Question

can you please answer 2a b c and d

2. Use the table below to calculate the answers to the questions that follow it. Be sure to write how you are calculating your answer. You must include the numbers from the table when showing your work. The following table shows a sample of CSI whether they signed up for the Winter session since attending CSI and whether they live in Staten Island. Results of a survey follow: Residence/Winter Staten Island:Staten Island: No Total Yes Ever done a Winter 19 30 120 150 session: Yes Ever done a Winter 56 64 session: No Total 75 75 a. What proportion of students have done Winter sessions? b. What proportion of Staten Island residents have done Winter sessions? What proportion of non-Staten Island residents have done Winter sessions? Which tell? session is 0.25. What is the sampling distribution of the proportion of students type of student is more likely to have taken a winter session? How can you c. Assume that the proportion of all CSI students who have taken a winter who have ever taken a Winter session based on a sample of 150 students? d. How likely is it that we would get a sample proportion less than what we obtained in part a?

Answer 1

(a)

P(Winter session) = 30/150 = 0.2

So,

Answer is:

0.2

(b)

(i)

P(Staten Island AND Winter session) = 19/150 = 0.1267

So,

Answer is:

0.1267

(ii)

P(Non-Staten Island AND Winter session) = 11/150 = 0.0733

So,

Answer is:

0.0733

(iii) Staten Island residents are more likely to have taken Winter session because their probability is more than Non-Staten Island residents.

(c)

n = 150

p = 0.25

q = 1- p = 0.75

So,'

SE = $\sqrt{pq/n}=\sqrt{0.25\times 0.75/150}=0.0354$

So,

The sampling distribution of the proportion of students who have even taken a Winter session is Normal Distribution with mean = 0.25 and standard deviation = 0.0354.

(d)

To find P($\hat{p}$<0.2):
Z = (0.2 - 0.25)/0.0354 = - 1.41

Table gives area = 0.4207

So,

P($\hat{p}$<0.2) = 0.5 - 0.4207 = 0.0793

So,

Answer is:

0.0793