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Write your tLecture 1

Descriptive statistics: Describe and summarize the data from the sample.

Probability models describe a model For the population which can be used to make probability statements about what Can be seen in the sample.

Statistical Inference goes in the Other direction. It uses the data in the sample to make inference about Features of  the population with Associated confidence and/or error bounds.

SPSS: Rows are subjects/units which are also Referred to as cases.

Columns are Things which are being measured on the cases. These are called variables.


Obtaining Frequency tables:

Analyze -> Descriptive Stats -> Frequencies

Bar Graph: Graphs -> Legacy Dialog -> Bars -> Simple -> Category Axis (zyg)

Pie Chart: Graphs -> L egacy Dialog -> Pie Charts  -> Define Slices By (zyg)

Descriptive Methods for continuous data

Frequency distribution and Histogram: Graphs -> Legacy Dialog -> Histogram (add to ‘Variable’ slot)

OR better for table data as Well: Analyze -> Descriptive -> Explore -> (Plots option)-choose Histogram

Measures of Spread

Range= Largest – smallest observation

Variance/St Dev.- take each number and subtract it from the average Number, then add up the differences = 0, then square them, then add them =100, SD= 100/divided by the number of numbers present –minus 1. B/c the differences Always add up to zero, so you already know the difference of the last one, so That’s why u minus by 1, but don’t know the first 6 differences. SD= sqrt of Variance

IQR-Analyze, descriptive stats, frequencies, stats options, check Quartiles and ect. And done. 

IQR= range from Quartile 3 (-) Quartile 1 = (range of middle 50%)

IQR Min max median IQR stdev variance, but no individual quartiles. 

Analyze -> Descriptive -> Explore

Box Plot

The Middle line is drawn at the median

The Box is drawn at the first quartile or 25th percentile (bottom) and The third quartile or 75th percentile (top). The box in the middle Thus covers the middle 50% of the data

Fences Are then constructed (not shown in the plot)
Lower fence = 25th percentile (bottom of the box) – 1.5*IQR

Upper Fence = 75th percentile (bottom of the box)  +  1.5*IQR
When comparing two different variables Numerically, go to analyze compare means, means, and fill in blanks.

Side by Side Boxplot Comparison: analyze-descriptive-explore

Lecture 2: Probability

Properties of Probability

  1. For an event A, 0 £ Prob(A) £  1 .

  2.  If A and B Cannot happen at the same time (A and B are disjoint or mutually Exclusive), then
        Prob(A or B)  = Prob (A) + Prob (B)

Example 3.6 of text: Let A be the event that a person has normotensive Diastolic blood pressure (DBP) readings (DBP < 90)  and let B be the event of borderline DBP (90 £ DBP < 95). Then A and B are mutually exclusive and

            P(A or B) = P(DBP < 95) = P(DBP < 90) + P(90 £ DBp < 95) = P(A) + P(B)

If C = {DBP ³ 90}  and D= {75 £ DP £ 100} , the C and D are not mutually exclusive

·The event C or D is also denoted as  CÈD (called C Union D) = {DBP ³ 75}

The event C and D is also denoted as  CÇD (called C Intersection D) = {90 £ DBP £ 100}

Independent Event
Two events A and B are called Independent (statistically or probabilistically) if
       P(A and B) = P(AÇ B) =   P(A)x P(B)

Independence means That the occurrence of the event A does not alter the chance of occurrence of The event B.

If two events are not Independent (that is, if the equality above does not hold), then they are Dependent.

Note that the definition of  “independence” is in terms of the probabilities of the events, not just In terms of the events themselves.

“Mutually exclusive” and “Independence” are two separate ideas/definitions.
Going back to Example 3.12, if we assume that {mother’s DBP ³ 95} and {father’s  DBP ³ 95} are two independent events (which is a reasonable assumption), the

 P(mother’s DBP ³ 95 and father’s  DBP ³ 95)

= P(AÇ B) =   P(A)x P(B)  (since we assume they are independent)
= P(mother’s DBP ³ 95)x P(father’s  DBP ³ 95)

= 0.1x0.2 = 0.02

Addition Rule of Probability

We mentioned before that if A and B are Mutually exclusive (disjoint), then

   P(A or B) = P(AÈB) = P(A) + P(B)

What if they are not Disjoint?

