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.
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
Find the common denominator of 5/6 and 3/4 then add
The diameter is equal to the radius multiplied by 2.
Answer:
11
Step-by-step explanation:
Given that in a sample of n=6, five individuals all have scores of X=10 and the sixth person has a score of x=16
i.e. the data entries are
10,10,10,10,10,16
Sum = 66
No of items n = 6
Average = sum/no of entries
=
Average is 11.
mean of this sample is 11
