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
... [Show More] 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.
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 Less]