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Question 1 A laboratory test is 95 % correct in detecting a certain disease when the disease is actually present. However, the test also yields a "false" result for 1 percent of the healthy people tested. If 0.5 % of the population has the disease, what is the probability a person has the disease given that his test result is positive? (6 marks) (a) It was reported that the U.S airlines average about 3.8 fatalities per month Assume that the probability distribution for x, the number of fatalities per month, can be approximated by a Poisson probability distribution. (i What is the probability that one fatality will occur during a month? (ii) Find E(r) and the standard deviation ofx. (3 marks) (2 marks) A machine that produces stampings for automobile engines is malfunctioning and (c) producing 10 % defectives. The defective and non-defective stampings proceed from the machine in a random manner (i) If the next five stampings are tested, find the probability that three of them are (2 marks) (i) If the next five stampings are tested, what is the probability that at least three (4 marks) (iii) Determine the mean and standard deviation of the random variable denoting (2 marks) defectives. of them are defectives? the number of defectives in this sample of five stampings. (d) Suppose the random variable X has a geometric distribution withp 0.2 (i) Write down the probability mass function for X. Find the mean and variance of X (3 marks) (ii) Determine the probabilities P(X = 4) and P(X > 2). (4 marks)

Answer 1

Solution:- Let B1 be the event that a person has the disease,

let B2 be the event that a person does not have the disease

and let A be the event that the test result is positive.

Then P(B1) = 0.005

P(B2) = 1−P(B1) = 0.995

P(A|B1) = 0.95

P(A|B2) = 0.01

so P(B1|A) = P(A|B1)P(B1) / P(A|B1)P(B1)+P(A|B2)P(B2)

                  = (.95)(.005) / (.95)(.005)+(.01)(.995)

                  = 95 / 294

                  = 0.323

More generally, we could set p = P(B1)

                                              R = P(A|B2) / P(A|B1)

and obtain P(B1|A) = p / p+R(1−p)

Think of R as badness ratio; the higher R the worse the test is.

The badness ration of this test is
                                                     0.01 / 0.95 = 0.0105

a) Call the mean μ and the number of occurrences we're interested in x.

Then, for a Poisson Distribution:

P(x) = (e^-μμ^x) / (x!)

μ = 3.8 fatalities

Poisson Distribution's probability formula is:

P(x) = (e^-3.83.8^x) / (x!)

The probability that one fatality occurs is P(1). So, let x = 1 and solve for P(1):

P(1) = (e^-3.83.8¹) / (1!)

P(1) = (0.0223(3.8)) / 1

P(1) = 0.08474

In a Poisson Distribution, the expected value E(x) is equal to μ.

So: E(x) = 3.8

The variance is also equal to μ. Since standard deviation is the square root of the variance:

Standard deviation = √3.8 = 1.950

c) Let x equal the number of defectives in n = 5 trials

Then x is a binomial random variable with p, the probability that a single stamping will be defective, equal to 0.1, and

q = 1 - p = 1 - 0.1 = 0.9

The probability distribution for x is given by the expression

p(x) = (ⁿ_x)p^x q^n-x = (⁵_x)(0.1)^x (0.9)^5-x

       = (5! / x!(5-x)! ) * (0.1)^x (0.9)^5-x                    (x = 0,1,2,3,4,5)

To find the probability of observing x = 3 defectives in a sample of n = 5

substitute x = 3 into the formula for p(x) to obtain

p(3) = (5! / 3!(5-3)!) * (0.1)³ (0.9)^5-3

           p(3) = 0.0081

Similarly p(4) = 0.00045 and p(5)= 0.00001

p(atleast 3) = p(3) + p(4) + p(5) = 0.00856

Now, n = 5, p = 0.1

mean = np = 5(0.1)

                  = 0.5

standard deviation = √npq = √5(0.1)(0.9)

                               = 0.67

d) p = 0.2

Probability mass function of X = f(x) = (1-p)^x-1p     for x = 1,2,3.......

                                                        = (1-0.2)^x-10.2            for x = 1,2,3.......

                                                        = (0.8)^x-10.2            for x = 1,2,3.......

mean = 1 / p = 1 / 0.2 = 5

variance = (1 - p) / p² = (1 - 0.2) / (0.2)² = 20

p(X=4) = (0.8)^4-10.2 = 2.56

p(X>2) = 1 - p(X<=2) = 1 - (p(X=1) + p(X=2))

= 1 - 0.2 - 0.8*0.2

p(X>2) = 0.64