Example  A cytological test for cervical cancer

Test (T)

Cancer (C)
















Assumption: All patients are Equally likely to be test positive or Cancer=yes (based on the given Information)

P(a random selected patient will Test +) = P(T) = 516/24103 = 0.021

P(a random selected patient will have cancer=y) = P(C) = 379/24103 = 0.015

 P(a random selected patient will Test Positive or have cancer )

= P(T or C) = P(TÈC)
=  (516  + 379 – 154)/24103

 = P(T) + P(C) – P(T and C)

Addition rule:   P(AÈB)  = P(A) + P(B) – P(AÇB) = P(A) + P(B) – P(A and B)

Conditional Probability

Sensitivity of the test = 154/379 = 0.406

Sensitivity = P(a diseased patient will Test positive)
                   = P(a patient will Test positive given that the patient has cancer)
                   = P(Test positive | Patient has cancer)  = P(T|C)

Note that this is a conditional probability which is different From P(T) or P(C) or P(T and C)

Specificity = 23362/23734= 0.984

Specificity =  P(a Non-diseased patient will Test negative)
                   = P(Test negative | Patient does not have Cancer)
                    = P(T-complement | C-complement)  = P(Tc |Cc)

Positive Predictivity or Positive Predictive Value

Suppose a patient test Positive in the test. Then, the important question is

What is the probability of Cancer  given that Test was positive

Positive Predictive Value or Predictive value positive =
   = PV+   =  P(Cancer |Test positive)  = P(C |T)

In our example,   PV+  = positive predictive value = P(C| T) =  154/516 =  0.298

                            PV-  = Negative predictive value = P(Not Cancer|Test Negative)

                                                                                = P(C-complement| T-complement)
                                                                                =  23362/23587 = 0.99
Example 3.19 of book:

  • Suppose among 100,000 Women with negative mammogram results, 20 will be diagnosed with breast Cancer within 2 years.
    M = positive mammogram
    B = Breast Cancer

P(B|M-complement) = 20/100000 = 0.00002

  • Suppose 1 woman in 10 With positive mammograms will be diagnosed with breast cancer Within 2 years
    P(B|M)  = 1/10= 0.1

    Is this sensitivity or something Else?

  • The Relative Risk of Breast Cancer for Mammogram Positive to negative then is

     RR = P(B|mammogram positive) / P(B|mammogram negative)
            = P(B|M)/ P(B|M-complement)
            = 0.1/0.0002 = 500

    “Breast Cancer” and “Mammogram” are clearly Dependent since a mammogram positive result increases the risk of breast Cancer 500 times.

Suppose that 7% of the general women population will have a positive Mammogram. What is the chance of developing breast cancer over the next Two years among women in the general population.
P(B) = P(Breast Cancer)
  = P(Breast cancer |Mammogram positive)x P(Mammogram positive)
     +  P(Breast cancer |Mammogram negative)x P(Mammogram negative)

  = (0.1)x(0.07)  + (0.0002)(0.93) = 0.00719 

Positive Predictive value depends on the prevalence of the disease

Population A

Population B

Test (T)

Test (T)

Disease (D)




























Population A                                                    Population B

Sensitivity =                                                     Sensitivity =

P(T=+ | D=y) = 45000/50000 = 0.90                   P(T=+|D=y) = 9000/10000 = 0.90

Specificity =                                                     Specificity =

P(T=- | D=n) = 45000/50000 = 0.90                   P(T=-|D=n) = 81000/90000 = 0.90

Disease Prevalence                                           Disease Prevalence

= 50000/100000 = 0.50                         = 10000/10000 = 0.10

Positive Predictive value                                   Positive Predictive value

= P(D=y|T=+) = 45000/50000=0.90                    = P(D=y|T=+)   = 9000/18000 = 0.50

Sensitivity, Specificity and ROC (receiver operating characteristic) curve

Open SPSS File BiomarkerROCCurve.Sav

Attaching Value Labels

  “Variable view” tab at Left-bottom.   -> Values

SPSS -> Analyze -> ROC curve

        State variable  NodalStatus.  Value =1
        Test Variables  -  Biomarker1-3.
        Check Coordinate points and Diagonal reference line

Values of X















Chap4: Discrete Probability Distributions
Example 4.4: Hypertension control:

A Pharmaceutical company is marketing a new drug and based on their studies, they Claim that if 4 patients with untreated hypertension are treated with this Drug,  then the chance that 0 out of 4 Will have their hypertension  brought Under control is 0.008, 1 patient out of 4 is 0.076, 2 out of 4 is 0.265, 3 out Of 4 is o.411 and all 4 is 0.240.

