Answer :
Answer:
[tex] E(X) = 4*0.2 +3*0.5+ 2*0.2 +1*0.8+ 0*0.02= 2.78[/tex]
So then the best answer for this case would be:
C. 2.78
Step-by-step explanation:
For this case we have the following probabability distribution function given:
Score P(X)
A= 4.0 0.2
B= 3.0 0.5
C= 2.0 0.2
D= 1.0 0.08
F= 0.0 0.02
______________
Total 1.00
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
If we use the definition of expected value given by:
[tex] E(X) = \sum_{i=1}^n X_i P(X_I)[/tex]
And if we replace the values that we have we got:
[tex] E(X) = 4*0.2 +3*0.5+ 2*0.2 +1*0.8+ 0*0.02= 2.78[/tex]
So then the best answer for this case would be:
C. 2.78
Based on the information given the expected value is: C. 2.78.
Expected value
Since the standard GPA scale of 4-A to 0-F applies, thus the expected value is calculated as:
Expected value:
Expected value=(4×0.20)+ (3× 0.50) +(2× 0.20)+(1× 0.08)+(0× 0.02)
Expected value=0.8+1.5+0.4+0.08+0
Expected value=2.78
Inconclusion the expected value is: C. 2.78.
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