Wright et al. [A-2] 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 in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Usin these values for the mean and standard deviation for the U.S. population, find the probability that a randonm sample of size 50 will have a mean: (mg). They found a) Greater than 800 mg b) Less than 700 mg. c) Between 700 and 850 mg.

Answer:

0.10935

0.3718

0.9778

0.606

Step-by-step explanation:

μ = 721 ; σ = 454 ; n = 50

P(x > 800)

Zscore = (x - μ) / σ/sqrt(n)

P(x > 800) = (800 - 721) ÷ 454/sqrt(50)

P(x > 800) = 79 / 64.205295

P(x > 800) = 1.23

P(Z > 1.23) = 0.10935

2.)

Less than 700

P(x < 700) = (700 - 721) ÷ 454/sqrt(50)

P(x < 700) = - 21/ 64.205295

P(x < 700) = - 0.327

P(Z < - 0.327) = 0.3718

Between 700 and 850

P(x < 850) = (850 - 721) ÷ 454/sqrt(50)

P(x < 850) = 129/ 64.205295

P(x < 700) = 2.01

P(Z < 2.01) = 0.9778

P(x < 850) - P(x < 700) =

P(Z < 2.01) - P(Z < - 0.327)

0.9778 - 0.3718

= 0.606

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