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## Test2ans.dvi

Department of MathematicsMATH 130, Elements of Statistics I, Test 2October 22, 2009, 3:00PM-4:15PM
Please answer the following questions. Your answers will be evaluated on their correctness,completeness, and use of statistical concepts we have covered. Please show all work andwrite out your work neatly. Answers without supporting work will receive no credit. Thepoint values of the problems are listed in parentheses.

1. (3 points each) During the 2004 season, the St. Louis Cardinals won 64.8% of their
games. Assume that the outcomes of the baseball games are independent and that thepercentage of wins next season will be the same as in 2004.

(a) What is the probability that the Cardinals will win two games in a row?
Using the Multiplication Rule for independent events,
= P (win 1st)P (win 2nd)= (0.648)(0.648)≈ 0.420
(b) What is the probability that the Cardinals will lose at least one of their next six
P (lose at least 1 of 6) = 1 − P (win 6) = 1 − (0.648)6 ≈ 0.926
(c) What is the probability that the Cardinals will win exactly three of their next
five games?Using the formula for binomial probability

2. (3 points each) The following table lists the marital status of Americans 25 years old
or older (in thousands) and their level of education in 2003.

(a) What is the probability that a randomly selected individual who has never married
is a college graduate?According to the table there are
people in the survey who have never been married. Since 8321 of them are collegegraduates, the probability that a randomly selected individual who has nevermarried is a college graduate is
(b) What is the probability that a randomly selected individual who is a college
graduate has never married?According to the table there are
8321 + 34693 + 1190 + 614 + 1746 + 4490 = 51054
people in the survey who are college graduates. Since 8321 have never beenmarried, the probability that a randomly selected individual is a college graduateand has never married is

3. (4 points) Suppose you draw three cards without replacement from a standard 52-card
deck. What is the probability that all three cards are aces?
Using the Multiplication Rule for dependent events
P (3 aces) = P (1st ace) P (2nd ace | 1st ace) P (3rd ace | 1st and 2nd aces)
4. (4 points) In the Pennsylvania Cash 5 lottery, balls are numbered 1 to 39. Five balls are
selected at random without replacement. The order in which the balls are selected doesnot matter. To win, your numbers must match the five numbers selected. Determinethe probability of winning the Pennsylvania Cash 5 lottery with one ticket.

There are 39C5 = 575757 combinations of numbers that may appear on the winningticket. Thus the probability of winning with one ticket is
5. (4 points each) Suppose the life of refrigerators is normally distributed with mean
µ = 15 years and standard deviation σ = 2.7 years.

(a) On the bell-shaped curve below label the locations of the mean µ and µ − σ and
(b) Shade the region under the graph that represents the proportion of refrigerators
6. (4 points) The U.S. Senate Appropriations Committee has 29 members. Suppose they
wish to form a subcommittee by randomly selecting 5 members of the AppropriationsCommittee. How many different subcommittees can be formed?
There are 29C5 = 118755 subcommittees which can be formed since order of member-ship in the committee does not matter.

7. (4 points) Five people are told to sit down in a row of 12 chairs. In how many different
Since seating order matters when the people are different, there are 12P5 = 95040different permutations in which they can be seated.

8. (4 points each) Clarinex-D is a medication whose purpose is to reduce the symptoms
associated with a variety of allergies. In clinical trials of Clarinex-D, 5% of the patientsin the study experienced insomnia as a side effect.

(a) If 250 users of Clarinex-D are randomly selected, how many would we expect to
experience insomnia as a side effect?The number of patients experiencing insomnia is a binomial random variable.

Thus
(b) Would it be unusual to observe 25 patients experiencing insomnia as a side effect
in 250 trials of the medication? Justify your answer.

The variance in the number of patients experiencing insomnia is
σ2 = np(1 − p) = (250)(0.05)(1 − 0.05) = 11.875 > 10.

Thus we may apply the Empirical Rule in this case. The usual number of patientsexperiencing insomnia will be in the range
µX − 2σX = 5.6 ≤ X ≤ 19.4 = µX + 2σX
Since 25 is not in this range, it would be unusual.

9. (4 points each) A company is testing a new medicine for migraine headaches. In the
study, 150 women received the new medicine and an additional 100 women received aplacebo. Each participant was directed to take the medicine when the first symptomsof a migraine occurred and then to record whether the headache went away within 45minutes or lingered. The results are recorded in the following table.

(a) If a study participant is selected at random, what is the probability she received
(b) If a study participant is selected at random, what is the probability her headache
(c) If a study participant is selected at random, what is the probability she received
P (receive placebo and headache lingered) =

(d) If a study participant is selected at random, what is the probability she received
the medicine, given that her headache went away in 45 minutes?
P (received medicine | headache went away) =
10. (4 points each) In the following probability distribution, the random variable X rep-
resents the number of activities a parent of a student in grades 6 through 8 is involvedin.

(a) Verify that this is a probability distribution.

Since 0 ≤ P (x) ≤ 1 for each value of x and
P (x) = 0.073 + 0.117 + 0.258 + 0.322 + 0.230 = 1.000
this is a probability distribution for a discrete random variable.

(b) Compute the mean of random variable X.

(c) Compute the standard deviation of random variable X.

(d) What is the probability that a randomly selected parent is involved in fewer than
P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2) = 0.073 + 0.117 + 0.258 = 0.448
11. (4 points each) According to a study 80% of adult smokers started smoking before
turning 18 years old. Suppose 10 smokers 18 years old or older are randomly selectedand the number of smokers who started before age 18 is recorded.

(a) Explain why this is a binomial experiment.

• There are a fixed number of trials (n = 10).

• The trials are independent.

• There are only two outcomes on a single trial (smoker before age 18 or non-
• The probability of being a smoker before age 18 is the same across trials
(b) Find the probability that exactly 6 of them started smoking before age 18.

Using the results listed in Table II (Binomial Probability Distribution)
(c) Find the probability that fewer than 5 of them started smoking before age 18.

Using the results listed in Table III (Cumulative Binomial Probability Distribu-tion)
P (X < 5) = P (X ≤ 4) = 0.0064.

12. (3 points each) The local sporting good store sells a bag of 36 used golf balls. The bag
contains 22 Titleists, 7 Maxflis, and 7 Top-Flites. Two golf balls will be selected atrandom from the bag without replacement.

(a) What is the probability that the two randomly selected golf balls are both Ti-
tleists?Using the Multiplication Rule for dependent events,
P (2 Titleists) = P (1st Titleist) P (2nd Titleist | 1st Titleist)
(b) What is the probability that the first ball selected is a Titleist and the second is
a Maxfli?Using the Multiplication Rule for dependent events,
P (1st Titleist and 2nd Maxfli) = P (1st Titleist) P (2nd Maxfli | 1st Titleist)
(c) What is the probability that one ball selected is a Top-Flite and the other is a
Maxfli?Using the Multiplication Rule for dependent events,
P (Top-Flite and Maxfli) = P (1st Top-Flite and 2nd Maxfli)
= P (1st Top-Flite) P (2nd Maxfli | 1st Top-Flite)
+ P (1st Maxfli) P (2nd Top-Flite | 1st Maxfli)

Source: http://banach.millersville.edu/~BobBuchanan/math130/tests/fall2009/test2ans.pdf

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