Question

The following probabilities for grades in management science have been determined based on past
records:
Grade Probability
A 0.1
B 0.2
C 0.4
D 0.2
F 0.10
1.00
The grades are assigned on a 4.0 scale, where an A is a 4.0, a B a 3.0, and so on. Determine the
expected grade and variance for the course.

Answer 1

Answer on Question #55481 – Math – Statistics and Probability

The following probabilities for grades in management science have been determined based on past records:

Grade Probability

A 0.1

B 0.2

C 0.4

D 0.2

F 0.1

1.00

The grades are assigned on a 4.0 scale, where an A is a 4.0, a B a 3.0, and so on. Determine the expected grade and variance for the course.

Solution

Let $\gamma$ be a random variable, which takes on values 4, 3, 2, 1, 0 with the following probabilities:

$$P(4) = P(\gamma = 4) = P(\text{grade equals } 4) = 0.1$$ $$P(3) = P(\gamma = 3) = P(\text{grade equals } 3) = 0.2$$ $$P(2) = P(\gamma = 2) = P(\text{grade equals } 2) = 0.4$$ $$P(1) = P(\gamma = 1) = P(\text{grade equals } 1) = 0.2$$ $$P(0) = P(\gamma = 0) = P(\text{grade equals } 0) = 0.1$$

So the expected grade is given by

$$E(\gamma) = \sum_{k=0}^{4} kP(k) = 4 \cdot 0.1 + 3 \cdot 0.2 + 2 \cdot 0.4 + 1 \cdot 0.2 + 0 \cdot 0.1 = 0.4 + 0.6 + 0.8 + 0.2 = 2$$

Method 1

The variance for the course is equal to

$$Var(\gamma) = \sum_{k=0}^{4} (k - E(\gamma))^2 P(k) = \\ = (4 - 2)^2 \cdot 0.1 + (3 - 2)^2 \cdot 0.2 + (2 - 2)^2 \cdot 0.4 + (1 - 2)^2 \cdot 0.2 + (0 - 2)^2 \cdot 0.1 = 1.2.$$

Method 2

The variance for the course is equal to

$$Var(\gamma) = \sum_{k=0}^{4} k^2 P(k) - (E(\gamma))^2 =$$ $$= 4^{2} \cdot 0.1 + 3^{2} \cdot 0.2 + 2^{2} \cdot 0.4 + 1^{2} \cdot 0.2 + 0^{2} \cdot 0.1 - 2^{2} = 16 \cdot 0.1 + 9 \cdot 0.2 + 4 \cdot 0.4 + 0.2 + 0 - 4 = 1.2.$$

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