On December 17, 2007 baseball writer John Hickey
wrote an article for the Seattle P-I about increases to
ticket prices for Seattle Mariners games during the 2008 season.
The article included a data set that listed the average ticket
price for each MLB team, the league in which the team plays (AL or
NL), the number of wins during the 2007 season and the cost per win
(in dollars). The data for the 16 National League teams are shown
below.
| team | league | price | wins | cost/win |
| Arizona Diamondbacks | NL | 19.68 | 90 | 35.40 |
| Atlanta Braves | NL | 17.07 | 84 | 32.89 |
| Chicago Cubs | NL | 34.30 | 85 | 65.33 |
| Cincinnati Reds | NL | 17.90 | 72 | 40.32 |
| Colorado Rockies | NL | 14.72 | 90 | 26.67 |
| Florida Marlins | NL | 16.70 | 71 | 38.13 |
| Houston Astros | NL | 26.66 | 73 | 59.11 |
| Los Angeles Dodgers | NL | 20.09 | 82 | 34.64 |
| Milwaukee Brewers | NL | 18.11 | 83 | 35.37 |
| N.Y. Mets | NL | 25.28 | 88 | 46.56 |
| Philadelphia Phillies | NL | 26.73 | 89 | 48.69 |
| Pittsburgh Pirates | NL | 17.08 | 68 | 40.67 |
| San Diego Padres | NL | 20.83 | 89 | 38.15 |
| San Francisco Giants | NL | 24.53 | 71 | 56.00 |
| St. Louis Cardinals | NL | 29.78 | 78 | 61.91 |
| Washington Nationals | NL | 20.88 | 73 | 46.30 |
Compute the correlation between average 2007 price and
cost per win for these 16 teams. (Assume the correlation
conditions have been satisfied and round your answer to the nearest
0.001.)
In: Statistics and Probability
One out of 25 healthy people carries a single gene for cystic
fibrosis (CF), these people are called carriers and
healthy people without a CF gene are called non-carriers.
A uniformly-chosen random healthy person has probability 1/25 of
being a carrier.
A person with two CF genes is not healthy; they are sick
(with cystic fibrosis). The child of a carrier has probability 1/2
of inheriting a CF gene from that parent. The child of two carriers
inherits each of their parents CF genes independently, so the child
has probability 1/4 of having CF.
Ok so this is how I solved it (I have not included all my steps as
they are long):
for number 1, I figured the probability is 1/2500 because the
probability of a child having CF is = to the probability of the
father being a carrier * by the probability that the mother is a
carrier * by the probability that it inherits both CF genes... 1/25
* 1/25 * 1/4...
for number 2, I got 2401/2500 because pr of child is a healthy non
carrier is = (pr of father being a carrier * pr of mother being a
non carrier)
+ (pr of father is non carrier * pr mother being a carrier * pr of
inheriting no CF gene)
+ (pr of father is non carrier * mother being a non carrier)
+ (pr of father is carrier * pr of mother is carrier * inheriting
no CF gene)
... calculating these probabilities gave me the answer above.
for number 3, I got 1/25 * 1/4 because the probability of it
inheriting both genes is 1/4 * by the probability of the other
parent being a carrier is 1/25 ... I got 1/100
for number 4, I got 49/1250 (this calculation was similar to #2 but
a bit longer)
for number 5, here we have the probability of a baby having CF
given that the baby has one CF gene:
P(baby has CF) = 1 - P(healthy non carrier) - P(healthy
carrier)
= 1 - 2401/2500 - 98/2500
= 1 / 2500
and the P(has CF given at least one CF gene)
= 1/ 2500
-------------
1/2500 + 98/2500
=1/99
Please let me know if I computed these correctly, if not to try and
guide me towards the direction!
thanks!
In: Statistics and Probability
The number of fires in a certain forest follows a Poisson process at a rate of 0.40 per week.
A) What is the probability that there are no more than 3 fires in a given week?
B) What is the the probability that the time between fires is at least 3 weeks?
In: Statistics and Probability
suppose you roll two fair dice.
A) what is the probability that you will roll an even number on the first die AND a 5 on the second die
B) What is the probability that the sum of the numbers on the two dice is 9?
show all work.
In: Statistics and Probability
From a group of 9 freshmen and 11 sophomores five students will be selected at random. The random variable x represents the number of freshman selected.
A)give the probability distribution
B)give the probability that more sophomores than freshman are selected
In: Statistics and Probability
Two cards are drawn at random from a standard deck of 52 cards. The number of aces drawn is counted. Prepare a probability distribution for this random experiment. Hint: Find the probability that no aces are drawn, exactly one ace is drawn, etc.
In: Statistics and Probability
Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 5 or odd? (Enter your probability as a fraction.)
Two dice are rolled. Determine the probability of the following. ("Doubles" means both dice show the same number.)
rolling a 4 or doubles
Use the data in the table below, which shows the employment status of individuals in a particular town by age group.
| Age | Full-time | Part-time | Unemployed |
|---|---|---|---|
| 0—17 | 27 | 170 | 358 |
| 18—25 | 199 | 199 | 272 |
| 26—34 | 342 | 71 | 22 |
| 35—49 | 521 | 175 | 238 |
| 50+ | 350 | 165 | 303 |
If a person is randomly chosen from the town's population, what is the probability that the person is under 18 or employed part-time?
In: Statistics and Probability
). An expert statistical typist averages about one error for every four pages typed. If the number of errors follow a Poisson distribution, then (i). What is the average errors per page? ( 2 pts). (ii). What is the probability that the typist will make no errors at all typing the next page? ( 3 pts). (iii). What is the probability that the typist will make two errors typing the next page? ( 3 pts). (iv). What is the probability that the typist will make at most four errors typing the next page? ( 5 pts). (v). What is the probability that the typist will make between five and eight errors inclusive typing the next page? ( 5 pts). (vi). What is the expected number of errors to be made by the typist if he/she averages about 5 errors for every 10 pages typed.
In: Statistics and Probability
The probability of success of a given vaccine is 0.82, that is, if a vaccinated patient is exposed to the disease, the probability of getting it is 18%. If I select 15 vaccinated patients, calculate the probability that:
a) None suffer the disease (3 pts)
b) Everyone suffers the disease (5 pts)
c) Two of them contract the disease (5 pts)
d) Determine the expected number of patients who will contract the disease (2 pts)
e) To complete a clinical study, 14 vaccinated patients who contract the disease are needed. What is the expected number of vaccinated patients who must be exposed to the disease to achieve the 14 required? (5 pts)
f) If of the 15, it is known that at least 2 contracted the
disease, what is the probability that less than 6 of the 15
vaccinated will contract the disease?
In: Statistics and Probability
Two transmitters send messages through bursts of radio signals to an antenna. During each time slot each transmitter sends a message with probability 1/2. Simultaneous transmissions result in loss of the messages. Let X be the number of time slots until the first message gets through.
(a) Describe the underlying sample space S of this random experiment and specify the probabilities of its elementary events.
(b) Show the mapping from S to the range of X.
(c) Find the probabilities for the various values of X. Now suppose that terminal 1 transmits with probability 1/2 in a given time slot, but terminal 2 transmits with probability p.
(d) Find the pmf for the number of transmissions X until a message gets through.
(e) Given a successful transmission, find the probability that terminal 2 transmitted.
In: Statistics and Probability