STAT 200 Week 2 Homework Solutions Academic Essay

Solution

STAT 200 Week 2 Homework Solutions Academic Essay

1. (2 points) Let A denote the event of placing a $1 straight bet on the Florida Play 4 lottery and winning.  The chance of event A occurring is 1 in 10,000.  What is the value of  ?  What is the value of  ?

P(A) = 1/10,000 = 0.0001

P( = 9,999/10,000 = 0.9999 or 1 – 0.001 = 0.9999

•  (2 point)  MicroSort’s YSORT gender-selection technique is designed to increase the likelihood that baby born from in vitro fertilization (IVF) will be a boy.  In updated results, from a test of MicroSort’s YSORT gender-selection technique, 291 births consisted of 239 baby boys and 52 baby girls. 

1. (1 point) Based on these results, what is the probability of a boy born to a couple using MicroSort’s YSORT method? 

Probability of a boy born to a couple is 239/291

• (1 point) Does it appear that the technique is effective in increasing the likelihood that a baby will be a boy?

Yes the p-value of boy is greater than girl. The technique is effective in increasing the likelihood a baby will be a boy.

•  (2 points) In a New York Times / CBS News poll, respondents were asked if it should be legal or illegal to use hand-held phones while driving.  One hundred forty-one said that it should be legal, 663 said that it should be illegal. 

1. (1 point) What is the probability of randomly selecting someone who believes it should be legal to use a hand-held cell phone while driving? 

Total respondents = 804

Legal 141, illegal 663

P(L) = 141/804

• (1 point) Consider an event “unlikely” if its probability is less than or equal to 0.05.  Is it unlikely to randomly select someone who believes that it should be legal to use a hand-held cell phone while driving? 

No it is not unlikely to randomly select someone who believes that it should be legal to use a hand-held cell phone while driving.

• (3 points) Suppose that a family has 4 kids.  Assume that the probability of getting a boy or girl is equally likely.

1. (1 point) List the entire sample space of outcomes for this event.

2^4 = 16

•  (1 point) What is the probability of getting exactly three girls and one boy (in any order)?

P(G=3,B=1) = 4C3(1/2)^3 (1/2)^4-3

4C3/2^4 = 4/16 = 1/4

•  (1 point) What is the probability of having all four kids be the same gender?

P(G=4) P(B=4) = 4C4(1/2)^4 = 1/16

• (4 points) Men have XY (or YX) chromosomes and women have XX chromosomes.  X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective X chromosome without a paired X chromosome that is good.  In the following, represent a defective X chromosome with lowercase x, so a child with the xY pair of chromosomes will have the disease and a child with XX, XY, or Xx will not have the disease.  Each parent contributes one of the chromosomes to the child.

1. (1 point) If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease?

Father has chromosomes xY

Mother has chromosomes XX

Then the chromosomes of the son will be XY. So the probability that a son will inherit a disease is 0.

•  (1 point) If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a daughter will inherit the disease?

The chromosome of the daughter will be Xx. So the probability that daughter will inherit a disease is 0.

•  (1 point) If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease?

Father has chromosomes XY

Mother has chromosomes Xx

Then the chromosome of a son can be XY or xY. So the probability that a son will inherit a disease is (1/2) if he has xY.

•  (1 point) If a mother has one defective x chromosome and one good X chromosome and the father has good XY chromosomes, what is the probability that a daughter will inherit the disease?

The chromosomes of a daughter can be XX or Xx. So the probability that a daughter will inherit the disease is 0.

• (5 points) In a research experiment, on group of physicians was given bottles of epinephrine labeled with a concentration of “1 milligram in 1 milliliter of solution,” and another group of physicians was given bottles labeled with a ration of “1 milliliter of a 1:1000 solution.”  The two labels describe the exact same amount, and the physicians were instructed to administer 0.12 milligrams of epinephine.  The results were reported in the New York Times. The table below lists the numbers of correct and wrong dosage amounts calculated by the physicians.

	Correct Dosage Calculation	Wrong Dosage Calculation
Concentration Label (“1 milligram in 1 milliliter solution”)	11	3
Ratio Label (“1 milliliter of a 1:1000 solution”)	2	13

1.  (1 point) If one of the physicians is randomly selected, what is the probability of getting one who calculated the dose incorrectly?

If one of the physicians is randomly selected, the probability of getting one who calculated the dose incorrectly = 16/29

•  (1 point) If one of the physicians is randomly selected, what is the probability of getting one who made a correct dosage calculation or was given the bottle with a concentration label?

