A distribution of grades in a course where A=4, B=3, etc.

X=0, 1, 2, 3, 4

P(X)=0.1, 0.17, 0.21, 0.32, 0.2

a. Find the probability that a student has passed this class with at least a C (at least a 2)

b. Find the probability that a student has an A (4) given that he has passed the class with at least a C (2).

c. Find the expected grade in this class

d. Find the variance and standard deviation for the class grades.

e. Suppose a student knows they passed the course with a C, what should they expect their grade to be?

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akiran007 akiran007 · Answer 1 · 2020-03-08T06:47:10+01:00

Answer:

a. P(x= at least 2)= 1- P(no pass)= 1- 0.27= 0.73

b. 0.55

c. Expected Value = mean = 2.35 which is more than C grade that is grade B

d. variance= 1.5675

standard deviation =1.252

e. Their grade would be from (73- 100)

or 3 as it has the highest probability

Step-by-step explanation:

X P(X) X .P(X) X² X².P(X)

0 0.1 0 0 0

1 0.17 0.17 1 0.17

2 0.21 0.42 4 0.84

3 0.32 0.96 9 2.88

4 0.2 0.8 16 3.2

∑ 10 1 2.35 30 7.09

Let X represent an event that the student has passed with at least a 2.

The probability of not passing (or below 2) is 0.1+ 0.17= 0.27

Then using law of complementation

a. P(x= at least 2)= 1- P(no pass)= 1- 0.27= 0.73

b. Let Y represent the event that a student has an A (4) given that he has passed the class with at least a C (2)

P(x)= 0.73

P(A)= 0.4

P(Y)= P(A)/P(X)= 0.4/0.73=0.55

c. Expected value= mean = 2.35 which is greater than C , Hence grade B

First we find the mean ∑X.P(X)= 2.35

d. Variance = ∑X².P(X) - (∑X.P(X)²= 7.09- (2.35)² = 7.09- 5.5225=1.5675

Standard Deviation = square root of Variance = √1.5675= 1.252

e. P(all pass) = P(A) +P(B) +P(C)= 0.2+ 0.32+ 0.21= 0.73