Montgomery, Applied Statistics and Probability for Engineers, 7e Help | System A

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Montgomery, Applied Statistics and Probability for Engineers, 7e Help | System Announcements PRINTER VERSION BACK NEXT Reserve Problems Chapter 3 Section 2 Problem 1 Let us suppose that some article studied the probability of death due to burn injuries. The identified risk factors in this study are age greater than 60 years, burn injury in more than 40% of body-surface area, and presence of inhalation injury. It is estimated that the probability of death is 0.003, 0.03, 0.33, or 0.74, if the injured person has, zero, one, two, or three risk factors, respectively. Suppose that three people are injured in a fire and treated independently. Among these three people, two people have one risk factor and one person has three risk factors. Let the random variable X denote number of deaths in this fire. Determine the cumulative distribution function for the random variable. Round your answers to five decimal places (e.g. 98.76543). F(x)- F(x)- F(x)- F(x)- F(x)- Click if you would like to Show Work for this question: with x

Explanation / Answer

Answer to the question)

There are three persons

One had 3 risk factors , hence P(3 death) = 0.74

Two of them have one risk factor , P(1 death) = 0.03

.

For x < 0 , the F(x) = 0

.

For F(x) where x is between 0 and 1 :

Now for x = 0 implies no death

P(no death) = (1-0.74) * (1-0.03)^2

P(no death) = 0.24463

F(x) = 0.24463

.

For 1< = x < 2

P(1 death) = 0.74*(1-0.03)^2 + 2*(1-0.74)*(1-0.03)*0.03

P(1 death) = 0.696266 + 0.015132 = 0.711398

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For x between 2 and 3

P(2 death) = 0.74*0.03*0.97*2 + 0.26*0.03*0.03

P(2 death) = 0.043068 + 0.000234

P(2 deaths) = 0.043302

.

P(3 deaths) = 0.74 *0.03 *0.03 = 0.000666

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