Let X denote the distance m that an animal moves from its bi

Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for banner- tailed kangaroo rats, X has an exponential distribution with parameter A = 0.01387. (a) What is the probability that the distance is at most loo m? At most 200 m? Between 100 and 200 m? (Round your answers to four decimal places.) (b) What s the probability that distance exceeds the mean distance by more than 2 standard deviations? (Round your answer to four decimal places.) (c) What is the value of the median distance? (Round your answer to two decimal places.) If bolt thread length is normally distributed, what is the probability that the thread length of a randomly selected bolt is (a) Within 1.1 SDs of its mean value? (b) Farther than 1.4 SDs from its mean value? An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with mu = 45 and sigma = 4.5. (a) What is the probability that yield strength is at most 39? Greater than 63? (Round your answers to four decimal places.) at most 39 greater than 63 (b) What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.)

Solution

1) mu = 45 sigma =4.5

P(X<39) = P(Z<-1.33) = 0.5-0.4082

= 0.0918

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P(X>63) = P(Z>4) = 0

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z value for 75% top = 0.675

x = 45+4.5(0.675) = 45+3.0375

= 48.0375

2) bolt problem

P(-1.1 sigma < mu <1.1 sigma) = 0.3643+0.3643 =0.7286

b) P(|Z|>1,4sigma) = 1-2(0.4192) = 1-0.8384

= 0.1616

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Exponenti

P(X<100 m) = 0.9931

P(X<200 m) = 0.9983

P(100<x<200) = 0.52

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b) P(|x-Mu|>2/lemda^2) = P(|X-72.10| > 10396.25)\\

= P(10324.15<x<10468.35)

=49.98

=1-1 =0

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c) Median = ln2/lemda =