capellitad9

2022-07-15

An upper-level math course is open to both udergraduate students and graduate students. THe table below gives the perfomance of the two level of students in the class last semester on the final exam:

$\begin{array}{|cc|}\hline & \text{Score, S, on Final Exam}\\ & S<70& 70\le S<80& 80\le S<90& 90\le S\le 100& \text{Total}\\ \text{Undergraduate}& 16& 14& 17& 3& 50\\ \text{Graduae}& 4& 10& 19& 7& 40\\ \text{Total}& 2024& 36& 10& 90\\ \hline\end{array}$

1) Find the probability that a student scored 80 or above and was a graduate student

2) Given that a student scored below 80, find the probability that the student was a graduate student

3) Find the probability that an undergraduate student scored 90 or above

$\begin{array}{|cc|}\hline & \text{Score, S, on Final Exam}\\ & S<70& 70\le S<80& 80\le S<90& 90\le S\le 100& \text{Total}\\ \text{Undergraduate}& 16& 14& 17& 3& 50\\ \text{Graduae}& 4& 10& 19& 7& 40\\ \text{Total}& 2024& 36& 10& 90\\ \hline\end{array}$

1) Find the probability that a student scored 80 or above and was a graduate student

2) Given that a student scored below 80, find the probability that the student was a graduate student

3) Find the probability that an undergraduate student scored 90 or above

Julianna Bell

Beginner2022-07-16Added 19 answers

Step 1

Event A = student score 80 and wus a graduate studebt. Because, there are total 46 student which score 80 or more but among thse 46 students 26(19+7) students are graduate

$\therefore n(A)=26\phantom{\rule{0ex}{0ex}}\therefore P(A)=\frac{n(A)}{n(S)}=\frac{26}{90}=0.288888\cong 0.2889$

Step 2

2) Event B = student score below 80

$n(B)=20+24=44\phantom{\rule{0ex}{0ex}}\therefore P(B)=\frac{n(B)}{n(S)}=\frac{44}{90}=0.488888$

Event A = student is graduate student

$(A\cap B)=$ student is graduate and score below 80

$n(A\cap B)=10+4=14\phantom{\rule{0ex}{0ex}}\therefore P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{14}{90}=0.155555\phantom{\rule{0ex}{0ex}}P(\text{student is graduate}|\text{student score is below 80})\phantom{\rule{0ex}{0ex}}P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.155555}{0.488888}=0.318181\phantom{\rule{0ex}{0ex}}\therefore P(A|B)\cong 0.3182$

Step 3

3) Event c = undergraduate student score 90 or above

$n(c)=3\phantom{\rule{0ex}{0ex}}\therefore P(c)=\frac{n(c)}{n(s)}=\frac{3}{90}=0.3333$

Event A = student score 80 and wus a graduate studebt. Because, there are total 46 student which score 80 or more but among thse 46 students 26(19+7) students are graduate

$\therefore n(A)=26\phantom{\rule{0ex}{0ex}}\therefore P(A)=\frac{n(A)}{n(S)}=\frac{26}{90}=0.288888\cong 0.2889$

Step 2

2) Event B = student score below 80

$n(B)=20+24=44\phantom{\rule{0ex}{0ex}}\therefore P(B)=\frac{n(B)}{n(S)}=\frac{44}{90}=0.488888$

Event A = student is graduate student

$(A\cap B)=$ student is graduate and score below 80

$n(A\cap B)=10+4=14\phantom{\rule{0ex}{0ex}}\therefore P(A\cap B)=\frac{n(A\cap B)}{n(S)}=\frac{14}{90}=0.155555\phantom{\rule{0ex}{0ex}}P(\text{student is graduate}|\text{student score is below 80})\phantom{\rule{0ex}{0ex}}P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.155555}{0.488888}=0.318181\phantom{\rule{0ex}{0ex}}\therefore P(A|B)\cong 0.3182$

Step 3

3) Event c = undergraduate student score 90 or above

$n(c)=3\phantom{\rule{0ex}{0ex}}\therefore P(c)=\frac{n(c)}{n(s)}=\frac{3}{90}=0.3333$

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