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om/courses/50514/assignments/980646?module_item_id=1608249 Hews M Gmail Week 9 Assignment Score: 12/13 6/7 answered . Question 3 1 Score on last try: 0,67 of 2 pts. See Details for more. > Next question Try a similar question You can retry this question below The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of sixteen months. A sample of 225 of these smartphones is taken. What is the parameter 1? a) What is the standard deviation? (Round to 4 facimal places) b) What is the distribution for the length of time one battery lasts? Exponential t o c) What is the distribution for the mean length of time 225 batteries last? Normal (Round to 4 d) Find the probability that the sample mean ts between 0 and 9. decimal places) . (Round e) Find the 74th percentile for the mean length of time 225 batteries last. to 2 decimal places) Question Help: Written Example Message instructor Submit Question

Answer 1

If X follows exponential distribution with parameter 2 then the pdf of X is f(x)=re-* :x>0, 2 > 0 then E(x) = - ---(1) V(x) = - --(2) so SD(X) = _ 3 Given mean = 16, from the above results (1) and (3) we get a) i = = = 0.0625 and standard deviation is 16 b) As it is given, the distribution for the life length of one battery is exponential with parameter 1 = = = 0.0625 16

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Given that sample size n=225, so by central limit theorem, the sample mean 8 follows normal distribution with parameters mean 4 =- = 16 and variance 02=772 1137 d) P(0<8<9)=P 10719 <2<39719 = P(-14.9998<Z<-6.5624) 0 e) Let C be the 74th percentile, then we can determine C such that P(X<C)=0.74 C-16 1.1378 = 0.74 from standard normal tables we get C-16 V1.1378=0.6433 ..C=(0.6433* V1.1378 ) +16=16.69