Rabu, 03 April 2013

What are the various types of hyperlipidemias?

Q.

A. Type I or Buerger-Gruetz syndrome, or Primary hyperlipoproteinaemia, or Familial hyperchylomicronemia
Problems: Decreased lipoprotein lipase (LPL) or altered ApoC2
Treatment: Diet Control

Type IIa or Polygenic hypercholesterolaemia or Familial hypercholesterolemia
Problems: LDL receptor deficiency
Treatment: Bile Acid Sequestrants, Statins, Nicotinic acid

Type IIb or Combined hyperlipidemia
Problems: Decreased LDL receptor and Increased ApoB
Treatment: Statins, Nicotinic acid, Gemfibrozil

Type III or Familial Dysbetalipoproteinemia
Problems: Defect in ApoE synthesis
Treatment: Gemfibrozil

Type IV or Endogenous Hyperlipemia
Problems: Increased VLDL production and Decreased elimination
Treatment: Nicotinic acid

Type V or Familial Hypertriglyceridemia
Problems: Increased VLDL production and Decreased LPL
Treament: Nicotinic acid, Gemfibrozil

Source:

http://en.wikipedia.org/wiki/Hyperlipidemia

a study reported that the prevalence of hyperlipidemia is 30% in children 2 to 6 years of age. if 12 children?
Q. are analyzed: a) what is probability that at least 3 are hyperlipidemia? b) what is the probability that exactly 3 are hyperlipidemic? and c) how many would be expected to meet the criteria for hyperlipidemia?

A. a. ANSWER: PROBABILITY = 0.75 at least 3 are hyperlipidemia

Why???

BINOMIAL DISTRIBUTION, POPULATION PROPORTION
n = NUMBER OF TRIALS [ 12] (sample size)
k = NUMBER OF SUCCESSES [2] (from 0 up to and including k NUMBER OF SUCCESSES)
p = POPULATION PROPORTION [30%]

significant digits2

COMPUTATION OF BINOMIAL PROPORTION:
P(k => 3) = 1 - P(k ⤠2) = 1 - n!/[k!*(n - k)!] * p^k * (1 - p)^(n - k)
0.75 = 12!/[2!*(12 - 2)!] * 0.3^2 * (1 - 0.3)^(12 - 2)

ALTERNATIVE COMPUTATION USING EXCEL:
"Look-up" value of PROBABILITY = 0.75 = 1 - BINOMDIST ( 2 , 12 , 30/100 , TRUE )
"Using Excel function: BINOMDIST(number_s, trials, probability_s, cumulative)
Number_s is the number of successes in trials. [ 12 ]
Trials is the number of independent trials. [ 2 ]
Probability_s is the probability of success on each trial. [ 30]"

Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, then BINOMDIST returns the cumulative distribution function, which is the probability that there are at most number_s successes; if FALSE, it returns the probability mass function, which is the probability that there are number_s successes.



b. ANSWER: PROBABILITY = 0.24 exactly 3 are hyperlipidemic

Why???

BINOMIAL DISTRIBUTION, POPULATION PROPORTION
n = NUMBER OF TRIALS [ 12] (sample size)
k = NUMBER OF SUCCESSES [3] (Exactly 3 NUMBER OF SUCCESSES)
p = POPULATION PROPORTION [30%]

significant digits2

COMPUTATION OF BINOMIAL PROPORTION:
P(k = 3) = n!/[k!*(n - k)!] * p^k * (1 - p)^(n - k)
0.24 = 12!/[3!*(12 - 3)!] * 0.3^3 * (1 - 0.3)^(12 - 3)

ALTERNATIVE COMPUTATION USING EXCEL:
"Look-up" value of PROBABILITY = 0.24 =BINOMDIST ( 3 , 12 , 30/100 , FALSE )
"Using Excel function: BINOMDIST(number_s, trials, probability_s, cumulative)
Number_s is the number of successes in trials. [ 12 ]
Trials is the number of independent trials. [ 3 ]
Probability_s is the probability of success on each trial. [ 30]"

Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, then BINOMDIST returns the cumulative distribution function, which is the probability that there are at most number_s successes; if FALSE, it returns the probability mass function, which is the probability that there are number_s successes.



c. ANSWER: (approx) 4 children expected to be hyperlipidemic

Why???
SAMPLE SIZE * POPULATION PROPORTION = EXPECTED [12 * 0.3 = (approx) 4]

Hyperlipidemia is 30% in children 2-6 years of age. If 12 children are analyzed, what is the probabilty that?
Q. At least 3 are hyperlipidemic?

A. Well, four kids out of twelve should have it. But your question isn't very specific. If you have three kids, and you wanna know what their odds of getting it are, it depends.
Family history is an important factor, but if you're going by statistics alone, one of your kids probably has it.




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