Answer: A student, Amara, plans to use the survey data to create a visualization and short summary of students' plans for college. First she wants to learn more about how the data was collected. Of the following things she might learn about the survey, which are the most likely sources of bias in the results based how it was collected:
- She learns that the survey administrators only asked a representative sample of students, rather than even student in each state .
- She learns that responses were collected only by mobile app
.
- She learns the survey was available to complete in both digital and paper form
.
- She learns that the survey was only given to students with scores in the top 10% on the PSAT
.
Explanation:
Conditions for choosing the participants in surveys can affect responses. Collecting survey responses by using an app introduces inclusive bias into the survey. This is when a group is selected for convenience and is not at all a random selection. Using an app has limited the participants to people who use the app.
Asking only the students who scored highly in tests also skews the neutrality of the survey. This introduces an omission bias by excluding a majority of students.
For this particular problem there will be Infinite solutions
Answer:
women's club; educational; charitable.
Explanation:
The Women's club was the social movement started by middle-class women's. During American Industrialization, many women's started this club with the purpose that they had moral duty and responsibility towards the society. The focus of these clubs was educational, self-improvement, and opportunities for women's.
Over the period of time, this focus of these clubs had shifted to charitable activities, especially during the times of the Great Depression.
Answer:
p(w) = p = 1/1000 = 0.001
q = 1-p = 0.999
reward for winning = 600$
payoff for attempt = 0.6$
a.
on an avg. jose loose 0.4$ per hand,
so equation will be to find out one win per x hands,
-0.4 x = -0.6x + 600
so x = 3000.
Hence 1 win out of 3000,
P(losing 0.4$) = \binom{3000}{1}(0.001)^1(0.999)^{2999}
P(losing 0.4$) = 0.1492
b.
Probability of winning 0.4$ per winning
equation,
0.4 x = -0.6x + 600
so ,
x = 600
so out of 600 game he have to win 1 game.
P(winning 0.4$) = \binom{600}{1}(0.001)^1(0.999)^{599}
P(winning 0.4$) = 0.3295
c.
Probability of winning 0.6$ per winning
equation,
0.6 x = -0.6x + 600
so ,
x = 500
so out of 500 game he have to win 1 game.
P(winning 0.6$) = \binom{500}{1}(0.001)^1(0.999)^{499}
P(winning 0.6$) = 0.3034
Attached is the same solutions incase the above isn't understandable.
