EasyBib: Free Bilolography c W 63 Homework O e webassignnetwebstudent/Assignment

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EasyBib: Free Bilolography c W 63 Homework O e webassignnetwebstudent/Assignment Responsesastrdepa15s79807 points Previous Answers BBunderstall 063.030 My Notes o Ask Your Teacher Acorotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of months electronic components 29 before certain deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 5 months, and the distribution of lifetimes is normal (a) If Acootime guarantees a ful refund on any defective watch for years after purchase, what percentage of total production will the company expect to replace? (Round your answer to two decimal 2 places.) Accrotime does not want to make refunds on more than 11% of the watches it makes, how long should the nearest month)? the guarantee period be (to If Need Help? -1 points BEUnderstat 063037 Most tion shows open in the morning and close in the My Notes o Ask Your Teache doors and late evening. A study of Saturday arrival times showed that the average arrival time was 3 hours and 48 minutes after the opened, the standard deviation was estimated at about 45 minutes. that the arrival times follow a normal distribution. a) At what time after the doors open who are coming to the Saturday show have a nd answer to the nearest number of minutes.) minutes will s6% of the people your after doors open (b) At what time after the doors open will only 18% of the people who are coming to the Saturday show have arrived (Round your answer to the nearest numRir nf minitm

Explanation / Answer

let X: life time of watches made by the company

now, the company would give 100% refund if the watch detoriate within 2 years i.e,24 months

it is given that X~normal distribution rather say a bifoded normal distribution( at 0), as X cannot takes values <0

now we are to calculate the probability that a watch will not get detoriate within 24 months

it is given by P(X<24) with mean of x as 29 and sd 5 it turns out to be 0.16(i.e 16%)

and we can clearly observed that if we put the limit as 23 monts instead of 24 then the same probability turns out to be P(X<23)=0.12(apx)(i.e,12%) so the new limit would have been 23-1=22 months.

now, it is observed that the aforementioned ans. leads to a cross in the answer so we will consider an normal( mean=29,sd=5) distribution instead of bifolded one altough it would be an hypothetical situation,

still it comes out to be the same ..... exactly equal..

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