Download Econ 351 Test 3 Answer all Questions Due 12/10/13 at the

Econ 351 Test 3 Answer all Questions Due 12/10/13 at the beginning of the class
NB: Show all steps involved in getting the answer
Write answers in the space provided.
1.
Given the finite population (N) of five observation
X1=3;
a.
X2=5;
X3=7;
X4=9
X5=11
Identify all the possible sample of two from the finite population of 5 and list them.
This question can be construed in a number of ways. I will assume that we have a population of elemetns
such that every element has value 3, 5, 7, 9 or 11. If we draw a pair of samples from this population, and if
we further assume there are at least two of each type of sample, the following are the possible pairs we
could draw
(3,3)
(5,3)
(7,3)
(9,3)
(11,3)
b.
(3,5)
(5,5)
(7,5)
(9,5)
(11,5)
(3,7)
(3,9)
(5,7) (5,9)
(7,7) (7,9)
(9,7) (9,9)
(11,7) (11,9)
(3,11)
(5,11)
(7,11)
(9,11)
(11,11)
Proof that you have identified all the possible sample of 2 from the population using the one of the counting
rules explained in class.
Using the counting principle, there are 52 possible pairs (assuming order matters). This is exactly what we
got in (a) by enumerating all the possibilities.
c.
Find the means of the samples and construct a probability distribution of the sample means (or what the
book calls the relative frequency table).
d.
Graph the probability distribution of the sample mean.
e.
Use the probability distribution of the sample mean to find the expected value and the standard deviation of
the sample mean.
f.
Calculate the population mean and the standard deviation of the population.
2. The number of days between billing and payment of a charge account at a telephone company is
approximately normally distributed with a mean of 20 days and standard deviation of 5 days.
a)
What percent of the bills will be paid in 18 to 25 day.
18 - 20
25 - 20
= -0.4 and z25 =
= 1, so the probability of being between 18 and 25 days
5
5
is P = N(1) - N(-0.4) , where N is the cumulative normal distribution. Evaluating, we have
P = 0.8413- 0.3446 = 0.4968 , or 49.68 %
z18 =
b) What is the probability that a randomly selected bills will be paid in less than 15 days?
z15 =
c)
15 - 20
= -1, so P = N(-1) = 0.1587
5
What is the probability that a randomly selected bill will be paid in more than 30 days?
z30 =
30 - 20
= 2 , so P = 1- N(2) = 0.02275
5
(Show your answers graphically)
3. Application (Binomial approximation) Based on past experience, the police department of the city of Bowie
knows that 50% of the automobiles reported stolen are recovered. In a month in which 10 automobile are
stolen, what is the probability that:
a.
3 or more of the stolen automobile will be recovered.
B(k ³ 3) = 1- B(k £ 2) = 0.9893
(Here B is the cumulative binomial distribution for p = ½ and n = 10)
b.
Less than 2 automobiles will be recovered.
B(k £ 1) = 0.0107
c.
Between 1 and 2 automobiles will be recovered.
There is no integer exclusively between 1 and 2. However, if we want the probability that exactly 1 or
2 autos are recovered, this would be
B(k £ 2) - B(k £ 0) = 0.0537
4. Among adults in the U.S., 17% voted for George Bush in 1992. Assume a sample of 800 adults is taken.
a. What is the probability that the sample proportion is within two percentage points of the population
proportion?
15% of 800 is 120, and 19% of 800 is 152, so we’d need the resulting number to be between these two
values. Using the cumulative binomial distribution B(k,n) for N trials with a p = 0.17 probability of
success, the probability of being in this range is
P = B(k £ 152, N = 800) - B(k £ 119, N = 800) = 0.8798
b. If the sample size is doubled to 1600 what is the probability that the sample proportion is within two
percentage points of the population mean?
15% of 1600 is 240, and 19% of 1600 is 304, so
P = B(k £ 239, N = 1600) - B(k £ 304, N = 1600) = 0.9695
5. Answer the following questions
a. What general statement can you make about what happens to the sampling distribution of
as the
sample size is increased, is this logical, explain.
The distribution of the sample mean becomes narrower, remaining centered on the true population mean.
This does make sense. If we sample a population randomly many, many times, eventually, we will sample
the whole population! In that case the sample mean is exactly, by definition, the population mean.
b. Give 5 characteristics of the normal curve
1) it’s peak value occurs at the mean
2) the area under the curve is 1
3) the curve is uniformly decreasing as the distance from the mean increases.
4) it is symmetric about the mean
5) it extends from negative infinity to positive infinity
6) its value is positive everywhere
7) the areas within 1, 2 and 3 standard deviations of the mean are approximately 0.683, 0.954, and 0.997,
respectively
c. Distinguish between a Poisson probability distribution and a Uniform probability distribution
A Poisson probability distribution is a discrete distribution that has an asymmetric peak at the most
probable value, with a somewhat long tail extending to the right. This distribution is not defined for
negative values. Some examples are shown on the left below.
A uniform distribution is a flat distribution with probability p = 1 / (x2 – x1) within some range x1 to x2, and
zero everywhere else. It can be discrete or continuous. An example is shown on the right below.