Evaluate 5-\dfrac{t}35−
2 answers:
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Answer:
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:
And if we replace the values that we have we got:
So then the best answer for this case would be:
C. 2.78