Critical Thinking Assignment # 2 (50 points) INSTRUCTIONS: In this second assign
ID: 3242506 • Letter: C
Question
Critical Thinking Assignment # 2 (50 points)
INSTRUCTIONS:
In this second assignment, you will further explore the data you obtained in critical assignment 1 by applying the concepts we have covered in chapters 6, 7, and 8.
USA
NON-USA
Mean GPA
3.23
3.63
Standard Deviation
2.64
3.57
Mean Age
24.50
29.50
Standard Deviation
7.88
10.33
Mean Hours spent on homework
9.44
10.73
Standard Deviation
6.76
7.68
Answer each of the questions below. Please show your calculations.
Calculate the age of a student at the 90th percentile for (a) the USA born group and (b) the foreign born group?
Assume that Broward College has announced that it will give a scholarship to students born in the USA who has a GPA between 3.75 and 4.00; calculate the percentage of students who will be eligible for a scholarship from the USA born group.
Assume that Broward College has announced that it will give scholarships to students who have GPA’s that falls within the top 2% of all GPA’s. Calculate the GPA a student will need to earn a scholarship from the USA born group.
In order to encourage students to spend more time doing homework at home, Broward College has announced that it will give a $100 scholarship to students who spend more than 10 hours per week doing homework at home. Calculate the percentage of students who will receive the scholarship from the NON-USAgroup.
Construct a 95% confidence interval estimate of the mean GPA for the USA born group. The sample size (n) for the USA born group is 231, Give your interpretation of the confidence interval estimate.
In a survey of 385 students at Broward College, North Campus, results revealed that 154 students were born outside of the USA and 231 were born in the USA. Construct a 90% confidence interval estimate for the proportion of students who were born outside the USA. Give your interpretation of the confidence interval estimate.
Suppose you wanted to estimate the percentage of students at North Campus who are foreign born, how many students must you survey if you wanted to be 95% confident in your result, with a margin of error of no more than 4%. Assume you have no estimate of p-hat.
Suppose you wanted to estimate the mean age of students born in the USA. How many students must you survey if you wanted to be 99% confident in your result and your error is no more than 3 years? Use the standard deviation from the table above.
In a survey of 385 students at Broward College, North Campus, results revealed that 154 students were born outside of the USA and 231 were born in the USA. Using these results, test the claim at the .03 significance level that the proportion of students who are born outside of the USA is less than 35 percent.Be sure to show all the steps in the hypothesis testing procedure including the correct wording of the final conclusion.
In a survey of 385 students at Broward College, North Campus, the mean GPA was found to be 3.15 with a standard deviation of .20. Using these results, test the claim at the .05 significance level that the mean GPA of students at North Campus is equal to 3.00. Be sure to show all the steps in the hypothesis testing procedure including the correct wording of the final conclusion.
Chapter 5 Review
10 items or less: The number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution.
X
P(X)
0
0.10
1
0.25
2
0.30
3
0.20
4
0.10
5
0.05
Does the table above satisfy the requirements for a probability distribution?
(a) Yes (b) No
Find the probability of getting exactly 2 customers in line?
(a) .10 (b) .30 (c) .50 (d) .20 (e) .40
Find the probability of getting no more than one customer in line?
(a) .30 (b) .40 (c) .50 (d) .60 (e) .35
Find the probability of getting at least two customers in line?
(a) .30 (b) .65 (c) .35 (d) .40 (e) .80
Find the probability of getting no more than two customers in line?
(a) .10 (b) .65 (c) .20 (d) .90 (e) .80
Compute the mean in the probability distribution above?
(a) 1.20 (b) 1.30 (c) 3.10 (d) 2.10 (e) .80
Compute the standard deviation of the probability distribution above.
(a) 1.30 (b) 1.960 (c) 0.980 (d) 2.890 (e) 1.963
STUDENT SURVEY DATA FILE.
Student #
Gender
Age
Race
Major
Crd/Hrs Sem
Total Crd Hrs
GPA
On line Classes
First in Coll
Employ
Hrs Work/wk
DependChild
High Sch
COB
Country of Birth
Crse Anxiety
Crse Difficulty
Hrs spend on sch wrk
Political affiliation
1
M
23
W
BUS ADM
6
32
2.10
Y
N
Y
25
N
US
US
US
3
3
4
N
2
F
37
B
Early Ed.
3
72
2.42
Y
Y
Y
50
Y
O
US
US
6
4
15
D
3
F
29
B
Resp. T
6
53
2.00
N
Y
Y
40
Y
US
US
US
6
4
5
D
4
F
26
W
RN
11
51
3.81
N
N
N
Y
US
US
US
5
6
10
R
5
F
21
H
BSN
6
50
2.80
Y
Y
Y
35
N
US
US
US
5
5
6
D
6
F
46
B
RN
2
4
2.80
N
Y
Y
40
Y
US
US
US
6
5
12
N
7
F
23
W
RN
9
15
DK
N
Y
Y
50
N
US
US
US
3
3
10
N
8
M
23
W
Pub.Adm.
