Using the Central Limit Theorem Quiz

PDF page 430: 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. NOTE 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 431: 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), a + b 1 + 5

PDF page 431: The probability that the mean stress score is less than 2 is about zero.

PDF page 432: ⎛ 1.15⎞ normalcdf 1,2,3, = 0 ⎝ 75 ⎠ REMINDER The smallest stress score is one.

PDF page 432: ̄ Find k, where P( x < k) = 0.90. k = 3.2

PDF page 432: 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. ⎛ 1.15⎞ = 3.2 invNorm 0.90,3, ⎝ 75 ⎠ For problems (c) and (d), let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3), ( 75) (1.15)].

PDF page 432: c. The mean of the sum of 75 stress scores is (75)(3) = 225.

PDF page 433: The standard deviation of the sum of 75 stress scores is ( 75) (1.15) = 9.96. P(Σx < 200) = 0

PDF page 434: 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. ⎛ 1 ⎞ X ∼ Exp

PDF page 434: . From previous chapters, we know that μ = 22 and σ = 22.

PDF page 434: ⎛ ̄ 22 ⎞ P( x > 20) = 0.79199 using normalcdf 20,1E99,22, ⎝ 80⎠ 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 435: b. Find P(x > 20). Remember to use the exponential distribution for an individual. X ~Exp

PDF page 435: c. 1. P(x > 20) = 0.4029, but P( x > 20) = 0.7919

PDF page 435: ⎛ 22 ⎞ k = 26.0 using invNorm 0.95,22, = 26.0 ⎝ 80⎠

PDF page 436: The 95 percentile for the sample mean excess time used is about 26.0 minutes for a random sample of 80 customers who exceed their contractual allowed time. 95 percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.

PDF page 436: Solution 7.10 0.5 σ = = 0.05. Therefore,

PDF page 436: a. We have μx = μ = 2 and σx =