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Some Sect 7.5 Recitation Problems

# Some Sect 7.5 Recitation Problems - ASW ESBE(5e Chapter 7...

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ASW ESBE (5e) Chapter 7, Section 7.5 Solutions to Selected Problems 24. a. Show the sampling distribution of xbar (all values in dollars). 1. ( ) 4260 E x = 2. / 900/ 50 127.28 x n σ σ = = = 3. The sampling distribution of the sample mean is normal by CLT by CLT ( n = 50 > 30) b. Within ±\$250 of μ Want Prob ( μ – \$250 ≤ xbar ≤ μ +\$250) = Prob (4010 x 4510) 4510 4260 1.96 127.28 z - = = Prob (4010 x 4510) = Prob (-1.96 z 1.96) = Prob (z 1.96) – Prob (z -1.96) = 0.9750 – 0.0250 = 0.9500 c. Within ±\$100 Want Prob ( μ – \$100 ≤ xbar ≤ μ +\$100) = Prob (4160 x 4360) 4360 4260 .79 127.28 z - = = Prob (4160 x 4360) = Prob (-0.79 z 0.79) = Prob (z 0.79) – Prob (z -0.79) = 0.7852 – 0.2148 = 0.5704 Some Section 7.5 problem solutions 1

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New question: Suppose a sample of size n = 50 was chosen. The sample mean xbar was calculated to be \$4615. What is the chance of selecting a sample whose is \$4615 or larger? Want Prob (xbar ≥ \$4615) 79 . 2 28 . 127 \$ 4260 \$ - 4615 \$ - = = = x x z σ μ Prob (xbar ≥ \$4615) = Prob (z ≥ 2.79) = 1 – Prob (z ≤ 2.79) = 1 – 0.9974 = 0.0026 Note: 0.0026 < 0.01, so xbar ≥ \$4615 is a rare event. Possible explanations: 1. Bad luck. 2. Non-random sample. 3. Maybe μ ≠ 4260 as claimed, or σ ≠ 900 as claimed, or both. Some Section 7.5 problem solutions 2
25. Given: μ = 1020 σ = 100 n = 75 Show the sampling distribution of xbar. 1. E ( x ) = 1020 2. σ σ x n = = = / / . 100 75 1155 3. The sampling distribution of the sample mean is normal by CLT ( n = 75 > 30) b. Want Prob ( μ – 10 ≤ xbar ≤ μ +10) = Prob (1010≤ xbar ≤ 1030) 1030 1020 .87 11.55 z - = = 1010 1020 .87 11.55 z - = = - Prob (1010≤ xbar ≤ 1030) = Prob (-0.87≤ z ≤ 0.87) = Prob ( z ≤ 0.87) – Prob (z ≤ -0.87) = 0.8078 – 0.1922 = 0.6156 c. Want Prob ( μ – 20 ≤ xbar ≤ μ +20) = Prob (1000≤ xbar ≤ 1040) 1040 1020 1.73 11.55 z - = = 1000 1020 1.73 11.55 z - = = - Area = .4582 Prob (1000≤ xbar ≤ 1040) = Prob (-1.73≤ z ≤ 1.73) = Prob ( z ≤ 1.73) – Prob (z ≤ -1.73) = 0.9582 – 0.0418 = 0.9164 Some Section 7.5 problem solutions 3

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New question: Suppose a sample of size n = 75 was chosen. The sample mean xbar was calculated to be 1050. What is the chance of selecting a sample whose is 1050 or larger? Want Prob (xbar ≥ 1050) 60 . 2 55 . 11 1020 - 1050 - = = = x x z σ μ Prob (xbar ≥ 1050) = Prob (z ≥ 2.60) = 1 – Prob (z ≤ 2.60) = 1 – 0.9953 = 0.0047 Note: 0.0047 < 0.01, so xbar ≥ 1050 is a rare event. Possible explanations: 1. Bad luck. 2. Non-random sample. 3. Maybe μ ≠ 1020 as claimed, or σ ≠ 100 as claimed, or both. Some Section 7.5 problem solutions 4
26. Given: μ = \$939 σ = \$245 n = 30, 50, 100, and 400 Show the sampling distribution of xbar.

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Some Sect 7.5 Recitation Problems - ASW ESBE(5e Chapter 7...

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