9
N.S
3.00
Y
N
Y
30
N
US
US
US
3
4
8
R
9
M
30
W
S.C.Mgmt.
9
60
2.33
Y
N
Y
25
N
US
US
US
5
5
10
I
10
F
45
B
Crim.Just.
11
64
3.70
Y
N
Y
40
N
US
US
US
4
6
15
D
11
M
30
W
NonDeg.
3
0
N/A
N
Y
Y
40
N
US
US
US
3
5
10
I
12
M
39
W
S.C.Mgmt.
12
12
3.50
N
N
Y
8
N
US
US
US
5
5
24
I
13
F
19
O
RN
9
24
2.64
N
N
Y
25
N
US
US
US
5
4
15
I
14
M
19
W
Pol. Sci.
16
29
3.89
Y
N
N
N
US
US
US
2
4
4
I
15
F
19
B
Sociology
11
35
3.70
N
Y
Y
16
N
US
US
US
5
6
35
N
247
F
23
B
Soc.Work
6
21
3.00
N
Y
Y
32
N
US
Haiti
OTHER
4
2
3
N
248
F
21
B
RN
10
N/A
N/A
N
N
Y
35
N
US
Haiti
OTHER
5
2
21
R
249
F
19
O
Med.
3
110
3.29
N
Y
N
Y
O
Jamaica
OTHER
7
2
20
D
250
F
24
O
RN
9
78
3.07
N
Y
Y
30
Y
US
Haiti
OTHER
5
10
N
251
M
20
H
MIS/Fin
15
24
4.00
Y
Y
Y
40
N
US
Colombia
OTHER
5
4
15
D
252
M
20
B
RN
4
46
N
N
Y
30
N
US
Haiti
OTHER
5
3
2
N
253
F
20
B
RN
7
12
3.00
N
Y
Y
26
Y
US
Haiti
OTHER
5
4
4
N
254
F
26
B
Nutrition
12
48
2.90
Y
N
Y
32
N
US
Jamaica
OTHER
3
2
25
R
255
F
27
O
Bio.
13
N/A
N/A
Y
N
N
Y
US
Trini
OTHER
6
6
8
R
256
F
38
B
RN
6
51
3.41
Y
N
Y
35
Y
O
Jamaica
OTHER
4
3
12
D
257
F
51
B
BSN
9
35
3.00
N
N
Y
40
Y
US
Haiti
OTHER
7
7
10
D
258
M
43
B
H.C.Adm.
12
12
2.69
N
Y
Y
40
Y
O
Haiti
OTHER
3
4
36
D
259
F
48
H
Logistic
12
24
3.00
N
Y
Y
25
N
O
Peru
OTHER
7
4
30
I
260
M
24
W
Bus.Adm.
13
48
3.10
Y
N
Y
40
N
US
Brazil
OTHER
5
5
10
R
261
M
45
H
Mus.Ed.
6
34
N
N
Y
50
N
US
Argent.
OTHER
6
6
15
I
262
F
42
B
RN
12
41
3.27
Y
Y
Y
20
Y
O
Jamaica
OTHER
7
7
20
N
263
F
38
B
Bus.
6
6
3.00
N
Y
Y
20
Y
O
Bahama
OTHER
1
6
30
D
264
F
34
B
RN
10
22
2.76
Y
N
Y
42
Y
O
Jamaica
OTHER
6
6
10
N
265
M
49
B
Nursing
11
12
3.50
N
Y
N
Y
OU
NIGERIA
OTHER
7
1
15
N
USA
NON-USA
Mean GPA
3.23
3.63
Standard Deviation
2.64
3.57
Mean Age
24.50
29.50
Standard Deviation
7.88
10.33
Mean Hours spent on homework
9.44
10.73
Standard Deviation
6.76
7.68
Explanation / Answer
From the given table
Does the table above satisfy the requirements for a probability distribution?
Correct Answer: Option (A) Yes, Since sum of probabilities are equal to 1
Find the probability of getting exactly 2 customers in line?
Correct Answer: Option (B) 0.30
Find the probability of getting no more than one customer in line?
Correct Answer: Option (E) 0.35
since P(X<=1) = P(X=0) + P(X=1) = 0.1+ 0.25 = 0.35
Find the probability of getting at least two customers in line?
Correct Answer: Option (B) 0.65
since P(X>=2) = 1- P(X<2) =1 - 0.35 = 0.65
Find the probability of getting no more than two customers in line?
Correct Answer: Option (B) 0.65
since P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 0.1+ 0.25+ 0.3 = 0.65
Compute the mean in the probability distribution above?
Correct Answer; Option (D) 2.10
since
Compute the standard deviation of the probability distribution above.
Correct Answer: Option (A)
since
Variance = 6.1 - 2.12 = 1.69
SD = Sqrt(1.69) = 1.3
x P(X) 0 0.1 1 0.25 2 0.3 3 0.2 4 0.1 5 0.05Related Questions
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