Question

An exam has 5 questions and each of them has 4 possible answers. A student gets 3 points
for each correct answer and loses 1 point for each wrong answer. Consider a student who
answers all questions completely at random. Let X denote the number of correct answers and
Y the number of points of this student at the end of the test. (A negative score is possible).
(a) Compute the mean and the standard deviation of Y , µY and σY .
(b) Compute P(µY − σY ≤ Y ≤ µY + σY ) and P(µY − 2σY ≤ Y ≤ µY + 2σY ).
(c) What is the probability that the student above gets a positive score?

Answer 1

(a) X is the number of correct answers , X takes values from 0 to 5

When X=0 , that means 0 correct answers , 5 wrong answers

For 5 wrong answers 5 points will be deducted , Y=-5

For X=1(1 correct , 4 wrong) , Y=3-4 =-1

For finding probability , we use binomial probability law

n= 5, p= 1/4 (probability of selecting correct answer)

We get the following table

	X	Y	P(Y)	Y.P(Y)	Y^2P(Y)
	0	-5	0.2373	-1.1865	5.9326
	1	-1	0.3955	-0.3955	0.3955
	2	3	0.2637	0.7910	2.3730
	3	7	0.0879	0.6152	4.3066
	4	11	0.0146	0.1611	1.7725
	5	15	0.0010	0.0146	0.2197
sum		30	1	-0.0001	15

Mean is

= -0.0001

Standard deviation of Y is

=

(b)

=

=

= P(Y=-5) +P(Y=-1) +P(Y=3)

= 0.8965

=

=

= P(Y=-5) +P(Y=-1) +P(Y=3) +P(Y=7)

= 0.9844

(c)

P( getting positive score) =P(Y >0)

=P(Y=3) +....+P(Y=15)

= 0.3672