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Chapter 5: Q 5.4 (page 212)

The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by

f(x)=10x2x>100x鈮�10

(a)FindP{X>20}

localid="1646589462481" (b)What is the cumulative distribution function of localid="1646589521172" X?

localid="1646589534997" (c)) What is the probability that of6such types of devices, at least localid="1646589580632" 3will function for at least localid="1646589593287" 15hours? What assumptions are you making?

Short Answer

Expert verified

(a)P{X>20}=0.5

(b)The cumulative distribution function (COD) ofXis0x鈭�(-鈭�,10]1-10xx鈭�(10,鈭�)

(c)P(A3)=鈭�i=166i23i136-i

Step by step solution

01

Part a Step 1  Given information.

The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by

f(x)=10x2x>100x鈮�10

02

Part a Step 2 explanation

We find as follows

P{X>20}=鈭�20鈭�f(x)dx=10鈭�20鈭�x-2dx=10limt鈫�鈭�鈭�20tx-2dx=10limt鈫�鈭�-1x20t=10limt鈫�鈭�-1t+120=1020=0.5

03

Part b Step 1 Explanation

We have that

F(x)=P(X鈮�x)=鈭�-鈭�xf(t)dt=10鈭�10xt-2dt=10-1t10x=10-1x+110=1-10x

We can express the CDF ASF(x)0x鈭�(-鈭�,10]1-10xx鈭�(10,鈭�)

04

part c Step 1 Explanation.

Let Abe the event that a single device works for at least 15hours. We thus have thatP(A)=1-F(15)=1-1-1015=1015=23

If we letA3denote the event that at least three will work for at least15hours, the we have

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Most popular questions from this chapter

Evidence concerning the guilt or innocence of a defendant in a criminal investigation can be summarized by the value of an exponential random variable X whose mean 渭 depends on whether the defendant is guilty. If innocent, 渭 = 1; if guilty, 渭 = 2. The deciding judge will rule the defendant guilty if X > c for some suitably chosen value of c.

(a) If the judge wants to be 95 percent certain that an innocent man will not be convicted, what should be the value of c?

(b) Using the value of c found in part (a), what is the probability that a guilty defendant will be convicted?

With 桅(x)being the probability that a normal random variable with mean 0and variance 1is less than x, which of the following are true:

(a)桅(-x)=桅(x)

(b)桅(x)+桅(-x)=1

(c)桅(-x)=1/桅(x)

Let X be a continuous random variable having cumulative distribution function F. Define the random variable Y by Y = F(X). Show that Y is uniformly distributed over (0, 1).

The median of a continuous random variable having distribution function F is that value m such that F(m) = 12. That is, a random variable is just as likely to be larger than its median as it is to be smaller. Find the median of X if X is

(a) uniformly distributed over (a, b);

(b) normal with parameters 渭,蟽2;

(c) exponential with rate 位.

Let X and Y be independent random variables that are both equally likely to be either 1, 2, . . . ,(10)N, where N is very large. Let D denote the greatest common divisor of X and Y, and let Q k = P{D = k}.

(a) Give a heuristic argument that Q k = 1 k2 Q1. Hint: Note that in order for D to equal k, k must divide both X and Y and also X/k, and Y/k must be relatively prime. (That is, X/k, and Y/k must have a greatest common divisor equal to 1.) (b) Use part (a) to show that Q1 = P{X and Y are relatively prime} = 1 q k=1 1/k2 It is a well-known identity that !q 1 1/k2 = 蟺2/6, so Q1 = 6/蟺2. (In number theory, this is known as the Legendre theorem.) (c) Now argue that Q1 = "q i=1  P2 i 鈭� 1 P2 i  where Pi is the smallest prime greater than 1. Hint: X and Y will be relatively prime if they have no common prime factors. Hence, from part (b), we see that Problem 11 of Chapter 4 is that X and Y are relatively prime if XY has no multiple prime factors.)
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