STAT 200 Week 3 Homework Problems PDF

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Week 3 Homeowrk for STATS at UMGC...

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STAT 200 Week 3 Homework Problems 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 84

Reliability 62

Cost 46

Performance 34

Comfort 47

Looks 27

Find the probability a person chooses a car for each of the given reasons. 1. 2. 3. 4. 5. 6. 7. 4.2.2

84+62+46+34+47+27=300 cars= 100% Safety: 84/300=0.28= 28% Reliability: 62/300=0.2067=21% Cost: 46/300=0.1533= 15% Performance: 34/300=0.1133= 11% Comfort: 47/300=0.1567=15% Looks: 27/300=0.09= 9%

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(S). 5865+1992/ 25891= 0.30 P(S)= 0.30= 30 b.) Find the probability of picking a lens that is the wrong PD or was lost in lab(L). 1398+976/ 25891= 0.09 P(L)=0.09= 9 c.) Find the probability of picking a lens that is not scratched P(N). 1-(5865/25891) = 0.77 P(N)=0.77= 77 d.) Find the probability of picking a lens that is not the wrong shape P(W). 1-(1485-25891) =0.94 P(W)= 0.94= 94 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(W) if you pick the number 7 and it comes up on the wheel. P(7) = 1/ 38 = 0.03 = 3 b.) Find the odds against winning(W2) if you pick the number 7. 37/38=0.97 97 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?

$20 x 0.97- $19.40 4.4.6 Find

10

P6

nPr= n! /(n-r)! 10P6= 10! / (10-6)! 10! = 10x9x8x7x6x5x4x3x2x1 10! = 3,628,800 (10-6)! = 4! 4! = 4x3x2x1 4! =24 10P6= 3,628,800/24 10P6= 151,200

4.4.12 How many ways can you choose seven people from a group of twenty? 20C7 20C7= 20!/7!(20-7!)= 20!/7! 13! = 77,520

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. The number of heads in 3-coin flips. b.) Write the probability distribution for the number of heads. TTT, TTH, THT, THH, HTT, HTH, HHT, HHH Number of Heads 0 1/8 1 3/8 2 3/8 3 1/8 c.) Draw a histogram for the number of heads.

d.) Find the mean number of heads. 0 x 1/8 + 1 x 3/8 + 2 x 3/8 + 3x 1/8 =1.5 e.) Find the variance for the number of heads. 0.75 f.) Find the standard deviation for the number of heads. 0.866

g.) Find the probability of having two or more number of heads. 3/8 + 1/8 = 0.5 50% h.) Is it unusual to flip 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, therefore it is not unusual. 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?

the value of warranty = 20%×$800 - $112.10 = $160 - $112.10 = $47.10 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. Formula: C(n,x)=n!/x!*(n-x)! Q=1- 0.30= 0.70 a.) P ( x=1 ) 0.30 b.)P ( x=5 ) 0.01 c.) P ( x=3) 0.019 d.)P ( x£3) 0.0302+ 0.324+0.05 = 0.676 e.)P ( x³5 ) 1- 0.75=0.25 f.) P ( x£4 ) 0.320+0.324+0.05+0.06 = 0.75 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= number of brown m&m’s b.) Argue that this is a binomial experiment. This is a binomial experiment because m= total number of trails (52). P= the probability of success in a single trail (0.14) The trails are independent. Find the probability that: c.) Six M&M’s are brown. 11.6% d.) Twenty-five M&M’s are brown.48.08% e.) All 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? It would be unusual because the people who make M&Ms make the packages with 5 other colors. If it were to

happen, I think it would be a promotional thing for M&M. Or I decided to spend my time handpicking all the brown ones out of the bag... 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 people left-handed b.) Write the probability distribution. 0- 0.20589 1- 0.34315 2- 0.26690 3- 0.12851 4- 0.04284 5- 0.01047 6- 0.00194 7- 0.00028 8- 0.00003 9- 0.00000 10- 0.00000 11- 0.00000 12- 0.00000 13- 0.00000 14- 0.00000 15- 0.00000 c.) Draw a histogram.

d.) Describe the shape of the histogram. The graph descends to the right showing that it is less likely that more people out of the 15-person group is left handed. e.) Find the mean. 1.500 f.) Find the variance. 1.350 g.) Find the standard deviation. 1.162...