Question

A psychologist determined that the number of sessions required to obtain the trust of a new patient is either 1, 2, or 3. Let x be a random variable indicating the number of sessions required to gain the patient’s trust. The following probability function has been proposed.

f(x) = x/6 at x= 1,2,or 3 Does this probability distribution satisfy equation 5.1?

Answer 1

The probability function f(x) = x/6 for x = 1, 2, or 3 is a valid probability distribution function because the sum of all probabilities equals 1 and each probability is between 0 and 1.

Step-by-step explanation:

The student is asking if the probability function f(x) = x/6 for x = 1, 2, or 3 satisfies the properties of a probability distribution function (PDF). To verify this, we need to check two conditions: first, that the sum of all probabilities is 1, and second, that each individual probability is between 0 and 1, inclusive. For the given function f(x), we have three probabilities: f(1) = 1/6, f(2) = 2/6, and f(3) = 3/6. Summing these, we get 1/6 + 2/6 + 3/6 = 6/6 = 1, thus satisfying the first condition. Each probability f(x) is also within the range of 0 to 1, satisfying the second condition. Hence, this function is a valid PDF.

As an example, let's take a child psychologist interested in the number of times a newborn baby's crying wakes its mother after midnight, we denote this with random variable X. If X takes on values of 0, 1, 2, 3, 4, 5 with given probabilities, to be a PDF, these probabilities must add up to 1, and each P(x) must be between 0 and 1.