Math/Stat2330 Elementary Statistics and Probability Lecture 14: Sample Variability Gejun Zhu Department of Mathematics University of Texas - Pan American Edinburg, Texas 78539 [email protected] June 24th, 2013 Last Class Review What did we learn in last class? Applications of Normal Distribution Notations z(α) Normal Approximation of Binomial Distribution G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 2 / 19 Applications of Normal Distribution Example 1: Given that x is normally distributed random variable with a mean of 60 and a standard deviation of 10, find the following probabilities: a. P(x > 60) d. P(65 < x < 82) b. P(60 < x < 72) e. P(38 < x < 78) c. P(57 < x < 78) f. P(x < 38) G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 3 / 19 Applications of Normal Distribution Example 1: Given that x is normally distributed random variable with a mean of 60 and a standard deviation of 10, find the following probabilities: a. P(x > 60) d. P(65 < x < 82) b. P(60 < x < 72) e. P(38 < x < 78) c. P(57 < x < 78) f. P(x < 38) Solution: a. 0.5000 d. 0.2946 b. 0.3849 e. 0.9502 c. 0.6072 f. 0.0139 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 3 / 19 Notations Example 2: z(0.05) (read “z of 0.05”) is the algebraic name for the z such that the area to the right and under the standard normal curve is exactly 0.05, as shown below: G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 4 / 19 Notations Example 2: z(0.05) (read “z of 0.05”) is the algebraic name for the z such that the area to the right and under the standard normal curve is exactly 0.05, as shown below: Using z-table, z(0.05) = 1.65 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 4 / 19 Normal Approximation of Binomial Distribution Rule The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 − p) both equal or exceed 5. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 5 / 19 Normal Approximation of Binomial Distribution Rule The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 − p) both equal or exceed 5. Example 3: Find the normal approximation for the binomial probability P(x ≤ 8), where n = 14 and p = 0.4. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 5 / 19 Normal Approximation of Binomial Distribution Rule The normal distribution provides a reasonable approximation to a binomial probability distribution whenever the values of np and n(1 − p) both equal or exceed 5. Example 3: Find the normal approximation for the binomial probability P(x ≤ 8), where n = 14 and p = 0.4. Using z-table, np = 14 × 0.4 = 5.6, n(1 − p) = 14 × 0.6 = 8.4 µ= √ √np = 14 × 0.4 = 5.6 √ σ = npq = 14 × 0.4 × 0.6 = 3.36 = 1.83 8.5 − 5.6 P(x ≤ 8) = P(z ≤ ) = 1.58 1.83 P(x ≤ 8) = 0.9430 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 5 / 19 Application of Normal Distribution Example 4: Find the normal approximation for the binomial probability P(x ≥ 8), where n = 14 and p = 0.4. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 6 / 19 Application of Normal Distribution Example 4: Find the normal approximation for the binomial probability P(x ≥ 8), where n = 14 and p = 0.4. np = 14 × 0.4 = 5.6, n(1 − p) = 14 × 0.6 = 8.4 µ = np = 14 × 0.4 = 5.6 √ √ √ σ = npq = 14 × 0.4 × 0.6 = 3.36 = 1.83 7.5 − 5.6 7.5 − 5.6 P(x ≥ 8) = P(z ≥ ) = 1 − P(z ≤ ) = 1 − P(z < 1.04) = 0.1492 1.83 1.83 Using z-table, P(x ≥ 8) = 0.1492 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 6 / 19 Sample Variability What will we cover in this chapter? We are going to use the sample mean x¯ to estimate the population µ. Actually, we don’t expect that the value of sample mean x¯ is exactly equal to the value of the population mean µ, but we will satisfied with our sample results if the sample mean is “close” to the value of the population mean. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 7 / 19 Sample Variability What will we cover in this chapter? We are going to use the sample mean x¯ to estimate the population µ. Actually, we don’t expect that the value of sample mean x¯ is exactly equal to the value of the population mean µ, but we will satisfied with our sample results if the sample mean is “close” to the value of the population mean. What is “close”? How to determine the closeness? G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 7 / 19 Sample Variability What will we cover in this chapter? We are going to use the sample mean x¯ to estimate the population µ. Actually, we don’t expect that the value of sample mean x¯ is exactly equal to the value of the population mean µ, but we will satisfied with our sample results if the sample mean is “close” to the value of the population mean. What is “close”? How to determine the closeness? Sampling distribution of a sample statistic The distribution of values for a sample statistic obtained from repeated samples, all of the same size and all drawn from the same population. