A distribution of grades in an introductory statistics course (where A 4, B 3, e
ID: 3072690 • Letter: A
Question
A distribution of grades in an introductory statistics course (where A 4, B 3, etc) is 0 P(X)0.11 0.170.24 0.32 0.16 Part a: Find the probability that a student has passed this class with at least a C (the student's grade is at least a 2) Part b: Find the probability that a student has an A (4) given that he has passed the class with at least a C rt e: Find the expected grade Part d: Find the variance and standard deviation for the class grades. BONUS: Suppose a student knows he/she has passed the course with a C, what should they expect their Pa in this class grade to be?Explanation / Answer
Solution:
We are given
X
P(X)
X*P(X)
0
0.11
0
1
0.17
0.17
2
0.24
0.48
3
0.32
0.96
4
0.16
0.64
Total
1
2.25
Part a
Here, we have to find P(X2)
P(X2) = P(X=2) + P(X=3) + P(4)
P(X2) = 0.24 + 0.32 + 0.16
P(X2) =0.72
Required probability = 0.72
Part b
Here, we have to find P(X=4|X2)
P(X=4|X2) = P(X=4)/P(X2)
P(X=4) = 0.16
P(X2) = P(X=2) + P(X=3) + P(4)
P(X2) = 0.24 + 0.32 + 0.16
P(X2) =0.72
P(X=4|X2) = P(X=4)/P(X2) = 0.16/0.72 = 0.222222
Required probability = 0.222222
Part c
Expected grade = X*P(X) = 2.25
X
P(X)
X*P(X)
0
0.11
0
1
0.17
0.17
2
0.24
0.48
3
0.32
0.96
4
0.16
0.64
Total
1
2.25
Expected grade = 2.25
Part d
Formulas for variance and standard deviation are given as below:
Variance = (X - mean)^2*P(X)
SD = Sqrt[(X - mean)^2*P(X)]
Calculation table is given as below:
X
P(X)
XP(X)
(X - mean)^2
(X - mean)^2*P(X)
0
0.11
0
5.0625
0.556875
1
0.17
0.17
1.5625
0.265625
2
0.24
0.48
0.0625
0.015
3
0.32
0.96
0.5625
0.18
4
0.16
0.64
3.0625
0.49
Total
1
2.25
1.5075
Variance = (X - mean)^2*P(X) = 1.5075
Standard deviation = sqrt(1.5075) = 1.227802916
*Bonus
If student knows he/she has passed the course with a C, they should expect their grade as greater than 2 and less than 3.
X
P(X)
X*P(X)
0
0.11
0
1
0.17
0.17
2
0.24
0.48
3
0.32
0.96
4
0.16
0.64
Total
1
2.25
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