Question

Fill in the P(X-x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -5 ,3 , 4, 5 , and 6.

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To create a legitimate probability distribution, we need to ensure that the sum of all the probabilities is equal to 1.

Given the values of X: -5, 3, 4, 5, and 6, we can assign probabilities P(-5), P(3), P(4), P(5), and P(6) to each value respectively.

Let's denote the probability values as follows:

P(-5) = a

P(3) = b

P(4) = c

P(5) = d

P(6) = e

To create a probability distribution, we need to assign valid probabilities to these values. This means that each probability must be greater than or equal to 0, and the sum of all probabilities must equal 1.

a + b + c + d + e = 1

Now, we need to fill in the values of P(X-x) for each value of x. P(X-x) represents the probability of X taking on the value x.

For -5:

P(X-(-5)) = P(X+5) = a

For 3:

P(X-3) = b

For 4:

P(X-4) = c

For 5:

P(X-5) = d

For 6:

P(X-6) = e

Therefore, the probability distribution for the discrete random variable X, with possible values -5, 3, 4, 5, and 6, is:

P(X+5) = a

P(X-3) = b

P(X-4) = c

P(X-5) = d

P(X-6) = e

We just need to find values of a, b, c, d, and e that satisfy the condition a + b + c + d + e = 1.

Answer:

To create a legitimate probability distribution for the discrete random variable X, whose possible values are -5, 3, 4, 5, and 6, we assign probabilities as follows:

P(X+5) = a

P(X-3) = b

P(X-4) = c

P(X-5) = d

P(X-6) = e

The values of a, b, c, d, and e need to satisfy the condition a + b + c + d + e = 1.

Given the values of X: -5, 3, 4, 5, and 6, we can assign probabilities P(-5), P(3), P(4), P(5), and P(6) to each value respectively.

Let's denote the probability values as follows:

P(-5) = a

P(3) = b

P(4) = c

P(5) = d

P(6) = e

To create a probability distribution, we need to assign valid probabilities to these values. This means that each probability must be greater than or equal to 0, and the sum of all probabilities must equal 1.

a + b + c + d + e = 1

Now, we need to fill in the values of P(X-x) for each value of x. P(X-x) represents the probability of X taking on the value x.

For -5:

P(X-(-5)) = P(X+5) = a

For 3:

P(X-3) = b

For 4:

P(X-4) = c

For 5:

P(X-5) = d

For 6:

P(X-6) = e

Therefore, the probability distribution for the discrete random variable X, with possible values -5, 3, 4, 5, and 6, is:

P(X+5) = a

P(X-3) = b

P(X-4) = c

P(X-5) = d

P(X-6) = e

We just need to find values of a, b, c, d, and e that satisfy the condition a + b + c + d + e = 1.

Answer:

To create a legitimate probability distribution for the discrete random variable X, whose possible values are -5, 3, 4, 5, and 6, we assign probabilities as follows:

P(X+5) = a

P(X-3) = b

P(X-4) = c

P(X-5) = d

P(X-6) = e

The values of a, b, c, d, and e need to satisfy the condition a + b + c + d + e = 1.

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