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
... [Show More] 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. Total = 84+62+46+34+47+27=300
P(Safety) = 84/300=0.28
P(Reliability) = 62/300=0.27 P(Cost) = 46/300=0.15
P(Performance) = 34/300=0.11 P(Comfort) = 47/300=0.16 P(Looks) = 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
a.) Find the probability of picking a lens that is scratched or flaked.
Scratch + Flaked = 5865+1992=7857 7857/25891=0.303
b.) Find the probability of picking a lens that is the wrong PD or was lost in lab.
Wrong PD + Lost in lab = 1398+976=2374 2374/25891=0.091
c.) Find the probability of picking a lens that is not scratched.
P(Scratch) = 5865/25891=0.227
P(Not scratch) = 1-0.227=0.773
d.) Find the probability of picking a lens that is not the wrong shape.
P(Wrong shape) = 1485/25891=0.057 P(Not wrong shape) = 1-0.057=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.
a.) Find the probability of winning if you pick the number 7 and it comes up on the wheel.
P(7)= 1/38=0.026
b.) Find the odds against winning if you pick the number 7.
P(Not 7)= 37/38=0.974
Odds against 7 = P(Not 7)/P(7) = 37/38 / 1/38 = 37:1
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?
The actual odds is 37:1, meaning the casino would pay $37 on every $1 bet. The casino actually pays $20 for every $1 bet. Therefore, the profit is $37-$20=$17
4.4.6 Find = 10!/(10-6)! = 10!/4! = 151200
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!x13!) = 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. x=number of heads.
b.) Write the probability distribution for the number of heads.
7 successful trials out of 8 total. P(X=1) = 0.000003337860107421875
P(X=2) = 0.00008177757263183594
P(X=3) = 0.0011448860168457031
P(X=4) = 0.010017752647399902
P(X=5) = 0.05609941482543945
P(X=6) = 0.19634795188903809
P(X=7) = 0.39269590377807617
c.) Draw a histogram for the number of heads.
d.) Find the mean number of heads.
Mean = 7*(⅞) = 6.125
e.) Find the variance for the number of heads.
Variance = 7*0.875*0.125 = 0.766
f.) Find the standard deviation for the number of heads.
Standard Deviation = √0.766 = 0.875
g.) Find the probability of having two or more number of heads.
P(x≥2)-P(x=8)=P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)+P(x=7)-P(x=8)
= 0.656 - 0.344
= 0.312
h.) Is it unusual to flip two heads?
P(x≤2)=P(x=2)+P(x=1)
= 0.0002
Yes, it is 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 warrantee costs $112.10 on a dishwasher. What is the expected value of the extended warranty assuming it is replaced in the first 2 years?
Chances to be replaced = 20% chances not to be replaced = 80% EV = 0.20*800 + 0.80*0
EV= 160
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.) = 6C1*(0.30^1){0.70^(6-1)}
= [6!/{1!(6-1)!}]*0.30*(0.70^5)
= [6!/5!]*0.30*0.17
= 6*0.30*0.17
= 0.306
b.) =6C5*(0.30^5){0.70^(6-5)}
= [6!/{5!(6-5)!}]*0.002*0.70
= 6*0.002*0.70
= 0.008
c.) =6C3*(0.30^3){0.70^(6-3)}
= [6!/{3!(6-3)!}]*0.027*0.343
= 20*0.027*0.343
=0.185
d.) = P(x=0)+P(x=1)+P(x=2)+P(x=3)
=6C0*(0.30^0){0.70^(6-0)} + 0.306 + 6C2*(0.30^2){0.70^(6-2)} + 0.185
= 1*1*0.118 + 0.306 + 15*0.09*0.24 + 0.185
= 0.118 + 0.306 + 0.324 + 0.185
=0.993
e.) = P(x=5)+P(x=6)
= 0.008 + 6C6*(0.30^6){0.70^(6-6)}
= 0.008 + 1*0.001*1
= 0.008 + 0.001
= 0.009
f.) = P(x≤3)+P(x=4)
= 0.993 + 6C4*(0.30^4){0.70^(6-4)
= 0.993 + 15*0.008*0.49
= 0.993 + 0.059 = 1.052
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 n=52 p=0.14
Find the probability that
c.) Six M&M’s are brown.
P(x=6)= 52C6*(0.14^6){0.86^(52-6)}
= 20358520*0.00000753*0.00097035
= 0.149
d.) Twenty-five M&M’s are brown.
P(x=25)= 52C25*(0.14^25){0.86^(52-25)}
= 4.78E+14*4.50E-22*0.01703955245
=3.66E-09
e.) All of the M&M’s are brown.
P(x=52)= 52C52*(0.14^52){0.86^(52-52)}
= 1*3.97E-45*1
=3.97E-45
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?
one major reason could of been human error.
5.3.4
Approximately 10% of all people are left-handed. Consider a grouping of fifteen people. a.) State the random variable.
x = left handed people
b.) Write the probability distribution.
Calculation Taken from online calculator.
P(X=0) = 0.8179069375972307
P(X=1) = 0.16523372476711728
P(X=2) = 0.01585576146755166
P(X=3) = 0.0009609552404576765
P(X=4) = 0.00004125312900954671
P(X=5) = 0.0000013334344730358534 P(X=6) = 3.367258770292559e-8 P(X=7) = 6.802542970287998e-10 P(X=8) = 1.1165790229008079e-11
P(X=9) = 1.50381013185294e-13 P(X=10) = 1.6709001465032665e-15 P(X=11) = 1.5343435688735235e-17 P(X=12) = 1.162381491570851e-19 P(X=13) = 7.225370576974987e-22 P(X=14) = 3.649177059078276e-24 P(X=15) = 1.4744149733649604e-26 P(X=16) = 4.654087668450001e-29 P(X=17) = 1.10614086e-31
P(X=18) = 1.8621900000000004e-34 P(X=19) = 1.9800000000000006e-37 P(X=20) = 1.0000000000000001e-40
c.) Draw a histogram.
d.) Describe the shape of the histogram.
The graph represents the chances that X will be equal that value. e.) Find the mean.
Mean = 15*0.10 = 1.5
f.) Find the variance. q = 1 - 0.10 = 0.90
Variance = 15*0.10*0.90 = 1.35 g.) Find the standard deviation.
Standard Deviation = √1.35 = 1.162 [Show Less]