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 7 / 19 Sampling Distribution of the Sample Mean Example: Heights of Starting Players Suppose that the population of interest consists of five starting players on a men’s basketball team, who we will call A, B, C, D, and E. Further suppose that the variable of interest is height, in inches. Here is the data of the players. Player Height A 76 B 78 C 79 D 81 E 86 a. Obtain the sampling distribution of the sample mean from samples of size 2. b. Make some observations about sampling error when the mean height of a random sample of two players is used to estimate the population mean. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 8 / 19 Sampling Distribution of the Sample Mean Solution: a. Obtain the sampling distribution of the sample mean from samples of size 2. Sample A, B A, C A, D A, E B, C B, D B, E C, D C, E D, E Height 76, 78 76, 79 76, 81 76, 86 78, 79 78, 81 78, 86 79, 81 79, 86 81, 86 x¯ 77.0 77.5 78.5 81.0 78.5 79.5 82.0 80.0 82.5 83.5 The population is so small that we can list all the possible samples of size 2. The first column of the above table gives the 10 possible samples, the second column the corresponding heights and the third column the sample means. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 9 / 19 Sampling Distribution of the Sample Mean b. Make some observations about sampling error when the mean height of a random sample of two players is used to estimate the population mean. P 76 + 78 + 79 + 81 + 86 xi Population Mean: µ = = = 80 inches. n 5 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 10 / 19 Sampling Distribution of the Sample Mean b. Make some observations about sampling error when the mean height of a random sample of two players is used to estimate the population mean. P 76 + 78 + 79 + 81 + 86 xi Population Mean: µ = = = 80 inches. n 5 From the table before, we can see that mean height of the two players selected isn’t likely to the population mean of 80 inches. In fact, only 1 of 10 samples has 1 , or 10%, that x¯ will a mean of 80 inches. The chances are, therefore, only 10 equal µ. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 10 / 19 Sampling Distribution of the Sample Mean b. Make some observations about sampling error when the mean height of a random sample of two players is used to estimate the population mean. P 76 + 78 + 79 + 81 + 86 xi Population Mean: µ = = = 80 inches. n 5 From the table before, we can see that mean height of the two players selected isn’t likely to the population mean of 80 inches. In fact, only 1 of 10 samples has 1 , or 10%, that x¯ will a mean of 80 inches. The chances are, therefore, only 10 equal µ. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 10 / 19 Sampling Distribution of the Sample Mean c. Find the probability that, for a random sample of size 4, the sampling error made in estimating the population mean by the same mean will be 1 inch or less; that is determine the probability that x¯ will be within 1 inch of µ. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 11 / 19 Sampling Distribution of the Sample Mean c. Find the probability that, for a random sample of size 4, the sampling error made in estimating the population mean by the same mean will be 1 inch or less; that is determine the probability that x¯ will be within 1 inch of µ. Solution: Sample Height x¯ A, B, C, D 76, 78, 79, 81 78.50 A, B, C, E 76, 78, 79, 86 79.75 A, B, D, E 76, 78, 81, 86 80.25 A, C, D, E 76, 79, 81, 86 80.50 B, C, D, E 78, 79, 81, 86 81.00 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 11 / 19 Sampling Distribution of the Sample Mean c. Find the probability that, for a random sample of size 4, the sampling error made in estimating the population mean by the same mean will be 1 inch or less; that is determine the probability that x¯ will be within 1 inch of µ. Solution: Sample Height x¯ A, B, C, D 76, 78, 79, 81 78.50 A, B, C, E 76, 78, 79, 86 79.75 A, B, D, E 76, 78, 81, 86 80.25 A, C, D, E 76, 79, 81, 86 80.50 B, C, D, E 78, 79, 81, 86 81.00 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 11 / 19 Sampling Distribution of the Sample Mean c. Find the probability that, for a random sample of size 4, the sampling error made in estimating the population mean by the same mean will be 1 inch or less; that is determine the probability that x¯ will be within 1 inch of µ. Solution: Sample Height x¯ A, B, C, D 76, 78, 79, 81 78.50 A, B, C, E 76, 78, 79, 86 79.75 A, B, D, E 76, 78, 81, 86 80.25 A, C, D, E 76, 79, 81, 86 80.50 B, C, D, E 78, 79, 81, 86 81.00 4 The probability is , or 5 0.8, that the sampling error made in estimating µ by x¯ will be 1 inch or less. