Using the Central Limit Theorem Quiz

PDF page 402: It is important for you to understand when to use the central limit theorem. If you are being asked to find the probability of the mean, use the clt for the means. If you are being asked to find the probability of a sum or total, use the clt for sums. This also applies to percentiles for means and sums.

PDF page 402: If you are being asked to find the probability of an individual value, do not use the clt. Use the distribution of its random variable.

PDF page 402: Law of Large Numbers The law of large numbers says that if you take samples of larger and larger sizes from any population, then the mean of the samples tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for is .) This means that the sample mean must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. Central Limit Theorem for the Mean and Sum Examples EXAMPLE 7.8

PDF page 402: Law of Large Numbers The law of large numbers says that if you take samples of larger and larger sizes from any population, then the mean of the samples tends to get closer and closer to μ. From the central limit theorem, we know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller the standard deviation gets. (Remember that the standard deviation for is .) This means that the sample mean must be close to the population mean μ. We can say that μ is the value that the sample means approach as n gets larger. The central limit theorem illustrates the law of large numbers. Central Limit Theorem for the Mean and Sum Examples EXAMPLE 7.8

PDF page 402: Let X = one stress score. Problems (a) and (b) ask you to find a probability or a percentile for a mean. Problems (c) and (d) ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Because the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (see Continuous Random Variables for an explanation of a uniform distribution),

PDF page 402: Let X = one stress score. Problems (a) and (b) ask you to find a probability or a percentile for a mean. Problems (c) and (d) ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75. Because the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (see Continuous Random Variables for an explanation of a uniform distribution),

PDF page 403: The probability that the mean stress score is less than 2 is about zero.

PDF page 403: th The 90 percentile for the mean of 75 scores is about 3.2. This tells us that 90 percent of all the means of 75 stress scores are at most 3.2, and that 10 percent are at least 3.2.

PDF page 404: c. The mean of the sum of 75 stress scores is (75)(3) = 225.

PDF page 405: Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract. The analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes. Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract. Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. X ∼ Exp

PDF page 405: . From previous chapters, we know that μ = 22 and σ = 22.

PDF page 405: The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.

PDF page 406: c. 1. P(x > 20) = 0.4029, but P( > 20) = 0.7919