Problem 4: House Prices
Use the “Fairfax City Home Sales” dataset for parts of this problem.
a) Use StatCrunch to construct an appropriately titled and labeled relative frequency histogram of Fairfax home closing prices stored in the “Price” variable. Copy your histogram into your document.
b) What is the shape of this distribution? Answer this question in one complete sentence.
c) Assuming the population has a similar shape as the sample with population mean $510,000 and population standard deviation $145,000; calculate the probability that in a random sample of size 10, the mean of the sample will be greater than $600,000. You may assume a random sample was taken and the sample came from a big population. However, be sure to check the central limit theorem condition of a large sample size before completing this problem using one complete sentence. If this condition is not met, you cannot complete the problem.
d) Assuming the population has a similar shape as the sample with population mean $510,000 and population standard deviation $145,000; calculate the probability that in a random sample of size 36, the mean of the sample will be greater than $600,000. You may assume a random sample was taken and the sample came from a big population. However, be sure to check the central limit theorem condition of a large sample size before completing this problem using one complete sentence. If this condition is not met, you cannot complete the problem.
Data:
Price Year, Days, TLArea, Acres
369900 1922 44 1870 0.39
373000 1952 0 1242 0.27
375000 1952 8 932 0.15
375000 1950 2 768 0.19
379000 1952 31 816 0.21
380000 1941 53 1092 0.19
385000 1951 5 984 0.27
387700 1953 5 975 0.36
395000 1954 18 957 0.29
395000 1951 12 1105 0.22
399900 1954 29 1206 0.28
399900 1951 6 1226 0.18
400000 1954 31 957 0.27
410000 1949 6 1440 0.2
410000 1954 17 1344 0.23
412500 1954 4 1008 0.25
415000 1953 17 1371 0.28
420000 1954 2 957 0.25
426000 1952 3 1694 0.25
430000 1953 19 975 0.23
434900 1950 5 1128 0.18
435000 1954 32 1252 0.24
440000 1960 3 1161 0.26
440000 1954 2 1036 0.28
440000 1955 12 1645 0.28
440000 1960 5 1746 0.31
441000 1952 133 1062 0.23
442000 1961 4 1414 0.32
443000 1951 26 962 0.2
444900 1955 4 1122 0.19
446500 1953 3 962 0.26
450000 1952 2 1488 0.15
450000 1955 49 1122 0.23
450000 1979 0 1092 0.28
450000 1951 70 962 0.2
450000 1957 23 1300 0.51
451000 1947 12 1325 0.34
455000 1952 7 2267 0.81
455000 1962 4 1050 0.31
460000 1955 5 997 0.3
460000 1954 10 1125 0.17
465000 1954 77 1288 0.46
465900 1947 21 1309 0.19
469000 1963 153 1149 0.27
474000 1959 5 1319 0.32
475000 1955 4 1530 0.28
475000 1953 29 1008 0.2
475000 1955 6 1530 0.28
475000 1956 116 1345 0.5
475000 1956 1 1530 0.28
480000 1960 27 1236 0.27
480000 1959 133 1527 0.24
485000 1955 4 1008 0.24
485000 1956 74 977 0.24
488000 1960 11 1972 0.33
500000 1963 0 2145 0.25
500000 1953 14 1758 0.54
500500 1955 6 1630 0.28
510000 1959 5 1680 0.34
512000 1963 0 1968 0.22
519000 1961 1 1312 0.29
520000 1954 15 1492 0.25
520000 1958 80 1443 0.33
520000 1963 122 1822 0.32
530000 1962 6 1393 0.29
540000 1962 12 1414 0.25
543600 1962 4 1414 0.24
560000 1967 5 1530 0.28
560000 1961 16 1438 0.53
565000 1947 6 1510 0.25
565500 1967 5 1217 0.26
589000 1954 32 2368 0.3
593000 1954 9 2044 0.25
610000 1978 140 2091 0.09
655000 1976 180 2728 0.24
660000 1947 10 2635 0.22
665000 1950 37 2645 0.57
685000 1982 120 2752 0.09
795000 2002 259 3402 0.12
852000 2000 4 3215 0.11
895000 2000 63 3230 0.11
930000 2015 135 3175 0.15
940000 1860 42 3038 0.57
968500 1850 74 3630 0.34
1100000 2004 161 3640 0.19
In: Math
Can someone please answer these 3 by Friday?