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 11 / 19 The Sampling Distribution of Sample Means Sampling Distribution of sample means (SDSM) If all possible random samples, each of size n, are taken from any population with mean µ and standard deviation σ, then the sampling distribution of sample means will have the following: 1. A mean µx¯ equal to µ σ 2. A standard deviation σx¯ equal to √ n Furthermore, if the sampled population has a normal distribution, then the sampling distribution of x¯ will also be normal for samples of all sizes. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 12 / 19 The Sampling Distribution of Sample Means Sampling Distribution of sample means (SDSM) If all possible random samples, each of size n, are taken from any population with mean µ and standard deviation σ, then the sampling distribution of sample means will have the following: 1. A mean µx¯ equal to µ σ 2. A standard deviation σx¯ equal to √ n Furthermore, if the sampled population has a normal distribution, then the sampling distribution of x¯ will also be normal for samples of all sizes. Central Limit Theorem (CLT) The sampling distribution of sample means will more closely resemble the normal distribution as the sample size increases. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 12 / 19 The Sampling Distribution of Sample Means If the sampled distribution is normal, then the sampling distribution of sample means is normal, as stated previously, and the central limit theorem is not needed. But if the sampled population is not normal, the CLT tells us that the sampling distribution will still be approximately normally distributed under the right conditions. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 13 / 19 The Sampling Distribution of Sample Means If the sampled distribution is normal, then the sampling distribution of sample means is normal, as stated previously, and the central limit theorem is not needed. But if the sampled population is not normal, the CLT tells us that the sampling distribution will still be approximately normally distributed under the right conditions. By combining the preceding information, we can describe the sampling distribution of x¯ completely: (1) the location of the center (mean), (2) a measure of spread indicating how widely the distribution is dispersed (standard error of the mean), and (3) an indication of how it is distributed. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 13 / 19 The Sampling Distribution of Sample Means Example: Let’s consider all possible samples of size 2 that could be drawn from a population that contains the three numbers 2, 4, 6. First let’s take a look at the population itself. Construct a histogram to picture its distribution, calculate the mean µ and standard deviation, σ. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 14 / 19 The Sampling Distribution of Sample Means Example: Let’s consider all possible samples of size 2 that could be drawn from a population that contains the three numbers 2, 4, 6. First let’s take a look at the population itself. Construct a histogram to picture its distribution, calculate the mean µ and standard deviation, σ. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 14 / 19 The Sampling Distribution of Sample Means Example: Let’s consider all possible samples of size 2 that could be drawn from a population that contains the three numbers 2, 4, 6. First let’s take a look at the population itself. Construct a histogram to picture its distribution, calculate the mean µ and standard deviation, σ. x 2 4 6 P P(x) 1/3 1/3 1/3 1 xP(x) 2/3 4/3 6/3 12/3 = 4 x 2 P(x) 4/3 16/3 36/3 56/3 = 18.66 µ = 4.0 σ= G. Zhu (UTPA) p √ 18.66 − (4.0)2 = 2.66 = 1.63 Math/Stat2330 Lecture-14 June 2013 14 / 19 The Sampling Distribution of Sample Means Here lists all the possible samples and the means of size 2 that can be drawn from this population. (One number is drawn, observed, and then returned to the population before the second number is drawn.) Sample 2, 2 2, 4 2, 6 4, 2 4, 4 4, 6 6, 2 6, 4 6, 6 G. Zhu (UTPA) x¯ 2 3 4 3 4 5 4 5 6 Math/Stat2330 Lecture-14 June 2013 15 / 19 The Sampling Distribution of Sample Means Here lists all the possible samples and the means of size 2 that can be drawn from this population. (One number