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?

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