Question

5. Nineteen people move out of a neighborhood; four are minorities. Of the nineteen, eight move onto a block with new housing, and one of these eight is a minority. How likely is it that, if there were no discrimination, less than two people out of the eight people on this new block would be minorities? If the resulting probability is less than 0.05, evidence for discrimination exists. Does such evidence exist in this case?

6. There is an average of four accidents per year at a particular intersection.   What is the probability of more than one accident there next month? Hint: Use 1 month = 1/12 of a year to first get the number of accidents that are expected next month.   

Please Help I cant figure this out !!!

Answer 1

5. Nineteen people move out of a neighborhood; four are minorities. Of the nineteen, eight move onto a block with new housing, and one of these eight is a minority. How likely is it that, if there were no discrimination, less than two people out of the eight people on this new block would be minorities? If the resulting probability is less than 0.05, evidence for discrimination exists. Does such evidence exist in this case?

Here we can consider a binomial model, where we will have a sample size of eight people that move out. They can be minority or they won't be so we have only 2 possible out comes.

P(being a minority) = 4 / 19 .........Since 4 out of 19 are minorities.

= 0.2105

X;denote the no. of people who are minorities

P( X =x) =

We need to find 'p' for lessthan 2 people beingminorities.

P( X <2) = P(X=0) + P( X =1)

=

= 0.1509 + 0.3219

P(X < 2) = 0.4728

since P( X < 2) > 0.05

We have evidence that discrimination does not exist.

6. There is an average of four accidents per year at a particular intersection.   What is the probability of more than one accident there next month?

We have been given an average value for a rare event. So we can model this using poisson distribution.

4 accidents in a year.

per year

Therefore 4/12 = 0.333 accidents in a month

per month

P( X= x) =

P( X > 1) = 1 - P( X <=1)

= 1 - [ P( X=0) + P( X = 1)]

= 1 - ()

= 1 - 0.9554

P( X > 1) = 0.04463