is drawn, observed, and then returned to the population before the second number is drawn.) Sample 2, 2 2, 4 2, 6 4, 2 4, 4 4, 6 6, 2 6, 4 6, 6 G. Zhu (UTPA) x¯ 2 3 4 3 4 5 4 5 6 Math/Stat2330 Lecture-14 June 2013 15 / 19 The Sampling Distribution of Sample Means x 2 3 4 5 6 P P(x) 1/9 2/9 3/9 2/9 1/9 1.0 G. Zhu (UTPA) xP(x) 2/9 6/9 12/9 10/9 6/9 4.0 x 2 P(x) 4/9 18/9 48/9 50/9 36/9 17.33 Math/Stat2330 Lecture-14 June 2013 16 / 19 The Sampling Distribution of Sample Means x 2 3 4 5 6 P P(x) 1/9 2/9 3/9 2/9 1/9 1.0 G. Zhu (UTPA) xP(x) 2/9 6/9 12/9 10/9 6/9 4.0 x 2 P(x) 4/9 18/9 48/9 50/9 36/9 17.33 σx¯ = Math/Stat2330 Lecture-14 p µx¯ = 4.0 17.33 − (4.0)2 = 1.15 June 2013 16 / 19 The Sampling Distribution of Sample Means P(x) xP(x) x 2 P(x) 1/9 2/9 4/9 p µx¯ = 4.0 2/9 6/9 18/9 σx¯ = 17.33 − (4.0)2 = 1.15 3/9 12/9 48/9 2/9 10/9 50/9 1/9 6/9 36/9 1.0 4.0 17.33 Let’s now check the truth of the three facts about the sampling distribution of sampling means: 1. The mean µx¯ of the sampling distribution will equal to the mean µ of the population: both µ and µx¯ have the value 4.0. 2. The standard error of the mean σx¯ for the sampling distribution will equal the standard deviation σ of the population divided by the square root of the σ 1.63 sample size, n: σx¯ = 1.15 and σ = 1.63, n = 2, √ = √ = 1.15; they are n 2 equal. 3. The distribution will become approximately normally distributed: the histogram very strongly suggests normality. x 2 3 4 5 6 P G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 16 / 19 Application of the Sampling Distribution of Sample Means Example: Consider a normal distribution with µ = 100 and σ = 20. If random sample size of 16 is selected, what is the probability that this sample will have a mean value between 90 and 110? That is, what is P(90 < x¯ < 100)? G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 17 / 19 Application of the Sampling Distribution of Sample Means Example: Consider a normal distribution with µ = 100 and σ = 20. If random sample size of 16 is selected, what is the probability that this sample will have a mean value between 90 and 110? That is, what is P(90 < x¯ < 100)? Solution: Since the population is normally distributed, the sampling distribution of x¯0 s is normally distributed. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 17 / 19 Application of the Sampling Distribution of Sample Means Example: Consider a normal distribution with µ = 100 and σ = 20. If random sample size of 16 is selected, what is the probability that this sample will have a mean value between 90 and 110? That is, what is P(90 < x¯ < 100)? Solution: Since the population is normally distributed, the sampling distribution of x¯0 s is normally distributed. σ 20 µx¯ = µ = 100, σx¯ = √ = √ = 5 n 16 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 17 / 19 Application of the Sampling Distribution of Sample Means Example: Consider a normal distribution with µ = 100 and σ = 20. If random sample size of 16 is selected, what is the probability that this sample will have a mean value between 90 and 110? That is, what is P(90 < x¯ < 100)? Solution: Since the population is normally distributed, the sampling distribution of x¯0 s is normally distributed. σ 20 µx¯ = µ = 100, σx¯ = √ = √ = 5 n 16 x¯ − µx¯ Using the standard transformation z = : σx¯ 90 − 100 = −2.00 5 110 − 100 z-score for x¯ = 110: z = = 2.00 5 z-score for x¯ = 90: z = G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 17 / 19 Application of the Sampling Distribution of Sample Means Example: Consider a normal distribution with µ = 100 and σ = 20. If random sample size of 16 is selected, what is the probability that this sample will have a mean value between 90 and 110? That is, what is P(90 < x¯ < 100)? Solution: Since the population is normally distributed, the sampling distribution of x¯0 s is normally distributed. σ 20 µx¯ = µ = 100, σx¯ = √ = √ = 5 n 16 x¯ − µx¯ Using the standard transformation z = : σx¯ 90 − 100 = −2.00 5 110 − 100 z-score for x¯ = 110: z = = 2.00 5 z-score for x¯ = 90: z = Therefore, P(90 < x¯ < 110) = P(−2.00 < z < 2.00) = 0.9773 − 0.0228 = 0.9545 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 17 / 19 Application of the Sampling Distribution of Sample Means Example: Kindergarten children have heights that are approximately normally distributed about a mean of 39 inches and a standard deviation of 2 inches. A random sample of size 25 is taken, and the mean x¯ is calculated. What is the probability that this mean value will between 38.5 and 40.0 inches? G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 18 / 19 