If one of the physicians is randomly selected, the probability of getting one who made a correct dosage calculation or was given the bottle with a concentration label = 16/29

• (1 point) For the physicians given the bottles labeled with “Concentration,” find the percentage of wrong dosage calculations.  What is the probability associated with this percentage?

The probability associated with this percentage = (3/14)(100)=21.43%

•  (1 point) For the physicians given the bottles labeled with “Ratio,” find the percentage of wrong dosage calculations.  What is the probability associated with this percentage?

The probability associated with this percentage = (13/15)(100)=86.67%

•  (1 point) Does it appear that either group did worse?  What do these results imply about drug labels?

• (7 points) A “1-Panel-THC” test for marijuana had the following results: Among 143 subjects with positive test results, there were 24 false positive results.  Among 157 negative results, there were 3 false negative results.   

1.  (1 point) If one of the test subjects is randomly selected, find the probability that the subject tested positive or used marijuana.

Probability that the subject tested positive or used marijuana = 143/300 = .4767

•  (1 point) If one of the test subjects is randomly selected, find the probability that the subject actually used marijuana.  Do you think that the result reflects the marijuana use rate in the general US population?

Probability that the subject actually used marijuana = (199+3)/300=.4067

•  (1 point) What is the probability of a correct test result?

Probability of a correct test result 273/300 = .91

•  (2 points) If 2 subjects are randomly selected without replacement, what is the probability that they both had incorrect test results?  Consider an event “unlikely” if its probability is less than or equal to 0.05.  Is the event of 2 incorrect results unlikely?

Probability that they both had incorrect test results=27/300(26/299)=.0078; unlikely event as <.05.

• (2 points) If 3 of the subjects are randomly selected without replacement, what is the probability that they all had true negative results?  Consider an event “unlikely” if its probability is less than or equal to 0.05.  Is the event of 3 true negative results unlikely?

Probability that they all had true negative results = 154/300(153/299)(152/298)=.1340

• (3 points) My mother and oldest sister were both born on the 4^th of July.  For the following questions, ignore leap years and assume that the 365 different days are equally likely.

1. (1 point) What is the probability that a randomly selected person is born on the 4^th of July?

Probability that a randomly selected person born on the 4^th of July = 1/365=.00274

•  (1 point) What is the probability that two randomly selected individuals were born on the same day?

Probability of two randomly selected individuals born on the same day 365/(365)^2 = 1/365 = .00274

•  (1 point) What is the probability that two randomly selected people were both born on the 4^th of July?

Probability of two ransomely selected people were both born on the 4^th July = 1/(365)^2 = .00001

.00000751.

• (3 points) According to FBI data, 12.4% of burglaries are cleared with arrests.  A new detective is assigned to five different burglaries.

1. (1 point) What is the probability that at least one of them is cleared with an arrest?

P(at least 1 of the 5 burglaries cleared) = 1-P(none of the five burglaries cleared)

=1- ((1-0.124)(5))

=1- ((.876)(5))

=1- .516

=.484

•  (1 point) What is the probability that the detective clears five burglaries with arrests?

P(All of the 5 burglaries cleared with arrests) = (.124)^5 =.00003

• (1 point) What should we conclude if the detective actually clears five burglaries with arrests?

If the detective clears all the five burglaries with arrests then the probability calculated in part b is highly impossible. It can be concluded that he is incredibly efficient.

1. (2 points) According to research by Pinhero, et.al. of Google, there is a 2% rate of catastrophic disk drive failure in a year.   

1. (1 point) If all of your computer data is stored on a hard drive with a backup stored on an external hard drive, what is the probability that, during a year, you can avoid catastrophe with at least 1 working hard drive?

P(at least one drive will work) = 1-P(No drive will work)

= 1-(.02)(.02)

= 1- .0004

= .9996

• (1 point) If copies of all your computer data are stored on three independent hard drives, what is the probability that during a year, you can avoid catastrophe with at least one working drive?

P(at least one drive will work) = 1- P(no drive will work)

= 1- (.02)(.02)(.02)

= 1- .000008

= .999992

1. (3 points) Answer the following questions about the Illinois Lottery.

1. (1 point) In the Illinois Little Lotto game, you win the jackpot by selecting five different whole numbers from 1 through 39 and getting the same five numbers (in any order) selected by the lottery commission.  What is the probability of winning a jackpot in this game?

There are C(39,5) ways to draw 5 numbers disregarding order the probability of winning a jackpot in this game is 1/C(39.5)

1/C(39,5) = COMBIN(39,5)

= 575,757

• (1 point) In the Illinois Pick 3 game, you win a bet by selecting three digits (i.e. 0-9) with repetition allowed and getting the same three digits in the exact same order that they are later drawn by the lottery commission.  What is the probability of winning this game?

