SOLUTION: Suppose that total carbohydrate intake in 12-14 years old males is normally distributed with mean 124 g/1000 cal and standard deviation of 20 g / 1000cal. a. What percent of boys

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Question 1175051: Suppose that total carbohydrate intake in 12-14 years old males is normally distributed with mean 124 g/1000 cal and standard deviation of 20 g / 1000cal.
a. What percent of boys in this age range have carbohydrate intake above 140 g/ 1000 cal?
b. What is the probability that a sample of 35 boys have a mean carbohydrate intake of greater than 120 g /100 cal?
Found 2 solutions by Boreal, ewatrrr:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
z=(x-mean)/sd
so for a (140-124)/20 or z > 0.8
that probability is 0.2119 from the calculator 2ndVARS ENTER 2 normalcdf(.8,6) ENTER. The 6 is the std deviations to essentially 100%. Some use 1E99, 6 works, and it is easier. 7 would and 8 would, but 5 has some error.
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for b z=(x bar-mean(/sigma/sqrt(n), since std error of sample is sigma/sqrt(n).
This is z> (120-124)/20/sqrt(35) or > -4*sqrt(35)/20= -1.18 and that probability is 0.8810

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi 
 
Normal Distribution:    μ = 124  and   σ = 20  
Continuous curve..
a) P(x > 140) = 1- P(x ≤ 140) =  1-0.7881 = .2119

b)P(x > 120) = 1-P(x ≤ 120) = 1 - .4207 = .5793

c) z =blue (x̄ - mu)/sigma/sqrt(n) = (120-124)/20/√35 =  -1.1832 
 invNorm(-1.832) = .1184
 P(z >-1.1834) = 1 - .1184 = .9916

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