In: Statistics and Probability
BIG Corporation produces just about everything but is currently interested in the lifetimes of its batteries, hoping to obtain its share of a market boosted by the popularity of portable CD and MP3 players. To investigate its new line of Ultra batteries, BIG randomly selects 1000 Ultra batteries and finds that they have a mean lifetime of 909 hours, with a standard deviation of 91 hours. Suppose that this mean and standard deviation apply to the population of all Ultra batteries. Complete the following statements about the distribution of lifetimes of all Ultra batteries.
?hours and ?hours. Round your answer to the nearest integer.)
approximately 68% of the lifetimes lie between ? hours and ? hours.
We have mean = 909 and standard deviation (Sd) = 91
(a) z score for 727 = (x-mean)/Sd = (727-909)/91 = -2
and z score for 1091 = (x-mean)/Sd = (1091-909)/91 = 2
So, number of standard deviations k = 2
By chebyshevs' theorem, percent =
answer is 75%
(b) We know that 36% of data lies within 1.25 standard deviations from the mean (using chebyshev's theorem)
Values = (mean - 1.25*sd) to (mean +1.25*sd)
=(909-1.25*91) to (909+1.25*91)
= 795.25 to 1022.75
= 795 hours to 1023 hours (answer)
(c) We know that by empirical rule, 68% of data lies within 1 standard deviation from the mean
Values = (mean - 1*sd) to (mean +1*sd)
=(909-1*91) to (909+1*91)
= 818 hours to 1000 hours (answer)
(d) In part (a), we found that 727 hours and 1091 hours are -2 and +2 standard deviations from the mean. Using empirical rule, we know that 95% of data lies between -2 and +2 standard deviations. So, answer is 95%