Question

I understand how to do part 2 and 3 of a) as well as part 2 of b). Can you please explain part 1 of a) and part 1 and 3 of b)?

(4 points) Sometimes frustrated students will turn to "homework" websites for assignment and exam solutions instead of trying to understand the concepts for themselves.

* Say the probability the University is able to track-down and charge a student who posted on these websites with cheating is 0.06.

* Say the number of solutions posted online from assignment 5, is 19 out of a total 36.

* Say a student using only online solutions typically was able to earn marks in any given assignment. (Consider only whole marks being possible)

(use at least five digits after the decimal point if rounding)


(a)
If there were 36 students, who had posted their solutions on these websites and given the University was able to charge at most 17 of them with cheating, what is the probablity the actual number of students charged with cheating is at least 7?

How many of the students who posted would you expect to be charged with cheating?

What is the variance for how many of the students who posted would you expect to be charged with cheating?

(b)
On Assignment 5 a student has completed 31 of 36 total problems. What is the probability that at least 16 of the completed problems were online?

Of the 31 completed problems how many would you expect to be able to find online?

Of the 31 completed problems what is the standard deviation for how many would you expect to be able to find online?

Answer 1

a)

1)Given that

Number of students who posted online=n=36

Probability that students charged with cheating=P=0.06

Let X is number of students charged with cheating

So X~Bin (36,0.06)

That is

we have to find that

P(X>​​​​7 |X<17)=?

Now

Now

so

b)

1)no.of students who completed their assignment=n=31

Probability that solutions posted online from assignment 5=P=19/36=0.53

Let X is number of solutions posted online out of 31 completed assignments .

So X~Bin (31,0.53)

We have to find P(X>16)

Now

3)

Var(X)=n*p*(1-p)=31*0.53*0.47=7.7221

So SD(X)=square root(var(X))=2.7789

sd(expectation)=SD/sqrt{n}=2.7789/sqrt{31} =2.0036