Application of the Sampling Distribution of Sample Means Example: Kindergarten children have heights that are approximately normally distributed about a mean of 39 inches and a standard deviation of 2 inches. A random sample of size 25 is taken, and the mean x¯ is calculated. What is the probability that this mean value will between 38.5 and 40.0 inches? Solution: What we are trying to find is P(38.5 < x¯ < 40.0). The values x¯, 38.5, x¯ − µ √ : 40.0 must be converted to z-scores using z = σ/ n x¯ = 38.5: z = 38.5 − 39.0 −0.5 x¯ − µ √ = √ = = −1.25 0.4 σ/ n 2/ 25 x¯ = 40.0: z = G. Zhu (UTPA) x¯ − µ 40.0 − 39.0 1 √ = √ = = 2.50 0.4 σ/ n 2/ 25 Math/Stat2330 Lecture-14 June 2013 18 / 19 Application of the Sampling Distribution of Sample Means Example: Kindergarten children have heights that are approximately normally distributed about a mean of 39 inches and a standard deviation of 2 inches. A random sample of size 25 is taken, and the mean x¯ is calculated. What is the probability that this mean value will between 38.5 and 40.0 inches? Solution: What we are trying to find is P(38.5 < x¯ < 40.0). The values x¯, 38.5, x¯ − µ √ : 40.0 must be converted to z-scores using z = σ/ n x¯ = 38.5: z = 38.5 − 39.0 −0.5 x¯ − µ √ = √ = = −1.25 0.4 σ/ n 2/ 25 x¯ = 40.0: z = x¯ − µ 40.0 − 39.0 1 √ = √ = = 2.50 0.4 σ/ n 2/ 25 Therefore, P(38.5 < x¯ < 40.0) = P(−1.25 < z < 2.50) = 0.9938 − 0.1057 = 0.8881 G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 18 / 19 Application of the Sampling Distribution of Sample Means Example: Use the heights of kindergarten children given previously. Within what limits the middle 90% of the sampling distribution of sample means for samples of size 100 fall? G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 19 / 19 Application of the Sampling Distribution of Sample Means Example: Use the heights of kindergarten children given previously. Within what limits the middle 90% of the sampling distribution of sample means for samples of size 100 fall? Solution: Using z-table, we find that the middle 90% is bounded by z = ±1.65. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 19 / 19 Application of the Sampling Distribution of Sample Means Example: Use the heights of kindergarten children given previously. Within what limits the middle 90% of the sampling distribution of sample means for samples of size 100 fall? Solution: Using z-table, we find that the middle 90% is bounded by z = ±1.65. x¯ − µ √ : use the formula z = σ/ n Then, we x¯ − 39.0 √ 2/ 100 x¯ − 39 = (−1.65)(0.2) x¯ = 39 − 0.33 = 38.67 z = −1.65: −1.65 = G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 19 / 19 Application of the Sampling Distribution of Sample Means Example: Use the heights of kindergarten children given previously. Within what limits the middle 90% of the sampling distribution of sample means for samples of size 100 fall? Solution: Using z-table, we find that the middle 90% is bounded by z = ±1.65. x¯ − µ √ : use the formula z = σ/ n x¯ − 39.0 √ 2/ 100 x¯ − 39 = (−1.65)(0.2) x¯ = 39 − 0.33 = 38.67 z = −1.65: −1.65 = G. Zhu (UTPA) Then, we x¯ − 39.0 √ 2/ 100 x¯ − 39 = (1.65)(0.2) x¯ = 39 + 0.33 = 39.33 z = 1.65: 1.65 = Math/Stat2330 Lecture-14 June 2013 19 / 19 Application of the Sampling Distribution of Sample Means Example: Use the heights of kindergarten children given previously. Within what limits the middle 90% of the sampling distribution of sample means for samples of size 100 fall? Solution: Using z-table, we find that the middle 90% is bounded by z = ±1.65. x¯ − µ √ : use the formula z = σ/ n x¯ − 39.0 √ 2/ 100 x¯ − 39 = (−1.65)(0.2) x¯ = 39 − 0.33 = 38.67 z = −1.65: −1.65 = Then, we x¯ − 39.0 √ 2/ 100 x¯ − 39 = (1.65)(0.2) x¯ = 39 + 0.33 = 39.33 z = 1.65: 1.65 = Thus, P(38.67 < x¯ < 39.33) = 0.90. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 19 / 19 Application of the Sampling Distribution of Sample Means Example: Use the heights of kindergarten children given previously. Within what limits the middle 90% of the sampling distribution of sample means for samples of size 100 fall? Solution: Using z-table, we find that the middle 90% is bounded by z = ±1.65. x¯ − µ √ : use the formula z = σ/ n x¯ − 39.0 √ 2/ 100 x¯ − 39 = (−1.65)(0.2) x¯ = 39 − 0.33 = 38.67 z = −1.65: −1.65 = Then, we x¯ − 39.0 √ 2/ 100 x¯ − 39 = (1.65)(0.2) x¯ = 39 + 0.33 = 39.33 z = 1.65: 1.65 = Thus, P(38.67 < x¯ < 39.33) = 0.90. Therefore, 38.67 inches and 39.33 inches are the limits that captures the middle 90% of the sample means. G. Zhu (UTPA) Math/Stat2330 Lecture-14 June 2013 19 / 19
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