Fill in the P=Xx values to give a legitimate probability distribution for the discrete random variable X, whose possible values are −1, 3, 4, 5, and 6.ValuexofX P=Xx−1 3 0.254 0.245 0.186
Question
Fill in the P=Xx values to give a legitimate probability distribution for the discrete random variable X, whose possible values are −1, 3, 4, 5, and 6.ValuexofX P=Xx−1 3 0.254 0.245 0.186
Solution
The sum of all probabilities in a probability distribution must equal 1. In this case, you already have the probabilities for when X equals 3, 4, and 5. These probabilities are 0.25, 0.24, and 0.18 respectively.
First, add these probabilities together: 0.25 + 0.24 + 0.18 = 0.67
Since the total probability must be 1, the combined probabilities for X equals -1 and X equals 6 must be 1 - 0.67 = 0.33
Without more information, we can't determine the individual probabilities for X equals -1 and X equals 6. However, if we assume they're equal, then each would be 0.33 / 2 = 0.165.
So, the complete distribution would be:
- For X = -1, P = 0.165
- For X = 3, P = 0.25
- For X = 4, P = 0.24
- For X = 5, P = 0.18
- For X = 6, P = 0.165
Please note that this is just one of many possible solutions. The actual values for P when X equals -1 and X equals 6 could be different, as long as they add up to 0.33.
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