X = # of patients whose hypertension brought Under control.
   This is a random variable. It is a discrete Random variable

This is the probability distribution of the random variable which describes its possible values and the Associated probabilities.
     The Expected value or mean of a Discrete random variable.
      mean(X) = E(X) = Sum over { value x Probability}
                                   = (0)x(0.008) + (1) x(0.076) + (2)x(0.265) + (3)x(0.411) + (4)x(0.240)
                                   = 2.80

      This represents the mean based on a Probability model (based on a model for the population) as opposed to mean or Average based on data from a sample.

    We will denote this population mean As  m  (pronounced as mu)  (see page 75 of text).

     Similarly, the variance of the Random variable X measures the variability in the values of X in terms of the Associated probabilities. We will denote this population variance as s2     (pronounced as sigma square)               

A Discrete Probability model: The Binomial distribution

The assumptions are
   a) Binary setting

   b) Cases/subjects are independent

   c) Probability of success is the same, that Is, does not vary from one case/subject to another

Example: A certain Drug is known to produce side effects (adverse effects) on 10% of patients. We Want to now that if the drug is given to 5 patients, how many patients may have AEs.

            X = # of patients having AEs

In this example, we are counting AEs. So, in terms of Binomial distribution
      success = having AE  and    Failure = not-having AE


  1. Binary Setting
  2. n = 5 Patients. We assume that these 5 patients are independent (as far as Having AE is concerned).

  3. We also Assume that each patient has p= 0.10 chance of having AE. That is, the Chance of having AE is constant and does not vary from patient to patient.

  Under these Assumptions, we can treat X as a binomial random variable with n=5 and
P= Prob(success) = Prob(AE) = 0.1

SPSS -   Transform -> Compute -> PDF.BINOM   & CDF.BINOM functions
                                                         Follow the explanation given On these functions

If we consider 50 patients Taking the drug, how many of them are expected to experience AE on the average.

Expected value or mean of a Binomial distribution = np  = (50)x(0.10) = 5  (p 200 of text)

It may not be exactly 5 in a specific group of 50 students since that is what We expect on the average. So, what kind of variability can we expect around the Expected value of 5.

Standard Deviation of a Binomial distribution = .
Example 4.4: Hypertension control (which we considered earlier):

A Pharmaceutical company is marketing a new drug and based on their studies, they Claim that if 4 patients with untreated hypertension are treated with this Drug,  then the chance that 0 out of 4 will Have their hypertension  brought under Control is 0.008, 1 patient out of 4 is 0.076, 2 out of 4 is 0.265, 3 out of 4 Is o.411 and all 4 is 0.240.

X = # of patients whose hypertension brought under control.

   This is a random variable. It is a Discrete random variable

Values of X














 Note that these probabilities are from a Binomial distribution with n=4 and

p = Prob(success) = Prob(Hypertension brought under control) = 0.7

Since this is a Binomial distribution, we can use the formula for mean of a Binomial distribution

   mean = np = (4) (0.7) = 2.8

   std deviation =

Lecture 3-4

Chap 5: Probability model for continuous data: The Normal distribution

probability Density function (pdf).

The probability of the random quantity falling in between two values is given By the AREA Under the curve in between those values.

The total are under a probability density Curve is, of course, 1  as it Covers all possibilities.

Technical point

One consequence of this is That for a continuous probability distribution, the probability of the random Quantity being exactly equal to a value is Always zero.

P(X=1)  = 0 And            P(X=1.5) =0   and        P(X= -2) = 0.

Why is that? Because P(X=1) means probability X is Exactly equal to 1, or    P(X=1.00000000000000000………) and that probability is zero.

P(0.9999999999 ≤ X ≤ 1.0000000001)  is not zero but is given by the area of the Curve over that interval.

We are going to focus on one Class of continuous probability distributions. The Normal probability Distribution has a density curve which is symmetric and bell-shaped with a Single-peak.

To speak specifically of any normal distribution, two quantities have to be specified: the mean m (a greek symbol pronounced mu) where the peak of the Density occurs, and the standard deviation s (pronounced Sigma) which indicates the spread or girth of the bell curve.

Thus, for a normal Distribution, almost all values lie within 3 standard deviations of the Mean.