• (1 point) In the Illinois Little Lotto game, you win the jackpot by selecting five different whole numbers from 1 through 39 and getting the same five numbers (in any order) selected by the lottery commission.  What is the probability of winning a jackpot in this game?

There are C(39,5) ways to draw 5 numbers disregarding order the probability of winning a jackpot in this game is 1/C(39,5)

1/C(39,5)=COMBIN(39,5)

= 575,757

1. (3 points) A clinical test on humans of a new drug is normally done in three phases.  Phase I is conducted with a relatively small number of healthy volunteers.  For example, Phase I of bexarotene involved only 14 subjects.  Assume that we want to treat 14 healthy humans with this new drug, and we have 16 suitable volunteers available.

1. (1 point) If subjects are selected and treated in sequence (so that the trial is discontinued if anyone displays any adverse side effects), how many different sequential arrangements are possible if 14 people are selected from the 16 that are available?

Number of different sequence = 16P14 = 16!/2! = 1.046 * (10)^13

• (1 point) If 14 subjects are selected from the 16 that are available, and the 14 selected subjects are all treated at the same time, how many different treatment groups are possible?

16C14 =240

•  (1 point) If 14 subjects are randomly selected and treated at the same time, what is the probability of selecting the 14 youngest subjects?

Probability of selecting 14 youngest = 1/240; as it is a single event out of 240 events.

1.  (7 points) The table below describes the results from 10 births from 10 different sets of parents.  The random variable x represents the number of girls among the 10 children. This data set is also available in the Microsoft Excel file “Homework 2 Data Sets.xlsx”

Number of Girls, x	P(x)
0	0.001
1	0.010
2	0.044
3	0.177
4	0.205
5	0.246
6	0.205
7	0.177
8	0.044
9	0.010
10	0.001

1. (1 point) Find the mean for the numbers of girls in 10 births.

(.001(.010)(.044)(.177)(.205)(.246)(.205)(.177)(.044)(.010)(.001))/10 = 5.6

• (1 point) Find the standard deviation for the numbers of girls in 10 births.

E(x^2) = (Sum) x^2 P(x) = -.372

• (1 point) Find the probability of getting exactly 1 girl in 10 births.

From the given probability distribution, the probability of getting exactly 1 girl in 10 births is .001.

• (1 point) Find the probability of getting 1 or fewer girls in 10 births.

P(1 or fewer girls) = P(0) + P(1)

= .001 +.010

= .011

•  (1 point) Which probability is relevant for determining whether 1 is an unusually low number of girls in 10 births: the result from part (c) or part (d)?

Since part c show the probability of getting exactly 1 girl in 10 births and part d shows the probability of getting 1 or fewer girls in 10 births and if we are trying to determine where 1 is an unusually low number of girls in 10 births then 0 girls in 10 births would also be unusually low. So, the result from part d is more relevant to determine whether 1 is an unusually low number of girls in 10 births.

• (1 point) Is 1 an unusually low number of girls in 10 births?  Why or why not?

P(1 or fewer girls) = .011 < .05

The probability of getting 1 or fewer girls in 10 births is .011 which is less than .05. So, 1 or fewer girls in 10 birhts is an unusually low number of girls in 10 births.

1. (5 points) The table below describes results of roadworthiness tests of Ford Focus cars that are 3 years old (based on data from the Department of Transportation).  The random variable x represents the number of cars that failed among the six that were tested for roadworthiness. This data set is also available in the Microsoft Excel file “Homework 2 Data Sets.xlsx”.

x	P(x)
0	0.377
1	0.399
2	0.176
3	0.041
4	0.005
5	0.001
6	0.001

1. (1 point) Find the mean for the numbers of cars that failed among the six cars that are tested.

Mean = Sum [xP(x)] = .894

•  (1 point) Find the standard deviation for the numbers of cars that failed among the six cars that are tested.

Variance = Sum[(x-mean)^2 * P(x)]

= .7939

= sqrt(.7939) = .8910

• (1 point) Find the probability of having exactly three cars fail among the six cars tested.

P(X=3) = .041

• (1 point) Find the probability of having three or more cars that fail among the six cars tested.

P(X>3) = P(X = 3) +P(x=4)+P(X=5)+P(X=6)

=.041+.005+.001+.001

=.048

•  (1 point) Is 3 an unusually high number of failures?  Why or why not?

Yes, 3 is an unusually high number failures, because the probability of having exactly three cars fail among the six cars tested is less than .05.