Problem 2 Let D, E, F, G, and H be events such that P(D) = 0.7,
P(E) = 0.6, P(F) = 0.8, P(G) = 0.9, and P(H) = 0.5. Suppose that
Dc, E, Fc, Gc, and H are independent.
(a) Find the probability that all of the events D, E, F, G, and H
occur.
(b) Find the probability that at least one of the events D, E, F,
G, and H occurs.
Problem 3 Five men and five women are ranked according to their scores on an examination. Assume that no two scores are alike, and all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a woman (for example, X = 2 if the top-ranked person was a man, and the next-ranked person was a woman). Find the probability mass function (pmf) of the random variable X, and plot the cumulative distribution function (cdf) of X.
Problem 4 Suppose there are three cards numbered 2, 7, 10, respectively. Suppose you are to be offered these cards in random order. When you are offered a card, you must immediately either accept it or reject it. If you accept a card, the process ends. If you reject a card, then the next card (if there is one) is offered. If you reject the first two cards, you have to accept the final card. You plan to reject the first card offered, and then to accept the next card if and only if its value is greater than the value of the first card. Let X be a number on the card you have accepted in the end. Find the pmf of X and plot the cdf of X.
In: Statistics and Probability
compare and contrast the county data to your state as a whole. Present the county and state data in the same table to make comparisons easier. Finally, discuss the overall patterns you find in social inequalities and any suggestions for social change to address these issues in your community.
|
Hudson County |
New Jersey State |
|||
|
N |
% |
N |
% |
|
|
Total Population |
679,756 |
- |
8,960,161 |
- |
|
Ages in Year |
||||
|
< 18 |
138,879 |
20.4% |
1,999,821 |
22.3% |
|
≥18 |
540,877 |
79.6% |
6,960,340 |
77.7% |
|
Race/Ethnicity |
||||
|
Hispanic/Latino |
293,465 |
43.2% |
1,764,520 |
19.7% |
|
American Indian/ Alaska Native |
5,542 |
0.8% |
60,528 |
0.7% |
|
White |
393,494 |
57.9% |
6,265,011 |
69.9% |
|
African American |
94,057 |
13.8% |
1,323,092 |
14.8% |
|
Asian |
110,255 |
16.2% |
920,332 |
10.3% |
|
Other |
99,315 |
14.6% |
630,313 |
7.0% |
|
Poverty Status |
||||
|
Total population for whom poverty status is determined |
672,241 |
- |
8,783,989 |
- |
|
Below poverty level |
115,254 |
17.1% |
938,252 |
10.7% |
|
Work Experience |
||||
|
Total population age ≥ 16 for whom poverty status is determined |
83,268 |
15.2% |
667,174 |
9.5% |
|
Below poverty and worked full-time |
8,585 |
20.2% |
62,101 |
11.2% |
|
Below poverty and worked part-time |
23,259 |
30.6% |
184,060 |
18.5% |
|
Health Insurance |
||||
|
Total civilian population |
287,254 |
76.55 |
3,795,828 |
85.8% |
|
Un-insured |
74,391 |
23.45 |
438,511 |
14.2% |
|
Language spoken at Home |
||||
|
Total population age ≥ 5 |
632,315 |
8,433,445 |
- |
|
|
Spanish or Spanish Creole |
242,261 |
38.3% |
1,360,981 |
16.1% |
|
Speaks English very well |
257,949 |
40.8% |
5,821,459 |
69.0% |
|
Speaks English less than very well |
159,413 |
25.2% |
1,028,372 |
12.2% |
In: Statistics and Probability
Garcia Real Estate is involved in commercial real estate ventures throughout the United States. Some of these ventures are much riskier than other ventures because of market conditions in different regions of the country.