Standard Normal Distribution or the Z curve

The normal curve with mean = 0 and sd =1 is called the standard normal curve or the z curve. The z curve Plays a special role since all other normal distributions can be transformed to The z-curve by using the z score    

Example If a single score is selected at random form a standard Normal distribution, what is the probability that 

 Z- standard normal distribution, which Means that the center is zero and s.D. Is one

A) the score is in between Z = -1.0 and Z = 1.0?  ans = 68%

b) The score is below Z = -1.0?

C) The score is below Z = 1.5? so want To find the area below 1.5, cdf.Norm function will be used to calculate this Probability bc this gives the left sided cumulative probability, cdf function Always gives the left sided area, not the right sided area. So using spss (CDF.NORMAL( x, 0,1) = 0.93 so = 1-0.93 = 0.07 or 7%.

SPSS -   Transform -> Compute -> CDF.NORMAL   function

D) the score is above z=1.5?

E) the score is above z=-2.08?

F) the score is in between z=0.4 and z=1.36? Trying to find area between 0.4 ad 1.36- so first find the 0.4 left sided Area using cdf function, then find the 1.36 left sided area using cdf function Then find the difference between the two. Ans = 0.25, because CDF.NORMAL for 1.36 – that for 0.4 = 0.91 – 0.66 = 0.25

g) What is  P(-1.0 < Z < 1.5)?

Example contd.

a)What is the z-value such that the probability of selection a score less Than or equal to z approximately 0.30  (or 30%)? So we are given the Area this time and have to figure out the point, we know the probability.

ans using spss IDF.NORMAL (prb, 0, 1) = -0.52

SPSS -   Transform -> Compute -> IDF.NORMAL   function I is for “Inverse” so using the inverse of the cdf

 b)  What is the z-value such that the probability of selection a score Greater than or equal to z (therefore a right sided area) approximately 0.65  (or 65%)? So must first take 1-0.65 And then do the same thing. So use 0.35


The total cholesterol value of a certain Target population is normally distributed with a mean of 200 (mg/100 ml) and a Standard deviation of 20 (mg/100 ml).

So no longer have z sigma is 20 u is 200, so Center is 200

·What is the probability that a person randomly selected from this population Will have cholesterol value between 180 and 220? Ans  = 68% because it is one Standard deviation on either side of the mean, one sd = 20

Between 160 and 240? Two sd either side, So ans =n

More than 200? 50% bc median is middle Value which is 200

Less than 230? Use cdf function and get number of everything to the left uysing 200 as mean 20 and s.D. And it is 1.5 sd away so 93 %

·Where is the median cholesterol value? For a normal distribution the median is equal to the mean - 200

·What is the 75th percentile of cholesterol values? IDF.NORMAL(0.75,200,20) = 213.49

Exercise 5.17:  People are Classified as hypertensive if their systolic blood pressure (SBP) is higher than a specified level for their age Group.

Suppose for the 1-14 yr age group, SBP Of subjects are normally distributed with mean = 105 (so center is 105) and Sd=5 and the specified hypertension level is 115. For the 15-44 yr age group, SBP of subjects are normally distributed with mean = 125 and sd=10 and the specified hypertension level is 140.

A) What proportion of 1-14 yr olds are hypertensive?  0.023 using Cdf function which gave left sided area of 0.977, then got o.023

b) What proportion of 15-44 yr olds are Hypertensive? 0.067 did the same calculation, there was a different normal distribution, Did the same calculation using cdf function.

The central limit theorem

Central Limit Theorem: If the sample size is Reasonably large, then the sampling distribution of the sample mean will be approximately a normal distribution.

So population does not have to look like a Normal distribution.

The mean of this sampling distribution will be the same as the population mean m

The standard deviation of the Sampling distribution will be

Example that we did in Chapter 5:

The total cholesterol value of a certain Target population is normally distributed with a mean of 200 (mg/100 ml) and a Standard deviation of 20 (mg/100 ml).

a)What is the probability that a person randomly selected from this Population will have cholesterol value between 195 and 205?

b)We take a random sample of n=30 From this target population. What is the probability that the average Cholesterol value of this sample will be between 195 and 205?  difference Now is that instead of for one single person, we are asking for a sample of 30 Ppl

Then the average will be the same as the population Average

So can still use a normal distribution with mean of 200, the mean will be the same but the s.D. Will become lower ( ) so if the original s.D was 20, then divide by sqrt or 30 to Get the sample sd


Suppose that a disease is Inherited via a dominant mode of Inheritance and that only one of the two parents is affected with the Disease.  The implications of this mode Of inheritance are that the probability is 1 in 2 that any particular offspring Will get the disease.

