STAT 200 Week 3 Homework Problems Kayla Johnson January 29, 2020 4.1.4 A project conducted by the Australian Federal Office of Road Safety asked people many questions about their cars. One question...

Some of the work is already done, I just need help with what I did not answer.


STAT 200 Week 3 Homework Problems Kayla Johnson January 29, 2020 4.1.4 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 table #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. Safety: 84/300 = 28% Reliability: 62/300 = 21% Cost: 46/300 = 15% Performance: 34/300 = 11% Comfort: 47/300 = 15% Looks: 27/300 = 9% 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 a.) Find the probability of picking a lens that is scratched or flaked. Scratched: 5865/25891 = 22% Flaked: 1992/25891 = 7% b.) Find the probability of picking a lens that is the wrong PD or was lost in lab. Wrong PD: 1398/25891 = 5% Lost: 976/25891 = 37% c.) Find the probability of picking a lens that is not scratched. Not Scratched: 5708/25891 = 22% d.) Find the probability of picking a lens that is not the wrong shape. Wrong Shape: 1485/25891 = 5% 4.2.8 In the game of roulette, there is a wheel with spaces marked 0 through 36 and a space marked 00. a.) Find the probability of winning if you pick the number 7 and it comes up on the wheel. 1/38 b.) Find the odds against winning if you pick the number 7. 1-1/38 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? 1/38 * 20 = $0.58 37/38 – 19*1/38 = $0.474 casino profit 4.4.6 Find npr=n!/(n-r)! 10p6=10!/(10-6)! N! = 10! = 3,628,800 (n-r)! = (10-6)! 4! = 4*3*2*1 = 24 10p6 = 3,628,800/24 = 151,200 4.4.12 How many ways can you choose seven people from a group of twenty? N = 20 r = 7 20c7 = 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. Coin b.) Write the probability distribution for the number of heads. Binomial distribution = p (x-r) = Cn/p ^ r q ^ n – r Cn/r = n!/r!n – r! c.) Draw a histogram for the number of heads. d.) Find the mean number of heads. e.) Find the variance for the number of heads. f.) Find the standard deviation for the number of heads. g.) Find the probability of having two or more number of heads. h.) Is it unusual to flip two heads? 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? 1 – 0.2 = .8 800*0.2 = 160 112.10*0.8 = $89.68 160 – 89.68 = $70.32 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. X = brown m&ms b.) Argue that this is a binomial experiment Find the probability that c.) Six M&M’s are brown. 6/52 = 11% d.) Twenty-five M&M’s are brown. 25/52 = 48% e.) All of the M&M’s are brown. 100% 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 because the average pack of M&Ms contains only 14% (7) brown M&Ms. 5.3.4 Approximately 10% of all people are left-handed. Consider a grouping of fifteen people. a.) State the random variable. X = number of left-handed people out of 15 b.) Write the probability distribution. Binomial distribution = p (x-r) = Cn/p ^ r q ^ n – r Cn/r = n!/r!n – r! c.) Draw a histogram. d.) Describe the shape of the histogram. e.) Find the mean. Mean = 1.5 f.) Find the variance. Variance = 1.35 g.) Find the standard deviation. Standard deviation = 1.162 P x = 5( ) Px=5 () P x = 3( ) Px=3 () P x ≤ 3( ) Px£3 () P x ≥ 5( ) Px³5 () P x ≤ 4( ) Px£4 () 10P6 10 P 6 P x = 1( ) Px=1 ()