1) If Garcia does not risk-adjust its discount rate for specific ventures properly, which of the following is likely to occur over time? Check all that apply.
A. The firm will increase in value.
B. The firm’s overall risk level will increase.
C. The firm could potentially reject projects that provide a higher rate of return than the company should require.
2) How do managers typically deal with within-firm risk and beta risk when they are evaluating a potential project?
A. Subjectively
B. Quantitatively
Consider the case of another company. Turnkey Printing is evaluating two mutually exclusive projects. They both require a $5 million investment today and have expected NPVs of $1,000,000. Management conducted a full risk analysis of these two projects, and the results are shown below.
|
Risk Measure |
Project A |
Project B |
|---|---|---|
| Standard deviation of project’s expected NPVs | $400,000 | $600,000 |
| Project beta | 0.9 | 0.7 |
| Correlation coefficient of project cash flows (relative to the firm’s existing projects) | 0.6 | 0.8 |
3) Which of the following statements about these projects’ risk is correct? Check all that apply.
A. Project B has more market risk than Project A.
B. Project B has more stand-alone risk than Project A.
C. Project A has more stand-alone risk than Project B.
D. Project A has more market risk than Project B.
In: Finance
Let z denote a random variable having a normal distribution with μ = 0 and σ = 1. Determine each of the probabilities below. (Round all answers to four decimal places.)
(a) P(z < 0.2) =
(b) P(z < -0.2) =
(c) P(0.40 < z < 0.86) =
(d) P(-0.86 < z < -0.40)
=
(e) P(-0.40 < z < 0.86) =
(f) P(z > -1.24) =
(g) P(z < -1.5 or z > 2.50) =
In: Statistics and Probability
Calculate the expected return and standard deviation of the following. T.Bills, HT, Coll, USR, MP
|
Economy |
Prob |
T.Bills |
HT |
Coll |
USR |
MP |
|
Recession |
0.1 |
5.5% |
-27.0% |
27.0% |
6.0% |
-17.0% |
|
Below Avg |
0.2 |
5.5% |
-7.0% |
13.0% |
-14.0% |
-3.0% |
|
Average |
0.4 |
5.5% |
15.0% |
0.0% |
3.0% |
10.0% |
|
Above Avg |
0.2 |
5.5% |
30.0% |
-11.0% |
41.0% |
25.0% |
|
Boom |
0.1 |
5.5% |
45.0% |
-21.0% |
26.0% |
38.0% |
In: Finance
Object E is dependent on Objects A and B.
P(A works) = 0.90
P(A fails) = 0.10
P(B works) = 0.90
P(B fails) = 0.10
If Object A works, then Probability of Object E working is 0.6
If Object A fails, then Probability of Object E working is 0.2
If Object B works, then Probability of Object E working is 0.6
If Object B fails, then Probability of Object E working is 0.2
What is the Probability of Object E working?
In: Statistics and Probability
Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the 1980s and 1990s is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, 70% of all arrests are of males aged 15 to 34 years†. Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 38 arrests last month, 24 were of males aged 15 to 34 years. Use a 10% level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from 70%.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: p = 0 .7; H1: p < 0.7H0: p = 0.7; H1: p > 0.7 H0: p < 0 .7; H1: p = 0.7H0: p = 0.7; H1: p ≠ 0.7H0: p ≠ 0.7; H1: p = 0.7
(b) What sampling distribution will you use?
The standard normal, since np > 5 and nq > 5.The Student's t, since np < 5 and nq < 5. The Student's t, since np > 5 and nq > 5.The standard normal, since np < 5 and nq < 5.
