SOLUTION: I am having a hard time in my Biostatistics class. I don't understand anything about it.Need someone to explain how to figure this out so i can complete the other questions.
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Question 348892: I am having a hard time in my Biostatistics class. I don't understand anything about it.Need someone to explain how to figure this out so i can complete the other questions.
Total cholesterol in children ages 10 to 15 years of age is assumed to follow a normal distribution with a mean of 191 and a standard deviation of 22.4.
a. What proportion of children 10 to 15 have a total cholesterol between 180 and 190?
b. What proportion of children 10 to 15 would be classified as hyperlipidemic(assume that hyperlipidemia is defined as a total cholesterol level over 200?
c. what is the 90th percentile of total cholesterol? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Total cholesterol in children ages 10 to 15 years of age is assumed to follow a normal distribution with a mean of 191 and a standard deviation of 22.4.
a. What proportion of children 10 to 15 have a total cholesterol between 180 and 190?
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Find the z-scores:
z(180) = (180-191)/22.4 = -0.4911
z(190) = (190-191)/22.4 = -0.0446
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The proportion between 180 and 190 = the probability
that z is between -0.4911 and -0.0446 = 17.05%
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b. What proportion of children 10 to 15 would be classified as hyperlipidemic(assume that hyperlipidemia is defined as a total cholesterol level over 200?
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z(200) = (200-191)/22.4 = 0.4018
Proportion over 200 = probability z is greater than 0.4018 = 34.49%
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c. what is the 90th percentile of total cholesterol?
Find the z-value corresponding to a left-tail of 0.90.
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invNorm(0.90) = 1.2816
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Use x = zs+u to find the total cholesterol.
x = 1.2816*22.4+191 = 219.71
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Cheers,
Stan H.
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