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Stat HW3 - Statistics 285 Solutions to HW#3 Problems 4.16 a...

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