THE FOLLOWING 24 OBSERVATION GIVE THE LENGTH OF TIME IN HOURS FOR WHICH A SPECIFIC FULLY CHARGED LAPTOP COMPUTER WILL OPERATE ON BATTERY BEFORE REQUIRING RECHARGING.. 1.2,1.4,1.5,1.6,1.7,1.7,1.8,1.8,1.9,1.9,2.0,2.0,2.1,2.1,2.1,2.2,2.3,2.4,2.4,2.5,3.1,3.6,3.7,4.5. THE OWNER OF THE THIS COMPUTER IS ABOUT TO WATCH A FILM ON HIS FULLY CHARGED COMPUTER . CALCULATE FROM THESE DATA THE LONGEST SHOWING TIME FOR A FILM THAT HE CAN WATCH . SO THAT THE PROBABILITY THAT THE BATTERY LIFE TIME WILL BE SUFFICIENT FOR WATCHING THE ENTIRE FILM IS 0.75?? ANS" LET t be the longest time p(T>t)=0.75= e^-lamda t now lambda is the expected value of these 24 observation expected value of observation is 53.5/24=2.229 e^-2.229 *t=.75 -2.226 * t = ln(.75) t=.1290 but ans is 1.75hr please correct me
You can't assume that these values came from an exponential distribution if it's not given in the question. You've to simply find the 1st quartile of the data.
You've to decipher it from this part of the question. If he watches a film whose length is equal to 1st quartile of this data, then the probability that the battery life of his laptop will be sufficient to watch the entire film is 0.75 Remember 1st quartile is the value which separates the bottom 25% of the sorted data from the top 75%
