Question

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 ?

Answer 1

Answer:

a)  0.31 = 31%

b) 0.03 = 3%

c) 0.36 = 36%

d) 2 times

Step-by-step explanation:

If $F_X(x)$ is the cumulative distribution function of the random variable X, then by definition the probability P of the random variable is given by

$P(X \leq x) = F_X(x)$

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

$P(X=a)=F_X(a)-\lim_{x \to a^-}F_X(x)$

a)

We are looking for P(X<2)

$P(X < 2) = P(X\leq 2)-P(X=2)=F_X(2)-\lim_{x \to 2^-}F_X(x)=0.84-0.53=0.31$

b)

In this case we want P(X>3)

$P(X >3) = 1-P(X\leq 3)=1-F_X(3)=1-0.97=0.03$

c)

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

$P(X =1) =F_X(1)-\lim_{x \to 1^-}F_X(x)=0.53-0.17=0.36$

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

$P(X =3) =F_X(3)-\lim_{x \to 3^-}F_X(x)=0.97-0.84=0.13$

$P(X =4) =F_X(4)-\lim_{x \to 4^-}F_X(x)=1-0.97=0.03$

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.

Answer 2

The probabilities for specified events and expectation are evaluated as:

• The probability that the process is recalibrated less than two times during the week is 0.53
• The probability that the process is recalibrated more than three times during the week is 0.03
• The probability that the process is recalibrated exactly once during the week is 0.36
• The expected number times that the process is recalibrated during the week is 1.47

What is cumulative distribution function?

Suppose that for a random variable X, its probability function be $f_X(x)$.

Then we have:

$CDF = F_X(x) = f_X(X \leq x)$

How to find the mean (expectation) of a random variable?

Supposing that the considered random variable is discrete, we get:

Mean = $E(X) = \sum_{\forall x_i} f(x_i)x_i$

As standard deviation is positive root of variance, thus,

where   $x_i; \: \: i = 1,2, ... ,n$ is its n data values

and $f(x_i)$ is the probability of   X = $x_i$

For the considered case, we're given that:

X = the number of times in a given week that the process is recalibrated

The cumulative distribution function of X is

• $F_X(x)$ : range

• 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

As we've $F_X(x+1)-F_X(x) = f_X(X \leq x+1) - f_X(X \leq x) = f_X(X =x)$ (if X takes integral values and x is an integer).

Therefore, we get the probability function for X as:

• x: f(x)
• 0: 0.17-0 = 0.17
• 1: 0.53 - 0.17 = 0.36
• 2: 0.84 - 0.53 = 0.31
• 3: 0.97 - 0.84 = 0.13
• 4: 1 - 0.97 = 0.03

Calculating the probabilities of the considered events:

a) The probability that the process is recalibrated less than two times during the week

$P(X < 2) = f_X(X \leq x= 2) -f_X(X = 2) =F_X(2) - f_X(X =2 )= 0.84 - 0.31\\P(X < 2) = 0.53$

Thus, the probability that the process is recalibrated less than two times during the week is 0.53

b) The probability that the process is recalibrated more than three times during the week

$P(X > 3) = 1 - P(X \leq 3) = 1 - f_X(X \leq 3) = 1 - F_X(3) = 1 - 0.97\\P(X > 3) = 0.03$

Thus, the probability that the process is recalibrated more than three times during the week is 0.03

c)  The probability that the process is recalibrated exactly once during the week

$P(X = 1) = f_X(X = 1) = 0.36$

Thus,  the probability that the process is recalibrated exactly once during the week is 0.36

d) The expected number times that the process is recalibrated during the week

It is the expected value of X.

It is evaluated as:

$E(X) = 0.f_X(X =0) + 1.f_X(X =1 )+ 2.f_X(X =2) \\+ 3. f_X(X =3) + 4.f_X(X =0)\\\\E(X) = 0.36 + 0.62 + 0.39 + 0.12 = 1.47$

Thus, the expected number times that the process is recalibrated during the week is 1.47

Thus, the probabilities for specified events and expectation are evaluated as:

• The probability that the process is recalibrated less than two times during the week is 0.53
• The probability that the process is recalibrated more than three times during the week is 0.03
• The probability that the process is recalibrated exactly once during the week is 0.36
• The expected number times that the process is recalibrated during the week is 1.47

Learn more about probability here:

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