A - The sample standard deviation is the population standard deviation over the square root of the sample size, 30 / sqrt(36). The mean of an approximately normal sample distribution is the same as the population mean, so the sample distribution is N(360, 5).
B - If you want the middle 80% of the data, you will need the z-scores of the bottom 10% and 90%. This can be done with either a calculator or a z-score table. The values of interest are +/- 1.28. You can then use the z-score formula to calculate the upper and lower bounds, with the sample distribution from (A).
C - From (A), we know N(360, 5). P(X < 350) = .023. (Note that this "X" is actually an x-bar since these are sample means.)
D - Because the probability of this event if low (less than 5%), it is unlikely that these batteries match the advertised standard.
E - This is a simple application of the product rule. .0227 * .0227 = .000518.
A firm produces batteries that have a lifetime which is normally distributed with a mean of 360 minutes and a standard deviation of 30 minutes. The firm needs to keep an eye on the production process to ensure that everything is working properly and that batteries are not being produced that do not meet the advertised standard. This is done by calculating the mean of the sample. To do this they regularly select a sample of 36 batteries in order to test the process.
(a) Describe the sampling distribution of the sample mean lifetime of batteries
(b) State a range within which you would expect the middle 80% of the sample means to lie.
(c) If the process were working correctly, what is the probability that a sample would produce a mean of less than 350 minutes?
(d) Based on your answer to part (c), what would you conclude about the process if the sample produced a mean life of batteries of 350 minutes?
(e) What is the probability that two samples in a row would have a mean life time of less than 350 minutes?
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