3.30What is the probability that in a family with two children, both Siblings are affected?

There is a 0.5 chance of being Affected. Therefore

25%: Since each child has a 0.5 chance of being affected, 0.5X0.5=0.25

3.31What is the probability that exactly one sibling is affected?

50%: Each child has a 0.5 Chance of inheriting the disease.

For exactly one child having The disease, the probability will be 0.25.

3.32What is the probability that neither sibling is affected?

25%:  Each child has a remaining 0.5 chance of NOT Being affected, so 0.5X0.5=0.25

These questions are from the Figure 3.14

The E4 allele of the gene encoding apolipoprotein E (APOE) is strongly associated With Alzheimer’s disease, but it’s value in making the diagnosis remains Uncertain.  A study was conducted among 2188 patients who were evaluated at autopsy for Alzheimer’s disease by Previously established criteria.  Patients were also evaluated clinically for the presence of Alzheimer’s Disease.  The data in Table 3.14 were Presented.

Suppose The pathological diagnosis is considered the gold standard for Alzheimer’s Disease.

3.93If the clinical diagnosis is considered a screening test for Alzheimer’s disease, then what is the sensitivity of the test?

1643/ 1643+127 = 92.8%

3.94What is the specificity of the test?

228/ 228+190 =54.5%

To possibly improve on the Diagnostic accuracy of the clinical diagnosis for Alzheimer’s disease, Information on both the APOE genotype as well as the clinical diagnosis were Considered.  The data are presented in Table 3.15.

Suppose we consider the Combination of both a clinical diagnosis for Alzheimer’s disease and the presence of greater than or Equal to 1 e4 allele as a screening test for Alzheimer’s disease.

Following questions refer to Table 3.15

3.95What is the sensitivity of this test?

1076/1770 = 60.8%

What Is the specificity of this test? ***

67+161+124/418 = 84.2%

Refer to table 3.18 for last Set of questions and table.

3.127    If an Abnormal ECG as determined by S-T segment depression is regarded as the gold Standard for the presence heart disease and an AAI <1.0 is regarded as a Possible test criterion for heart disease, then what is the sensitivity of the Test?

            20/33= 0.606

3.128    What is The specificity of the test?

            318/318+95 = 0.769

3.129    What is The PV+? (hint: assume the subjects in this study are a random sample from a General population in Japan).


Homework #3

Infectious Disease

Newborns were Screened for human immunodeficiency virus (HIV) or acquired immunodeficiency Syndrome (AIDS) in five Massachusetts hospitals. The data are shown in Table 4.14.

1.If 500 newborns are screened at The inner-city hospital, then what is the exact binomial probability of 5 HIV-positive tests?

If 500 newborns are screened at the inner-city Hospital, then for an exact binomial probability of 5 HIV-positive tests is 16%.

2.If 500 newborns are screened at The inner-city hospital, then what is the exact binomial probability of at least 5 HIV-positive tests?

SPSS entered: CDF.BINOM( 4, 500, 0.08)

1 - 0.63 = 0.37 that 5 or more will be HIV positive.

Cancer, Epidemiology

An experiment is Designed to test the potency of a drug on 20 rats. Previous animal studies have Shown that a 10-mg dose of the drug is lethal 5% of the time within the first Four hours; of the animals alive at 4 hours, 10% will die within the next 4 Hours.

3.What is the probability that 3 or More rats will die in the first 4 hours?

Using SPSS : CDF.BINOM(2, 20, 0.05)

Result: 0.92

1 – 0.92 = 0.08

4.Suppose 2 rats die in the first 4 Hours. What is the probability that 2 or fewer rats will die in the next 4 Hours?

Used SPSS: CDF.BINOM( 2, 18, 0.1)

There is a 73% probability that 2 rats will die in the first 4 hours

5.What is the probability that 0 Rats will die in the 8 hour period?

P(surviving 8 hours) = % 1st 4 hours x % 4 Hours later

                                             (.95) x (.90) = .855

SPSS entered: PDF.BINOM (20,20,0.855)

Thus, the Probability of 20 rats surviving in the 8 hour period is 4%.

