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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Statistics 285 Solutions to HW #3 Problems 4.16 a. = E ( x ) = ) ( x xp = 10(.05) + 20(.20) + 30(.30) + 40(.25) + 50(.10) + 60(.10) = .5 + 4 + 9 + 10 + 5 + 6 = 34.5 2 = E ( x- ) 2 = ) ( ) ( 2 x p x - = (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 = 174.75 = 13.219 b. c. 2 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. = 2 2 = ) ( ) ( 2 x p x - 2 Firm A: 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 = 661.44 Firm B: 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 = 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....
View Full Document

Page1 / 20

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online