Cholesterol is a fatty substance that is an important part o

Cholesterol is a fatty substance that is an important part of the outer lining (membrane) of cells in the body of all humans. Its normal range for an adult is 120 – 240 mg/ dl. The Food and Nutrition Institute of the Philippines found that the total cholesterol level for Filipino adults has a mean of 159.2 mg/dl and 84.1% of adults have a cholesterol level below 200 mg/dl. Suppose that the total cholesterol level is normally distributed.

A) Determine the standard deviation of this distribution.   

B) What are the quartiles ( the 25% and the 75% values ) of this distribution?   

C) What is the value of the cholesterol level that exceeds 90% of the population?

D) An adult is at moderate risk if the cholesterol level is more than one but less than two standard deviations above the mean. What percentage of the population is at moderate risk according to this criterion?

E)An adult is thought to be at high risk if his cholesterol level is more than two standard deviations above the mean. What percentage of the population is at high risk?                                                                                                                                                      

F)An adult has low risk if his cholesterol level is one standard deviation or more below the mean. What percentage of the population is at low risk?                                                                                                                                                                            

Solution

A)

The z score corresponding to a left tailed area of 0.841, by table/tchnology, is

z = 0.998576271

Thus, as

sigma = (x-u)/z

Then

sigma = (200-159.2)/0.998576271

= 40.85817096 [ANSWER]

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b)

FIRST QUARTILE:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    159.2      
z = the critical z score =    -0.67448975      
s = standard deviation =    40.85817096      
          
Then          
          
x = critical value =    131.6415825   [ANSWER, 1ST QUARTILE]  

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3RD QUARTILE:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    159.2      
z = the critical z score =    0.67448975      
s = standard deviation =    40.85817096      
          
Then          
          
x = critical value =    186.7584175   [ANSWER, 3RD QUARTILE]  

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c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    159.2      
z = the critical z score =    1.281551566      
s = standard deviation =    40.85817096      
          
Then          
          
x = critical value =    211.561853   [ANSWER]

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d)

z1 = lower z score =    1      
z2 = upper z score =     2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.841344746      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.135905122 or 13.59% [ANSWER]      

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