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 |
Delegate your assignment to our experts and they will do the rest.
Find the probability a person chooses a car for each of the given reasons.
Probability =
Total number of outcomes= 84+62+46+34+47+27
=300
P (Safety) =
=0.28 or 28%
P (Reliability) =
=0.207 or 20.7%
P (Cost) =
=0.153 or 15.3%
P (Performance) =
=0.113 or 11.3%
P (Comfort) =
=0.157 0r 15.7%
P (Looks) =
=0.09 or 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 |
Find the probability of picking a lens that is scratched or flaked.
n(s)= total number of defectives
=25891
P (scratched or flaked) =
=
=0.3035
Find the probability of picking a lens that is the wrong PD or was lost in lab.
P (wrong PD or was lost in lab) =
=
=0.0917
Find the probability of picking a lens that is not scratched.
P (not scratched) = 1-P (scratched)
= 1-
P (not scratched) =1-0.2265
=0.7735
Find the probability of picking a lens that is not not the wrong shape.
P (not the wrong shape) = 1-P (the wrong shape)
= 1-
=0.9426
4.2.8
In the game of roulette, there is a wheel with spaces marked 0 through 36 and a space marked 00.
Find the probability of winning if you pick the number 7 and it comes up on the wheel.
Probability =
= =0.0263
Find the odds against winning if you pick the number 7.
Odd against =
Probability event will not happen=1-
=
Odd against = *
= 37 odds
Odds against winning if you pick the number 7=37:1
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?
Profit= $38-$20
=$18
4.4.6 Find
Solution
n P r =
10p6 =
Numerator n! :
N! = 10!
10! = 10*9*8*7*6*5*4*3*2*1
10! = 3,628,800
Denominator:
(n-r)! = (10-6)!
(10-6)! = 4!
4! =4*3*2*1
4! =24
Permutation value:
10p6 =
= 151,200
4.4.12 How many ways can you choose seven people from a group of twenty?
Required 20 C 7
n C r =
n = 20, r = 7
=
20c7 = 77520
Therefore there are 77520 ways of choosing seven people from a group of twenty
5.1.2
Suppose you have an experiment where you flip a coin three times. You then count the number of heads.
State the random variable.
Random variable x=number of heads
Write the probability distribution for the number of heads.
Let H= Heads, T= Tails
Combinations: TTT, TTH, THT THH, HTT, HTH, HHT ,HHH
Probability distribution is as follows:
| Number of Heads X |
0 |
1 |
2 |
3 |
| Probability (p(x)) |
Draw a histogram for the number of heads.
Find the mean number of heads.
= Therefore mean number of heads =1.5
Find the variance for the number of heads.
Variance= E(X 2 )-(EX )2
Variance=3-(1.5) 2
=0.75
Find the standard deviation for the number of heads.
Standard deviation=
=
=0.8860
Find the probability of having two or more number of heads.
P (two or more number of heads) =
=0.5
Is it unusual to flip two heads?
P (two heads) = =0.375
I t is not unusual to flip two heads since 0.375>0.05
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?
There is 20% chance of replacement, then value of warranty = 20% * $800 - $112.10
= $160 - $112.10 = $47.10
When it is assumed that replacement is done in first two years, then the value of warranty increases to:
$800 - $112.10 = $687.90
Assuming it is definitely replaced in the first 2 years expected value of the extended warranty is $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.) = 0.3025
b.) = 0.0102
c.) = 0.1852
d.) = 0.9295
e.) = 0.0109
f.) = 0.9891
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.
State the random variable.’
Random variable = x = number of brown M&M's in a milk cholate packet
Argue that this is a binomial experiment
This is a Binomial Experiment because:
(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
Six M&M’s are brown.
P(X=6) = ) 0.86 46 *0.14 6
=0.1487
Twenty-five M&M’s are brown.
P(X=25) = ) 0.86 27 *0.14 25
= 0.000001
All of the M&M’s are brown.
P(X=52) = ) 0.86 0 *0.14 52
= 0.000001
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 for a package to have only brown M&M's because probability of such as event is almost 0.
If there were to happen, the reason may be:
We have come across a rare event, which will happen again after a large number of trials
The claim that the proportion of brown M&M's in a milk chocolate packet is approximately 14% could be wrong.
5.3.4
Approximately 10% of all people are left-handed. Consider a grouping of fifteen people.
State the random variable.
Random variable=x= number of left-handed people in a group of 15 people
Write the probability distribution.
P(X=x) = ) p x (1-p) n-x
| X | P(x=x) |
| 0 | 0.2059 |
| 1 | 0.3432 |
| 2 | 0.1285 |
| 3 | 0.1285 |
| 4 | 0.0428 |
| 5 | 0.0105 |
| 6 | 0.0019 |
| 7 | 0.0003 |
| 8 | 0.0000 |
| 9 | 0.0000 |
| 10 | 0.0000 |
| 11 | 0.0000 |
| 12 | 0.0000 |
| 13 | 0.0000 |
| 14 | 0.0000 |
| 15 | 0.0000 |
Draw a histogram.
Describe the shape of the histogram.
The histogram is skewed to the right
Find the mean.
Mean=np
=15*0.1
=1.5
Find the variance.
Variance =n*p*(1-p)
=15*0.1*0.9
=1.35
Find the standard deviation .
Standard deviation=
=
=1.1619
