Homework SourseThe serum cholesterol levels of a population of 12 to 14year
The serum cholesterol levels of a population of 12 to 14-year-olds follow a normal distribution with mean 155 mg/dl and standard deviation 27 mg/dl (as in Example 4.1.1).
What percentage of the 12- to 14-year-olds have serum cholesterol values between 145 and 165 mg/dl?
Suppose we were to choose at random from the population a large number of groups of nine 12- to 14-year-olds each. In what percentage of the groups would the group mean cholesterol value be between 145 and 165 mg/dl?
If Y bar represents the mean cholesterol value of a random sample of nine 12- to 14-year-olds from the population, what is Pr{145less than or equal to Ybar less than or equal to 165}?
Solution
a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 145
x2 = upper bound = 165
u = mean = 155
s = standard deviation = 27
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.37037037
z2 = upper z score = (x2 - u) / s = 0.37037037
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.355553274
P(z < z2) = 0.644446726
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.288893452 or 28.89% [ANSWER]
*******************
b)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 145
x2 = upper bound = 165
u = mean = 155
n = sample size = 9
s = standard deviation = 27
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -1.111111111
z2 = upper z score = (x2 - u) * sqrt(n) / s = 1.111111111
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.133260263
P(z < z2) = 0.866739737
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.733479474 or 73.35% [ANSWER]
*********************
c)
It is the same as part B, 0.733479474. [ANSWER]
