SOLUTION: A student takes a multiple choice exam with 10 questions, each with four possible selections for the answer. A passing grade is 60% or better. Suppose that the student was unable t

Question 252818: A student takes a multiple choice exam with 10 questions, each with four possible selections for the answer. A passing grade is 60% or better. Suppose that the student was unable to find time to study for the exam and just guesses at each question. Find the probability that the student
a. gets at least one question correct
b. passes the exam
c. receives an "A" on the exam (90% or better)
d. How many questions would you expect the student to get correct?
e. Obtain the standard deviation of te number of questions that the student gets correct
Found 2 solutions by richwmiller, edjones:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
D)
I would expect about 25% or 1/4 since there are four choices per question.
so 10*1/4=10/4= between two and three correct
A) There is good chances of getting at least one right.
B) slim to none that he will pass and
C)a better chance of winning the megalottery than getting 90%

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
Except for question e) these can be figured out by expansion of a binomial.
Let r = .25 probability of getting question right. and w = .75 prob getting it wrong.
(r+w)^10 since there are 10 questions will tell us what we want to know.
a)
1-w^10 doesn't get them all wrong.
1-.75^10
=1-.0563
=.9437
.
b)
6-10 answers right
r^10+10r^9w+45r^8w^2+120r^7w^3+210r^6w4
=.0197
.
c)
9 or 10 right
r^10+10r^9w
=.00003
.
d)
10 * .25 = 2.5 expected correct answers.
.
e)
Standard deviation for a binomial distribution is sqrt(n*p(1-p))
=sqrt(10*.25*.75)
=sqrt(1.875)
=1.369
.
Ed
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