Question

A student takes a multiple choice exam with 10 questions, each with four possible answers. 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 queston. Find the probability that the student:
a. gets at least one correct
b. passes the exam
c. receives an A (90% or higher)
d. How many do you expect him to get right?
e. Standard deviation of correct questions?

Answer 1

Answer:

Step-by-step explanation:

Let r = .25 probability of getting question right. and w = .75 prob getting it wrong.

$\left ( r+w \right )^{10}$ since there are 10 questions, this will tell us what we want to know.

a)  1-$\left (w\right )^{10}$ doesn't get them all wrong.

   1-$\left (0.75\right )^{10}$

  =1-.0563

  =.9437

b)  Probability of getting 6-10 answers right

$\left (r\right )^{10}+10\left ( r \right )^{9}w+45\left ( r \right )^{8}w^{2}+120\left ( r \right )^{7}w^{3}+120\left ( r \right )^{6}w^{4}$

=.0197

 

c) This means 9 or 10 answer should be right

$\left (r\right )^{10}+10\left ( r \right )^{9}w$

=.00003

d) $10\times 0.25$ = 2.5 expected correct answers.

e) Standard deviation for a binomial distribution is $\sqrt{n\ast p\left ( 1-p \right )}$

=$\sqrt{10\ast0.25\ast0.75}$

=$\sqrt{1.875}$

=1.369