In the 1999-2000 National Health Examination Survey the dietary intakes of 10 ke
ID: 3125486 • Letter: I
Question
In the 1999-2000 National Health Examination Survey the dietary intakes of 10 key nutrients were estimated. The researchers found that the mean sodium intake in men and women 60 years and older was 2940 mg with a standard deviation of 1476 mg. Use these values for the mean and standard deviation of the U.S. population and find the probability that a random sample of 75 people from the population will have a mean: a. Less than 2450 mg b. Over 3100 mg c. Between 2500 and 3300 d. Between 2500 and 2900 mg
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 2450
u = mean = 2940
n = sample size = 75
s = standard deviation = 1476
Thus,
z = (x - u) * sqrt(n) / s = -2.875016584
Thus, using a table/technology, the left tailed area of this is
P(z < -2.875016584 ) = 0.002020031 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 3100
u = mean = 2940
n = sample size = 75
s = standard deviation = 1476
Thus,
z = (x - u) * sqrt(n) / s = 0.938780926
Thus, using a table/technology, the right tailed area of this is
P(z > 0.938780926 ) = 0.173921617 [ANSWER]
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c)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 2500
x2 = upper bound = 3300
u = mean = 2940
n = sample size = 75
s = standard deviation = 1476
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -2.581647545
z2 = upper z score = (x2 - u) * sqrt(n) / s = 2.112257082
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.004916497
P(z < z2) = 0.982667798
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.9777513 [ANSWER]
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d)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 2500
x2 = upper bound = 2900
u = mean = 2940
n = sample size = 75
s = standard deviation = 1476
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -2.581647545
z2 = upper z score = (x2 - u) * sqrt(n) / s = -0.234695231
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.004916497
P(z < z2) = 0.407222645
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.402306147 [ANSWER]