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.

To find the probability mass function (PMF) for a random variable X, we need to calculate the probability that X takes on each possible value. In this case, since X is a continuous variable, we have a piecewise defined cumulative distribution function (CDF).

a. To find the PMF for X, we need to calculate the difference in probabilities between consecutive intervals of X.

For X < 0, the probability is 0, so P(X = 0) = 0.
For 0 <= X < 1, the probability is 1/2 - 0 = 1/2, so P(X = 1) = 1/2.
For 1 <= X < 2, the probability is 3/5 - 1/2 = 1/10, so P(X = 2) = 1/10.
For 2 <= X < 3, the probability is 4/5 - 3/5 = 1/5, so P(X = 3) = 1/5.
For 3 <= X < 3.5, the probability is 9/10 - 4/5 = 1/10, so P(X = 3.5) = 1/10.
For X >= 3.5, the probability is 1 - 9/10 = 1/10, so P(X = 4) = 1/10.

Therefore, the 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 = 4) = 1/10

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 X taking on values greater than or equal to 1.5.

P(X >= 1.5) = P(X = 2) + P(X = 3) + P(X = 3.5) + P(X = 4)
= 1/10 + 1/5 + 1/10 + 1/10
= 1/10 + 2/10 + 1/10 + 1/10
= 5/10
= 1/2

Therefore, the probability that a given observation of the value of X is greater than or equal to 1.5 is 1/2.