SOLUTION: the total cholesterol in children aged 10-15 is assumed to follow a normal distribution with a mean of 191 and a standard deviation of 22.4. what is the proportion of children 1

Question 514001: the total cholesterol in children aged 10-15 is assumed to follow a normal distribution with a mean of 191 and a standard deviation of 22.4.
what is the proportion of children 10-15 years of age have total cholesterol between 180 and 190?
Found 2 solutions by stanbon, drcole:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
the total cholesterol in children aged 10-15 is assumed to follow a normal distribution with a mean of 191 and a standard deviation of 22.4.
what is the proportion of children 10-15 years of age have total cholesterol between 180 and 190?
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z(180) = (180-191)/22.4 = -0.4911
z(190) = (190-191)/22.4 = -0.0446
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P(180 < x < 190) = P(-0.4911 < z < -0.0446) = 0.1705
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Cheers,
Stan H.
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Answer by drcole(72) (Show Source): You can put this solution on YOUR website!
We can compute the proportion using z-scores and a normal distribution table (or a calculator). We want to find the proportion of children 10-15 years of age that have total cholesterol between 180 and 190. The idea here is to figure out how many standard deviations 180 and 190 are from the mean. The number of standard deviations left or right of the mean is called the z-score of a number. The formula for z-score is:
z-score of x =
where is the mean of the normal distribution and is the standard deviation. In this case, and , so:
z-score of 180 =
z-score of 190 =
In other words, 180 is 0.491 standard deviations below the mean; 190 is only 0.045 standard deviations below the mean.
One we know the z-scores, we can figure out the proportions based on the probability tables of the standard normal distribution.
P(180 < x < 190) = P(-0.491 < z-score < -0.045) = P(z-score < -0.045) - P(z-score < -0.491)
The last change is necessary because most tables only give the probability that a z-score is less than some number. The probability that the z-score is between two numbers a and b equals the probability that the z-score is less than b minus the probability that the z-score is less than a (since we shouldn't count the times when the z-score is both below b and below a).
Looking these probabilities up in a table you should have, or using a calculator or computer (or even an internet applet --- google it), you should find that:
P(z-score < -0.045) = 0.4840 (so nearly 50%, since we are near the mean)
P(z-score < -0.491) = 0.3121
Thus P(180 < x < 190) = 0.4840 - 0.3121 = 0.1719
So 0.1719 of children 10-15 years of age have total cholesterol between 180 and 190.
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