STAT 200 Week 3 Homework Problems,100% CORRECT

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A project conducted by the Australian Federal Office of Road Safety asked people many questions about their cars. One question was the reason that a person chooses a given car, and that data is in ta... ble #4.1.4 ("Car preferences," 2013). Table #4.1.4: Reason for Choosing a Car Safety Reliability Cost Performance Comfort Looks 84 62 46 34 47 27 Find the probability a person chooses a car for each of the given reasons. Answer: Total reasons for choosing a car = 84+62+46+34+47+27 =300 Safety Probability 84/300= 0.28 Reliability Probability 62/300= 0.21 Cost Probability 46/300= 0.15 Performance Probability 34/300= 0.11 Comfort Probability 47/300= 0.16 Looks Probability 27/300= 0 .09 4.2.2 Eyeglassomatic manufactures eyeglasses for different retailers. They test to see how many defective lenses they made in a time period. Table #4.2.2 gives the defect and the number of defects. Table #4.2.2: Number of Defective Lenses Defect type Number of defects Scratch 5865 Right shaped – small 4613 Flaked 1992 Wrong axis 1838 Chamfer wrong 1596 Crazing, cracks 1546 Wrong shape 1485 Wrong PD 1398 Spots and bubbles 1371 Wrong height 1130 Right shape – big 1105 Lost in lab 976 Spots/bubble – intern 976 Answer: Sum of total Defects =25891 a.) Find the probability of picking a lens that is scratched or flaked. 5865+ 1992/25891 = .304 b.) Find the probability of picking a lens that is the wrong PD or was lost in lab.1398+976/25891 = .092 c.) Find the probability of picking a lens that is not scratched.1-5865/25891 = 0.773 d.) Find the probability of picking a lens that is not the wrong shape.1-1485/25891= 0.943 4.2.8 In the game of roulette, there is a wheel with spaces marked 0 through 36 and a space marked 00. Possible # of outcomes are 38, n=38 a.) Find the probability of winning if you pick the number 7 and it comes up on the wheel. Probability 1/38 = 0.02632 b.) Find the odds against winning if you pick the number 7. Odds against = P (Event will not happen)/ P (Event will happen) 1-0.02632/ 0.02632 = .97368/.02632 =36.99 then Odds against winning if you pick 7 =37 c.) The casino will pay you $20 for every dollar you bet if your number comes up. How much profit is the casino making on the bet? There are 38 slots on the wheel if you pay $1.oo for each slot then you pay $38. If you win then the casino will pay you $20.00 . Casino Profit = $38- $20 = $18.00 4.4.6 Find n=10 r =6 10!/ (10-6)! = 151200 4.4.12 How many ways can you choose seven people from a group of twenty? n = 20 r =7 20! / 7! (20-7)! = 20! / 7!13! = 77520 5.1.2 Suppose you have an experiment where you flip a coin three times. You then count the number of heads. a) State the random variable. Random variable X = number of heads Let, H= Heads, T = Tails b) Write the probability distribution for the number of heads. Combinations are TTT, TTH, THT, THH, HTT, HTH, HHT, HHH Number of Heads(X) 0 1 2 3 Probability(P) 1/8 3/8 3/8 1/8 c)Draw a histogram for the number of heads. d) Find the mean number of heads. Mean = E(X) = 0*1/8 + 1*3/8 + 2*3/8 + 3*1/8 = 1.5 e) Find the variance for the number of heads E(X^2) = 0*1/8 + 12*3/8 + 22*3/8 + 32*1/8 = 3 Variance = E(X^2) - (E(X))^2 = 3 - (1.5)2 = 0.75 f) Find the standard deviation for the number of heads. Standard deviation = sqrt(variance) = sqrt(0.75) = 0.866 g) Find the probability of having two or more number of heads. P(2 or more heads) = 3/8 + 1/8 = 0.5 h) Is it unusual to flip two heads? P (two heads) = 3/8 = 0.375 It is likely that out of 8 times of 3 flips, 3 times we can observe two heads out of 3 5.1.4 An LG Dishwasher, which costs $800, has a 20% chance of needing to be replaced in the first 2 years of purchase. A two-year extended warranty costs $112.10 on a dishwasher. What is the expected value of the extended warranty assuming it is replaced in the first 2 years? Value of Warranty = 20% x 800 -112.10 = 20/100 x 800 -112.10 = 160 - 112.10 = 47.9 Replacement infirst 2 yrs. value of the Warranty increases to 800- 112.10 = $687.90 5.2.4 Suppose a random variable, x, arises from a binomial experiment. If n = 6, and p = 0.30, find the following probabilities using technology. a.) b.) c.) d.) e.) f.) 5.2.10 The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s. a.) State the random variable. Random variable = x = number of brown M&M's in a milk cholate packet b.) Argue that this is a binomial experiment The characteristics of a binomial distribution are: there is n number ofindependent trials, there are only two possible outcomes on each trial-success (S) and failure (F), and the probability of success, p varies from trial to trial. (i)n = Total number of trials = 52 (ii) p = Probability of success in a single trial = p = 0.14 (iii) The trials are independent Find the probability that c.) Six M&M’s are brown. n = 52p = 0.14q = 1 - p = 0.86 x = 6 d.) Twenty-five M&M’s are brown. X= 25 e.) All of the M&M’s are brown. For x = 52 f.) Would it be unusual for a package to have only brown M&M’s? If this were to happen, what would you think is the reason? Yes it would be unusual for a package of M&M’s to contain only brown candy. As indicated above the probability of this occurring is almost 0.000. If this were to happen, I would think this was not a random occurrence, but instead a special promotion. 5.3.4 Approximately 10% of all people are left-handed. Consider a grouping of fifteen people. a.) State the random variable. The random variable X is the number of left handed people in a group of fifteen. b.) Write the probability distribution. The probability distribution is Binomial. c.) Draw a histogram. d.) Describe the shape of the histogram. The shape of the histogram is skewed to the right. e.) Find the mean. Mean = n*p Mean = 0.10*15 Mean = 1.50 f.) Find the variance. Variance = 1.35 g.) Find the standard deviation. Standard deviation = 1.162 [Show More]

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