Question

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Let X denote the datance (m) that animal meves from its birth ste te-nrt tenteral vacancy a encerten. Suppose fo. b.mertalad kangareo n s 300 and 200 m (Reund yeur ansuvers to feur decimal glaces.) a..pe ental doed ii.1. pa-e. 1437 (a) What is the prebability that the datance is at most 100 m Ac most 200 m? Be between 100 and 200 m (b) What is the probability that distance exceeds the mean distance by more than 2 standard deviatiens? (Round your ander to four decimal places) (c) what is the value of the median distance, Mint: Find·such that PCX")-0.50 (Round yw mirer to tie dennal places.) Only 5% of animals wil move farther than what distance? Hnt, Find a such that p Xsa). 035 (Aeund your annrer ee bro decimal places.)

Answer 1

a)P(at most 100) =P(X<100)=1-e^-100*0.01437 =0.7624

P(at most 200)=P(X<200)=1-e^-200*0.01437 =0.9435

P(100<X<200)=0.9435-0.7624=0.1812 ( please try 0.1811 if this comes wrong)

b)

P(X>1/$\lambda$+2/$\lambda$)=P(X>3/$\lambda$)=e^-3 =0.0498

c)

median =ln(2)/$\lambda$ =48.24

d)

for 95 th percentile =-ln(1-p)/$\lambda$ =-ln(1-0.95)/0.01437 =208.47