Suppose a random variable X has a cumulative distribution function given by
F(a) = {0 for a<0, 1/2 for 0<=a<1, 3/5 for 1<=a<2, 4/5 for 2<=a<3,
9/10 for 3<=a<3.5, 1 for 3.5<=a.
a. Find the probability mass function for X.
b. Find the probability that a given observation of the value of X is greater than or equal to 1.5.
I'm confused. I think to answer this question, we need to know the range of your variable a. Does a go from -infinity to +3.5?
Nevermind, I get it.
P(x) = 5/10 for 0<= x < 1
= 1/10 for 1<= x < 2
= 2/10 for 2<= x < 3
= 1/10 for 3<= x < 3.5
= 1/10 for 3.5<= x
P(x>1.5) = 5/10
a. To find the probability mass function (pmf) for X, we need to find the probability of each possible outcome. Recall that the pmf is defined as the derivative of the cumulative distribution function (CDF).
For X < 0, the probability is 0 since the CDF is 0 for a < 0.
For 0 ≤ X < 1, the probability is F(1) - F(0) = 1/2 - 0 = 1/2.
For 1 ≤ X < 2, the probability is F(2) - F(1) = 3/5 - 1/2 = 1/10.
For 2 ≤ X < 3, the probability is F(3) - F(2) = 4/5 - 3/5 = 1/5.
For 3 ≤ X < 3.5, the probability is F(3.5) - F(3) = 9/10 - 4/5 = 1/10.
For 3.5 ≤ X, the probability is F(∞) - F(3.5) = 1 - 9/10 = 1/10.
Therefore, the probability mass function (pmf) for X is:
P(X = 0) = 0
P(X = 1) = 1/2
P(X = 2) = 1/10
P(X = 3) = 1/5
P(X = 3.5) = 1/10
P(X = x) = 0 for other values of x.
b. To find the probability that a given observation of the value of X is greater than or equal to 1.5, we need to sum the probabilities of all values greater than or equal to 1.5.
P(X ≥ 1.5) = P(X = 1.5) + P(X = 2) + P(X = 3) + P(X = 3.5)
= 0 + 1/10 + 1/5 + 1/10
= 3/10
Therefore, the probability that a given observation of the value of X is greater than or equal to 1.5 is 3/10.
To find the probability mass function (pmf) for a random variable X, we need to determine the probabilities of each possible value that X can take.
a. To find the pmf for X, we need to determine the probability of X taking each integer value. Given the cumulative distribution function (CDF) F(a) provided, we can identify the jump points where the CDF changes its value.
From the CDF, we can determine the probabilities of X taking each integer value as follows:
P(X = 0) = F(0) - F(-1) = 1/2 - 0 = 1/2
P(X = 1) = F(1) - F(0) = 3/5 - 1/2 = 1/10
P(X = 2) = F(2) - F(1) = 4/5 - 3/5 = 1/5
P(X = 3) = F(3) - F(2) = 9/10 - 4/5 = 1/10
P(X = 4) = F(3.5) - F(3) = 1 - 9/10 = 1/10
For any other integer values, the probability is 0, since the CDF is constant and does not change.
The probability mass function for X can be summarized as follows:
P(X = 0) = 1/2
P(X = 1) = 1/10
P(X = 2) = 1/5
P(X = 3) = 1/10
P(X = 4) = 1/10
P(X = k) = 0 for any other integer value k.
b. To find the probability that a given observation of the value of X is greater than or equal to 1.5, we need to calculate P(X >= 1.5).
Since the values between 1 and 1.5 have a cumulative probability of 3/5 - 1/2 = 1/10, we can add this probability to the probability of X being 2 or greater:
P(X >= 1.5) = P(X = 2) + P(X = 3) + P(X = 4) = 1/5 + 1/10 + 1/10 = 3/10.
Therefore, the probability that a given observation of the value of X is greater than or equal to 1.5 is 3/10.