# Additional problems_3 on statistical distributions -...

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Additional problems_ 3 on statistical distributions.doc // Consider at first handouts #6b, 6a, the file // Binomial Probability Distribution.doc, // and HW #4 (part 1); then - handout #7 // and HW #4 (part 2); then - handout #8 // and HW #5. Apr 4’10 Sampling Distribution of the Sample Means //Use the program Three_Distributions.doc , sheet 3 In addition to HW #6 the following problems may be discussed. 5.3.4 , pg. 139, textbook . Wright et al. used the 1999-2000 National Health and Nutrition Examination Survey (NHANES) to estimate dietary intake of 10 key nutrients. One of those nutrients was calcium (mg). They found in all adults 60 years and older a mean daily calcium intake of 721 mg with a standard deviation of 454 mg. Using these values for the mean and standard deviation for the US population find the probability that a random sample of size 50 will have a mean: a ) Greater than 800 mg _ __ Solution. µ x_mean = µ = 721 G58, σ x_mean = σ/√n = 454/√50 = 64.2 G59 _ _ P(X > 800) = 1 – f(X ≤ 800) =

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Unformatted text preview: 0.109249 I60 b ) Less than 700 mg P(X ≤ 700) _ Solution. X = 700 G57 _ _ P(X ≤ 700) = f(X ≤ 700) = 0.371795 I59 c ) Between 700 and 850 mg P(700 ≤ X ≤ 850) _ _ _ I59 I59 P(700 ≤ X ≤ 850) = f(X ≤ 850) – f(X ≤ 700) = 0.97775 – 0.371795 = 0.605955 1 5.3.6 , pg. 140, textbook . Given a normally distributed population with a mean of 100 and a standard deviation of 20 , find the following probabilities based on a sample of size 16 . _ ( a ) P(X ≥ 100) _ _ Solution. P(X ≥ µ) = 1 – f(X ≤ µ) = 1/2- general result _ __ µ x_mean = µ = 100 G58, σ x_mean = σ/√n = 20/√16 = 5 G59 _ X = 100 G57 _ _ P(X ≥ 100) = 1 – f(X ≤ 100) = 0.5 I60 _ ( b ) P(X ≤ 110) _ Solution. X = 110 G57 _ _ P(X ≤ 110) = f(X ≤ 110) = 0.97725 I59 _ (c) P(96 ≤ X ≤ 108) _ _ _ I59 I59 P(96 ≤ X ≤ 108) = f(X ≤ 108) – f(X ≤ 96) = 0.945201 – 0.211855 = 0.733346 2...
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Additional problems_3 on statistical distributions -...

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