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?
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) The sampling distribution of the sample mean lifetime of batteries follows a normal distribution. The mean of the sampling distribution is equal to the population mean, which is 360 minutes in this case. The standard deviation of the sampling distribution, also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size. Since we have a sample size of 36 and a population standard deviation of 30 minutes, the standard error is 30 / √36 = 5 minutes.
(b) To determine the range within which we would expect the middle 80% of the sample means to lie, we can use the concept of confidence intervals. For a normal distribution, the middle 80% falls within approximately 1.28 standard deviations above and below the mean. Therefore, the range of sample means would be approximately 360 ± (1.28 * 5) minutes, or 354.4 to 365.6 minutes.
(c) To calculate the probability that a sample would produce a mean of less than 350 minutes, we need to standardize the value of 350 using the formula z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the population mean, and σ is the population standard deviation. Plugging in the values, we get z = (350 - 360) / 30 = -1/3. Using a standard normal distribution table or calculator, we can find the probability corresponding to a z-score of -1/3. Let's assume it is 0.3693.
(d) If the sample produced a mean life of batteries of 350 minutes, the probability of observing such a mean, assuming the process is working correctly, would be 0.3693. Since this probability is quite high, it suggests that the observed mean of 350 minutes is not statistically significantly different from the population mean of 360 minutes. Therefore, we would conclude that the process is working correctly.
(e) To calculate the probability that two samples in a row would have a mean lifetime of less than 350 minutes, we can assume that the mean of each sample is independent from the other. Therefore, we can multiply the probabilities. The probability of one sample having a mean lifetime of less than 350 minutes, as found in part (c), is 0.3693. Therefore, the probability of two samples in a row having means of less than 350 minutes would be 0.3693 * 0.3693, which is approximately 0.1362 or 13.62%.
(a) The sampling distribution of the sample mean lifetime of batteries is normally distributed. It follows a normal distribution because the population distribution of battery lifetimes is assumed to be normally distributed, and the sample mean is an unbiased estimator of the population mean.
(b) To determine the range within which we would expect the middle 80% of the sample means to lie, we can use the concept of the standard deviation. Approximately 80% of the sample means will fall within 1.96 standard deviations of the mean, assuming a normal distribution. So the range would be:
Mean ± (1.96 * standard deviation)
= 360 ± (1.96 * 30)
= 360 ± 58.8
= (301.2, 418.8)
Therefore, we would expect the middle 80% of the sample means to lie between 301.2 and 418.8 minutes.
(c) To calculate the probability that a sample would produce a mean of less than 350 minutes, we need to standardize the value and convert it to a z-score using the formula:
z = (x - mean) / standard deviation
Substituting the values, we have:
z = (350 - 360) / 30
= -10 / 30
= -1/3
Now, we can use a z-table or software to find the probability associated with a z-score of -1/3. Referencing the z-table, we can find that the probability is approximately 0.3693.
Therefore, the probability that a sample would produce a mean of less than 350 minutes is approximately 0.3693.
(d) If the sample produced a mean lifetime of batteries of 350 minutes, which is less than the expected mean of 360 minutes, we can conclude that the process is not working correctly. This is because the sample mean is significantly lower than the population mean, and it is less likely to occur by chance.
(e) To calculate the probability that two samples in a row would have a mean lifetime of less than 350 minutes, we can use the concept of the binomial distribution. Assuming each sample is independent, we multiply the probabilities of each sample producing a mean of less than 350 minutes.
The probability for each sample is 0.3693, as calculated in (c). So the probability for two samples in a row is:
0.3693 * 0.3693 = 0.1363
Therefore, the probability that two samples in a row would have a mean lifetime of less than 350 minutes is approximately 0.1363.