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

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )

just sum

10 + 10 = 20