The output of a chemical process is continually monitored to ensure that the concentration remains within acceptable limits. whenever the concentration drifts outside the limits, the process is shut down and recalibrated. let x be the number of times in a given week that the process is recalibrated. its cumulative distribution function is fx(x) = 0, x < 0 0.17, 0 ≤ x < 1 0.53, 1 ≤ x < 2 0.84, 2 ≤ x < 3 0.97, 3 ≤ x < 4 1, x ≥ 4 (a) what is the probability that the process is recalibrated less than two times during the week ? (b) what is the probability that the process is recalibrated more than three times during the week ? (c) what is the probability that the process is recalibrated exactly once during the week ? (d) what is the expected number times that the process is recalibrated during the week ?

a)  0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

If additionally the random variable is discrete (only has non-negative integers as outcomes as is the case in this problem) then

a)

We are looking for P(X<2)

b)

In this case we want P(X>3)

c)

Now, we are interested in P(X=1)

d)

The expected number of times that the process is recalibrated during the week is the expected value of the probability distribution:

P(X=1)+2P(X=2)+...+nP(X=n)+...

But it is easy to see that P(X=n) = 0 if n is an integer >4

So, the expected value is

P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)

We already have P(X=1) and P(X=2). Let's compute the rest

and the expected value is

0.36 + 2*0.53+3*0.13+4*0.03= 1.93 = 2 times rounding to the nearest integer.

homozygous. one is homozygous, while two is heterozygous.