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 (mg). They found in all adults 60 years or older a mean daily calcium intake of 721 mg with a standard deviation of 454. Using these values for the mean and standard deviation for the U. S. population, find and interpret the probability that a random sample of size 50 will a mean:

Complete question :

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.

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

answer:

sometimes

*i think