In the judicial case of United States vs. City of Chicago, discrimination was charged in a qualifying exam for the position of Fire Captain. In the table below Group A is a minority group and Group B is a Majority Group.
Group A 10 14
Group B 417 145
A) If one of the test subjects is randomly selected, find the probability of getting someone who passed the exam.
B) Find the probability of randomly selecting one of the test subjects and getting someone who is in Group B or passed.
C) Find the probability of randomly selecting two different test subjects and finding that they are both in Group A.
D) Find the probability of randomly selecting one of the test subjects and getting someone who is in Group A and passed the exam.
E) Find the probability of getting someone who passed, given that the selected person is in Group A.
Based on the results above, can we make a probability argument that discrimination is present based on p = 0.05? Why or why not. I am not interested in theory here, only the impact of the probabilities above!
Suppose that we know that the average income in Malvern, PA is $30,000 and that the standard deviation is $2,000. Assume that household incomes in Malvern are normally distributed. Using the Empirical Rule, please provide the range of incomes that we can expect 68% of the data to lie within. What about 95% and 99.73%?
Assume that we have just learned that the population of Malvern is bimodal and not normally distributed. Using Chebychev's Theorem, what percent of our population can we expect to lie within plus or minus two standard deviations? What percent of our population can we expect to lie within plus or minus three standard deviations?
What kind of differences do we see in the above distributions when we compare the Empirical Rule to Chebychev's Theorem. Which approach provides more precision? Why can't we use plus or minus one standard deviation with the Chebychev analysis?© BrainMass Inc. brainmass.com October 25, 2018, 5:08 am ad1c9bdddf
1. First, note that there are 10 + 14 + 417 + 145 = 586 candidates who wrote the exam.
A) Of the 586 candidates, 10 + 417 = 427 passed, hence the probability of passing is 427/586 = 0.7287.
B) We are interested in people who are in group B or passed, these could be Group B guys who passed (417), group B guys who didn't pass (145) or group A guys who passed (10). Thus are 572 candidates who fit in this group. Thus the probability of finding one is 572/586 = 0.9761.
C) There are 24 people in group A. If we make our first selection, the chance of finding someone from group A is 24/586 = 0.04096. Once we picked the first guy, there are 23 people left in group A and 585 people in total. The chance of finding a second person from group A, conditioned on finding a first guy from A, is 23/585 = 0.039312.
Thus, by the law of conditional probability, P(both event 1 and event 2 occuring) = P(event 1)P(event 2|event 1) = 0.04096 X 0.039312 = 0.00161.
D) There are ...
Statistics: Discrimination suit filed against the City of Chicago for Fire Captain exam
Sufficiency and Order Statistics
Let Y1<Y2<...<Yn be the order statistics of a random sample of size n from the uniform distribution over the closed interval [-theta, theta ]
having pdf f(x; theta ) = (1/2(theta))I[-theta , theta ](x).
Argue that the mle of theta; equals theta;hat= max(-Y1, Yn).
Demonstrate that the mle theta;hat is a sufficient statistic for theta;.
Define at least two ancillary statistics for this distribution
See attachment for better symbol representation.View Full Posting Details