HELP! STAT REVIEW QUESTIONS

Question

1) Forty percent of the english students at Parkway High school had Dr.Mima for history, and the rest had Dr. P. Thirty percent of Dr.Mima's students earn an A in probability, while only ten percent of Dr. P's students earn an A in probability.

a) What percent of Parkway High School history students earn an A?

A= have Dr.Mima=0.40

B= rest have Dr.P=0.60

C= Dr.Mima students earn A = 0.30

C= Dr.P students earn A = 0.10

P(C)=P(A)P(C|A) + P(B)P(C|B)

= (0.40)(0.30) + (0.60)(10)

P(C) = 0.19 -answer

b) Tiffany did not earn an A in english. What is the probability that Tiffany had Dr.P for history?

For each of the following, define the random variable(s) needed to answer the questions,tell what kind of distribution each has, and what probabilites need to be calculated to answer the question.

2) Patty has basketball practice everyday of the week at 4:00pm (M-F). Every morning, Patty has a 20% chance of going to class.

a)What is the probability that in a given week (5 days), Patty goes to class at least once.

b) What is the probability that the Exam day (22nd day) is the first day Patty goes to his class?

X=# of days he goes to class on first day

X~Geom(22)

P(X=22)

P(x)= (1-p)^(x-1)p

= (1-0.20)^(22-1)(0.20)

= 0.0018 -answer

c) What is the probability that the Final Exam day(84th day) is the third day Patty goes to his class?

d) If Patty goes to class at least two times in a given week, Patty will get himself a doughnut. What is the probability that he doesn't get a doughnut until the fourth week.

Answer 1

a) What percent of Parkway High School history students earn an A?

A= Student had Dr.Mima

B= Student had Dr.P

C= Student earned "A"

P(A)= 0.40 P(B)=0.60 P(C|A)= 0.30 P(C|B)=0.10

P(C)=P(A)P(C|A) + P(B)P(C|B)

= (0.40)(0.30) + (0.60)(0.10)

P(C) = 0.18

b) Tiffany did not earn an A in english. What is the probability that Tiffany had Dr.P for history?

-- if the question was about Dr. P teaching English:

C' = complement of event C = student not earning "A"

P(B| C') = P(C'|B)P(B) / P(C')

P(C')= 1-P(C) = 1-0.18= 0.82

P(C'|B)= 1-P(C|B) = 1-0.10=0.9

P(B| C') = 0.9*0.6 / 0.82 = calculate!

2a - use binomial. 2c-.Use negative binomial. 2d- use negative binomial but calculate the p using binomial.

Good luck!