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4.135 Suppose xhas an exponential distribution with =1. Find

the following probabilities:

a.P(x>1)b.P(x3)cP(x>1.5)d.P(x5)

Short Answer

Expert verified
  1. .The probability is 0.3679
  2. The probability is 0.9502
  3. The probability is 0.2231
  4. The probability is 0.9933

Step by step solution

01

Given Information

The random variable x has an exponential distribution with=1

02

The probability density function (PDF) of x

Here x is a random variable with parameters=1

The pdf of x is given by,

f(x,)=1exp-x,x>0

Here, =1

f(x)=exp(-x);x>0

03

Finding cdf of x

F(x)=P(Xx)=0xf(t)dt=0xexp(-t)dt=exp(-t)-10x=-exp(-x)+1=1-exp(-x)F(x)=1-exp(-x)

04

Finding the probability when P(x > 1)

a.P(x>1)=1-P(x1)=1-F(1)=1-1-exp(-1)=exp(-1)=0.37879=0.3679

Thus, the required probability is 0.3679.

05

Finding the probability when P(x≤3)

b.P(x3)=F(3)=1-exp(-3)=1-0.049787=0.950213=0.9502

Thus, the required probability is 0.9502.

06

Finding the probability when P(x>1.5)

c.P(x>1.5)=1-P(x1.5)=1-F(1.5)=1-1-exp(-1.5)=exp(-1.5)=0.22313=0.2231

The required probability is 0.2231.

07

Finding the probability when P(x≤5)

d.P(x5)=F(5)=1-exp(-5)=1-0.006738=0.993262=0.9933

The required probability is0.9933.

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Most popular questions from this chapter

The gender diversity of a large corporation鈥檚 board of directors was studied in Accounting & Finance (December 2015). In particular, the researchers wanted to know whether firms with a nominating committee would appoint more female directors than firms without a nominating committee. One of the key variables measured at each corporation was the percentage of female board directors. In a sample of 491firms with a nominating committee, the mean percentage was 7.5%; in an independent sample of 501firms without a nominating committee, the mean percentage was role="math" localid="1652702402701" 4.3% .

a. To answer the research question, the researchers compared the mean percentage of female board directors at firms with a nominating committee with the corresponding percentage at firms without a nominating committee using an independent samples test. Set up the null and alternative hypotheses for this test.

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Question: Refer to the Bulletin of Marine Science (April 2010) study of lobster trap placement, Exercise 6.29 (p. 348). Recall that the variable of interest was the average distance separating traps鈥攃alled trap-spacing鈥攄eployed by teams of fishermen. The trap-spacing measurements (in meters) for a sample of seven teams from the Bahia Tortugas (BT) fishing cooperative are repeated in the table. In addition, trap-spacing measurements for eight teams from the Punta Abreojos (PA) fishing cooperative are listed. For this problem, we are interested in comparing the mean trap-spacing measurements of the two fishing cooperatives.

BT Cooperative

93

99

105

94

82

70

86

PA Cooperative

118

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106

72

90

66

98


Source: Based on G. G. Chester, 鈥淓xplaining Catch Variation Among Baja California Lobster Fishers Through Spatial Analysis of Trap-Placement Decisions,鈥 Bulletin of Marine Science, Vol. 86, No. 2, April 2010 (Table 1).

a. Identify the target parameter for this study.b. Compute a point estimate of the target parameter.c. What is the problem with using the normal (z) statistic to find a confidence interval for the target parameter?d. Find aconfidence interval for the target parameter.e. Use the interval, part d, to make a statement about the difference in mean trap-spacing measurements of the two fishing cooperatives.f. What conditions must be satisfied for the inference, part e, to be valid?

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