What is the value of the sample test statistic? (Round your answer
to two decimal places.)
(c) Find the P-value of the test statistic. (Round your
answer to four decimal places.)
Sketch the sampling distribution and show the area corresponding to
the P-value.
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis? Are the data statistically
significant at level α?
At the α = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.10 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
(e) Interpret your conclusion in the context of the
application.
There is sufficient evidence at the 0.10 level to conclude that the true proportion of arrests of males aged 15 to 34 in Rock Springs differs from 70%.There is insufficient evidence at the 0.10 level to conclude that the true proportion of arrests of males aged 15 to 34 in Rock Springs differs from 70%.
In: Statistics and Probability
Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the 1980s and 1990s is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, 70% of all arrests are of males aged 15 to 34 years†. Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 36 arrests last month, 25 were of males aged 15 to 34 years. Use a 10% level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from 70%.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: p = 0.7; H1: p ≠ 0.7H0: p < 0 .7; H1: p = 0.7 H0: p = 0 .7; H1: p < 0.7H0: p = 0.7; H1: p > 0.7H0: p ≠ 0.7; H1: p = 0.7
(b) What sampling distribution will you use?
The Student's t, since np > 5 and nq > 5.The standard normal, since np < 5 and nq < 5. The standard normal, since np > 5 and nq > 5.The Student's t, since np < 5 and nq < 5.
What is the value of the sample test statistic? (Round your answer
to two decimal places.)
(c) Find the P-value of the test statistic. (Round your
answer to four decimal places.)
Sketch the sampling distribution and show the area corresponding to
the P-value.
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis? Are the data statistically
significant at level α?
At the α = 0.10 level, we reject the null hypothesis and conclude the data are statistically significant.At the α = 0.10 level, we reject the null hypothesis and conclude the data are not statistically significant. At the α = 0.10 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.10 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
(e) Interpret your conclusion in the context of the
application.
There is sufficient evidence at the 0.10 level to conclude that the true proportion of arrests of males aged 15 to 34 in Rock Springs differs from 70%.There is insufficient evidence at the 0.10 level to conclude that the true proportion of arrests of males aged 15 to 34 in Rock Springs differs from 70%.
In: Statistics and Probability
Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the 1980s and 1990s is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, 70% of all arrests are of males aged 15 to 34 years†. Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 39 arrests last month, 29 were of males aged 15 to 34 years. Use a 5% level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from 70%.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: p = 0.7;
H1: p ≠ 0.7H0: p
< 0 .7; H1: p =
0.7 H0: p = 0.7;
H1: p > 0.7H0:
p = 0 .7; H1: p <
0.7H0: p ≠ 0.7; H1:
p = 0.7
(b) What sampling distribution will you use?
The Student's t, since np > 5 and
nq > 5.The standard normal, since np > 5 and
nq > 5. The Student's t,
since np < 5 and nq < 5.The standard normal,
since np < 5 and nq < 5.
What is the value of the sample test statistic? (Round
your answer to two decimal places.)
(c) Find the P-value of the test statistic.
(Round your answer to four decimal places.)
Sketch the sampling distribution and show the area corresponding to the P-value.
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?
At the α = 0.05 level, we reject the null hypothesis
and conclude the data are statistically significant.At the α = 0.05
level, we reject the null hypothesis and conclude the data are not
statistically significant. At the α = 0.05
level, we fail to reject the null hypothesis and conclude the data
are statistically significant.At the α = 0.05 level, we fail to
reject the null hypothesis and conclude the data are not
statistically significant.
(e) Interpret your conclusion in the context of the application.
There is sufficient evidence at the 0.05 level to
conclude that the true proportion of arrests of males aged 15 to 34
in Rock Springs differs from 70%.There is insufficient evidence at
the 0.05 level to conclude that the true proportion of arrests of
males aged 15 to 34 in Rock Springs differs from
70%.
In: Statistics and Probability