statistics

HOMEWORK3.docx

Home >Mathematics homework help >Statistics homework help >statistics

Score=6.6

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	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	Reliability	Cost	Performance	Comfort	Looks
84/300=0.28	62/300=0.21	46/300=0.15	34/300=0.11	47/300=0.16	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.

P(s or f)

= p (s) + p (f)

=5865/25891 + 1992/25891

=0.23 +0.08

=0.31

b.) Find the probability of picking a lens that is the wrong PD or was lost in lab.

=0.05 +0.04

=0.09

c.) Find the probability of picking a lens that is not scratched.

=1- p(s)

=1- 0.23

=0.77

d.) Find the probability of picking a lens that is not the wrong shape.

=1 – p (ws)

=1- 0.06

=0.94

(0.0)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 x p 00

=1/37 x 1/2

=0.014

b.) Find the odds against winning if you pick the number 7.

=1-0.014

=0.986

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 0.986

=$19.72

(Total number of spaces in the wheel is 38.

a.) There is only one space with 7. Therefore, P(7) =

b.) odds against winning with a 7 =

c.) $1 gets you $20. If the casino had used the actual odds against, for every $1 you would get $37. So the casino makes a profit of $17 for every $1 bet.)

4.4.6 Find

nPr= n! / (n-r)!

= 10! /4!

=151200

4.4.12 How many ways can you choose seven people from a group of twenty?

= 77520

(0.6)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.

Heads

b.) Write the probability distribution for the number of heads.

X= 1 if Heads, 0 otherwise

c.) Draw a histogram for the number of heads.

Elementary event Value of X

TTT 0

TTH 1

THT 1

HTT 1

THH 2

HTH 2

HHT 2

HHH 3

d.) Find the mean number of heads.

=12/8

=1.5 (Mean needs not be an integer)

e.) Find the variance for the number of heads.

=(-4) ^2/ 7

=2.286

()

f.) Find the standard deviation for the number of heads.

=sqrt 2.286

=1.51

()

g.) Find the probability of having two or more number of heads.

= 4/8

=0.5

h.) Is it unusual to flip two heads?

No. there is a fifty-fifty change of getting two heads.

(0.0)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?

Ev = p (x) * n

=0.20 x 112.10

=$22.42

(Let the random variable x = net value of the extended warranty.

If the dishwasher does not need to be replaced in the first two years, then the consumer will not get the $112.10 back, which means the net value is -112.10.

If the dishwasher needs to be replaced in the first two years, the net value for the warranty is 800 – 112.10 = 687.90 as the consumer has paid $112.10 upfront. In summary, the probability distribution for the net value is shown in the table below:

Expected value =

So in terms of the LG dishwasher, it makes sense to purchase the extended warranty.)

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.

0.302526

0.010206

0.18522

0.92953

0.010935

0.989065

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 and p = 0.14

Find the probability that

c.) Six M&M’s are brown.

0.14874

d.) Twenty-five M&M’s are brown.

0.000001

e.) All of the M&M’s are brown.

0.000001

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 is very unusual to get a package with only brown M&M’s. The reason for this occurrence could be because of failure to add milk.

(0.0)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.

X= 1 if left handed, 0 otherwise

c.) Draw a histogram.

d.) Describe the shape of the histogram.

It is skewed to the right

e.) Find the mean.

=0.05326

f.) Find the variance.

=(0.5011^2) / 14

=0.01794

g.) Find the standard deviation.

=sqrt var

=0.1339

(

1.) n = 15 and p = 0.10

a.) Random Variable x = number of people who are left-handed

b.) Probability distribution is found using TI-83/84: binompdf(15,.10,x value) or using Excel: “=BINOM.DIST(15,0.10,x)”

x	p(x)
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.) Histogram

d.) This graph is skewed to the right.

e.) Mean =

f.) Variance =

g.) Standard deviation = )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 6.7000000000000002E-3 1.2200000000000001E-2 0.02 2.6700000000000002E-2 3.3300000000000003E-2 0.04 4.6699999999999998E-2 5.33E-2 0.06 6.6699999999999995E-2 7.3300000000000004E-2 0.08 8.6699999999999999E-2 9.3299999999999994E-2 0.1

P x = 1( )

Px=1

()

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