Download 10. a. if you roll a single die and count the number

10. a. if you roll a single die and count the number of dots on top, what is the sample
space of all possible outcomes?
b. assign probabilities to the outcomes of the sample space of part (a). Do the possibilities
add up to 1? should they add up to 1?
a) {1, 2, 3, 4, 5, 6}
b) All outcomes are equally likely with probability = 1/6. The probabilities do add up to
1, as they must.
10. you roll two fair dice, a green one and a red one.
b. find P(1 on green die and 2 on red die).
P = (1/6)2 = 1/36
c find P92 on green die and 1 on red die).
P = (1/6)2 = 1/36
d. Find P(( 1 on green die and 2 on red die).
P = (1/6)2 = 1/36
5.1
10. 25% of all state prison inmates released on parole become repeat offenders while on
parole. Suppose the parole board is examining five prisoners up for parole. Let x =
number of prisoners out of five on parole who become repeat offenders.
x: 0 1 2 3 4 5
P(x): 0.237 0.396 0.264 0.088 0.015 0.001
a. find the probability that one or more of the five parolees will be repeat offenders. How
does this number relate to the probability that none of the parolees will be repeat
offenders?
P(x ≥ 1) = 0.396 + 0.264 + 0.088 + 0.015 + 0.001 = 0.764
P(x ≥ 1) = 1 − P(0) = 1 − 0.237 = 0.763
The small discrepancy is likely the result of roundoff error.
b. find the probability that two or more of the five parolees will be repeat offenders.
P(x ≥ 2) = 0.264 + 0.088 + 0.015 + 0.001 = 0.468
c. find the probability that four or more of the five parolees will be repeat offenders.
P(x ≥ 4) = 0.015 + 0.001 = 0.0.016
d. compute the mean, the expected number of repeat offenders out of 5.
5
µ = x = ∑ xP(x) = 1(0.396) + 2(0.264) + 3(0.088) + 4(0.015) + 5(0.001) = 1.253
x=0
e. compute the standard deviation of the number of repeat offenders out of five.
σ=
x2 − x
2
=
5
∑ x 2 P(x) − µ 2 = 0.969 = 0.984
x=0
14. sara is a 60 year old anglo female and is reasonably good health. she wants to take out
a $50,000 term ( that is, straight death benefit) life insurance policy until she is 65. The
policy will expire on her 65th birthday. the probability of death is given year is provided
by the Vital Statistics section.
x=age: 60 61 62 63 64
P(death at this age):0.00756 0.00825 0.00896 0.00965 0.01035
a. Sara is applying to big rock insurance company for her term policy. what is the
probability that sara will die in her 60th year? Using the probability and the 50,000 death
benefit, what is the expected cost to big rock insurance?
P(60) = 0.00756
E(C) = ($50000)(0.00756) = $378
b. repeat part (a) for years 61, 62, 63, and 64. What would be the total expected cost to
big rock insurance over the years 60 through 64?
age
60
61
62
63
64
total
probability
of death
0.00756
0.00825
0.00896
0.00965
0.01035
0.04477
expected
cost
378.0
412.5
448.0
482.5
517.5
2238.5
c. If big rock insurance wants to make a profit of $ 700 above the expected total cost paid
out for sara's death, how much should it charge for the policy.
$2938.5 or $48.98/month.
d. If big rock insurance company charges $5000 for the policy, how much profit does the
company expect to make?
$2761.50
5.2
12.What makes up a trial? An single observation of a random event or process.
what is a success? A trial where the prescribed outcome is realized.
What is failure? A trial where the prescribed outcome is not realized.
What are the values of, n, p and q? n is number of trials, p is the probability of a
successful trial, and q = 1 – p is the probability of an unsuccessful trial.
Have you ever purchased an article of clothing worn it once and then returned the
purchase? this is called a ont time fling. about 10% of all adults deliberately do a one
time fling. In a group of seven adult friends what is that probability that:
a. no one has done a one-time fling?
P(0) = (0.9)7 = 0.478
b. at least one person has done a one-time fling?
P(x ≥ 1) = 1 − P(0) = 0.522
c. no more than two people have done a one-time fling?
Let p = 0.1 be the probability a person has a 1-time-fling
P(x ≤ 2) = P(0) + P(1) + P(2)
=
7! 0
7!
7!
p (1 − p)7 +
p(1 − p)6 +
p(1 − p)2 = 0.974
7!
6!1!
5!2!
What makes up a trial? what is a success? What is failure? What are the values of, n, p
and q?
A trial is a random sampling of one a person. Success here is the outcome that the person
has had a one-time-fling. Failure is the outcome that the person has not had a fling.
There are 7 trials. The probability of success is 0.1, and the probability of failure is 0.9.
16. What makes up a trial? what is a success? What is failure? What are the values of, n,
p and q? Trevor is interested in purchasing a sporting goods store. After examining his
accounting records for the past several years, he found that the store has been grossing
over $850 per day and about 60% of the business days it is open. Estimate that the
probability that the store will gross over $850.
a. at least 3 out of 5 business days.
p = 0.6
P(x ≥ 3) = P(3) + P(4) + P(5)
=
5!
5!
5!
(0.6)3 (0.4)2 +
(0.6)4 (0.4)1 +
(0.6)5 (0.4)0 = 0.683
2!3!
1!4!
0!5!
b. at least 6 out of 10 business days.
P(x ≥ 6) = 0.633
c. fewer than 5 out of 10 business days.
P(x ≤ 5) = 0.834
d. fewer than 6 out of the next 20 business days.
P(x < 6) = 0.0016
If this actually happened, might it shake you confidence in the statement p=0.60? Might it
make you suspect that p is less than 0.60? Explain.
Yes. This is an exceedingly rare occurrence, and unless I knew from other means that
p = 0.6, it would lead me to believe that p is indeed smaller than 0.6.
e. more than 17 out of the next 20 business days.
P(x > 17) = 0.003611
If this actually happened, might you suspect that p is greater than 0.60? explain.
Yes, this is also a very rare occurrence, and would indicate p > 0.6.
5.3
10. The mountain states office of state farm insurance company reports that
approximately 85% of all automobile damage liability claims were made by people under
25 years of age. A random sample of five automobile insurance claims is under study.
a. make a histogram showing the probability that r=0 to 5 claims are made by people
under 25 years of age?
probability (out of 5 random selections) 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 no. people under 25 b. find the mean and standard deviation of this probability distribution.
µ = Np = 4.25
σ = npq = 0.798
for sample size of 5, what is the expected number of claims made by people under 25
years of age?
µ = Np = 4.25
22. It is reported that approximately 45% of all university professors are extroverted,
suppose you have classes with six different professors.
a. what is the probability that all six are extroverts?
P(6) = (0.45)6 = 0.008304
b. What is the probability that non of your professors is an extrovert?
P(0) = (0.55)6 = 0.027681
c. What is the probability that ar least two of your professors are extroverts?
P(x ≥ 2) = 0.836433
d. In a group of six professors selected at random, what is the expected number of
extroverts? What is the standard deviation of the distribution?
µ = Np = 2.7
e. quota problem:
???
σ = Npq = 1.219