Question

One-hundred students were allowed to re-take an exam for their math course. The probability distribution shows how studying for the latest exam affected their grade when compared with the first time they took the exam. What is the probability that a student who studied for the exam saw an increase in their exam grade? Round to the nearest thousandth. Exam Grades Studied Did Not Study Totals Raise in Grade 0.52 0.06 0.58 No Raise in Grade 0.05 0.37 0.42 Totals 0.57 0.43 1

Answer 1

you need to format the data better so we can help you

Answer 2

right now it makes no sense

Answer 3

take a picture or draw it

Answer 4

I know that's why I am to but I go to online school and that is how it came up on the screen

Answer 5

can somebody please help me figure this out please??

Answer 6

sorry no, just take a screenshot. use the tool on windows, or on mac press command-shift-4 and drag over what you want to click

Answer 7

like that

Answer 8

how do I take a screenshot on windows???

Answer 9

programs >> accessories >> snipping tool

Answer 10

from what i remember, i don't know windows 8 if you have that

Answer 11

I use prt scrn > paint > paste > select the problem > crop > save > attach

Answer 12

here watch this video:

http://www.capture-screenshot.org/snipping-tool/

or read the short instructions

Answer 13

\(\large

\begin{array}{|c|c||c|}
\hline
\text{Exam Grades}&\text{Studied}&\text{Did Not Study}&\text{Totals}\\
\hline
\text{Raise in Grade}&0.52&0.06&0.58\\
\hline
\text{No Raise in Grade}&0.05&0.37&0.42\\
\hline
\text{Totals}&0.57&0.43&1\\
\hline

\end{array}
\)

Answer 14

something like this ?

Answer 15

that's what I thought it would be

Answer 16

Okay, and your question is :
`What is the probability that a student who studied for the exam saw an increase in their exam grade? `

Answer 17

yes

Answer 18

So only look at the "\(\text{Studied}\)" column

Answer 19

ok

Answer 20

I took the screen shot

Answer 21

i chose B 0.897 and it was marked wrong

Answer 22

Probability for \(\text{"Studied" AND "Raise in Grade"} = 0.52 \)
Probability for \(\text{"Studied" } = 0.57 \)

\(\large \implies \) Probability for \(\text{Raise in Grade}\) given that he \(\text{Studied} = \dfrac{0.52}{0.57}\)

Answer 23

oh wow your teacher sucks, what a terribly formatted question :( i'm sorry, you should mention that the homework is not formatted well

Answer 24

simplify

Answer 25

i did and they said there was nothing they could do

Answer 26

C .912

Answer 27

dang. well yeah @ganeshie8 sounds right, conditional probability problem

Answer 28

yeah bayee's thm.. in a convoluted format lol

Answer 29

thank you guys for taking time and helping me you don't even know how much i appreciate it :D thank you!

Answer 30

I did nothing :) maybe taught you about screenshots lol

Answer 31

lol hey you helped me learn something though

Answer 32

I found an answer of .912

Answer 33

yeah i had help already but thank you so much for helping still

Answer 34

Okay wesniki23

Answer 35

Hey just a tip to elaborate on how @ganeshie8 derived 0.52/0.57:

So, essentially you have two things relevant to the question. Students who studied, and students who raised their grade.

Let's calls students who studied: group A

Let's call students who raised their grade: group B

Baye's Theorm states that:

\[P(B|A) = \frac{P(AB)}{P(A)}\]

Meaning, the probability of a student who raised their grade, given that they studied ... is equal to... the probability of students who studided AND raised their grade... divided by.... the probability of students who studied

this is derived from:

P(AB) = P(A)P(B}A)

which is just the basic multiplication rule

to find the intersection of A and B (students who studied and raised their grade), you multiply the probability of those who studied by those who raised their grade, given that they studied.

algebraically manipulating this equation to solve for P(B}A) is what @ganeshie8 did

Answer 36

The P(AB) = P(A)P(B|A) is the same as P(AB) = P(A)*P(B) as long as they are independent events. The P(B|A) accounts for dependence