6.What is the probability that 1 Rat will die in the 8 hour period?

P(surviving 8 hours) = % 1st 4 hours x % 4 hours later

                                             (.95) x (.90) = .855

SPSS entered: PDF.BINOM (19,20,0.855)

Thus, the Probability of 19 rats surviving in an 8 hour period is 15%.

7.What is the probability that 2 rats Will die in the 8 hour period?

P(surviving 8 hours) = % 1st 4 hours x % 4 hours later

                                             (.95) x (.90) = .855

SPSS entered: PDF.BINOM (18,20,0.855)

Thus, the Probability that 18 rats will survive in an 8 hour period is 24%.

Infectious Disease

A study considered risk factors for HIV Infection among intravenous drug users. It was found that 40% of users that had ≤ 100 injections per month (light users) and 55% of users that had > 100 Injections per month (heavy users) were HIV positive.

8.What is the probability that Exactly 3 of 5 light users are HIV positive?

In SPSS entered: PDF.BINOM(3,5,0.4)

The probability that 3 of 5 lighter users will be HIV positive is 23%

9.What is the probability that at Least 3 of 5 light users are HIV positive?

In SPSS entered: CDF.BINOM (2,5,0.4)

The probability that 2 or fewer are HIV positive is 0.68

So, for at least 3 out of 5 = 1- 0.68 = 0.32

10.Is the distribution of the number Of HIV positive among the 20 users binomial? Why or why not?

No the Distribution is not binomial because though the data fulfills the two main Criteria for a binomial distribution, i.E. The data has binary setting, and the Subjects are independent, the probability of success varies among subjects.


Because serum cholesterol is related to age and sex, Some investigators prefer to express it in terms of z-scores.  If X=raw serum cholesterol, the Z = (X-µ)/(σ), where µ is the mean and σ is the standard deviation of serum Cholesterol for a given age-sex group. Suppose Z is regarded as a standard Normal random variable.

5.1) What is Pr(Z<0.5)?

CDF.NORMAL (0.5, 0, 1)

Answer: 0.6915

5.2) What is Pr(Z>0.5)?

1-0.69 = 0.31

Answer: 0.31

5.3) What is Pr(-1.0 < Z < 1.5)?

CDF.NORMAL (1.5, 0, 1) – CDF.NORMAL (-1. 0, 1)

Answer: 0.7745

Suppose a person is regarded as having high Cholesterol if Z>2.0 and borderline cholesterol if 1.5 < Z < 2.0?

5.4) What proportion of people have high cholesterol?

1-CDF.NORMAL (2, 0, 1)

1-.98 = 0.02

Answer: 0.0228

5.5) What proportion of people have borderline Cholesterol?

CDF.NORMAL (2, 0, 1) – CDF.NORMAL (1.5, 0, 1)

Answer: 0.0441

People are classified as hypertensive if their Systolic blood pressure (SBP) is higher than a specified level for their age Group according to the algorithm in Table 5.1.  Assume SBP is normally distributed with mean and standard deviation Given in Table 5.1 for age groups 1-14 and 15-44, respectively.  Define a family As a group of two people in age group 1-14 and two people in age group 15-44. A Family is classified as hypertensive if at least one adult and at least one Child are hypertensive.

5.17) What proportion of 1- to 14-year olds are Hypertensive? 

CDF.NORMAL (115, 105, 5)

Answer: 1-0.98 = 0.023

5.18) What proportion of 15- to 44-year olds are Hypertensive?

CDF.NORMAL (114, 125, 10)

Answer: 1-0.93 = 0.07 (0.067)

In pharmacologic research a variety of clinical Chemistry measurements are routinely monitored closely for evidence of side Effects of the medication under study.  Suppose typical blood-glucose levels are normally distributed, with mean = 90 mg/dL and standard deviation = 38 mg/dL.

5.31) If the normal range of 65-120 mg/dL, then what Percentage of values will fall in the normal range?

CDF.NORMAL (120, 90, 38) – CDF.NORMAL (65, 90, 38)

Answer: .53 = 53%

5.32) In some studies only values at least 1.5 times As high as the upper limit of normal are identified as abnormal.  What percentage of values would fall in this Range?

1- CDF.NORMAL (180, 90, 38)

Answer: 0.00893 = 0.893%

5.33) Answer Problem 5.32 for values 2.0 times the Upper limit of normal.

1-CDF.NORMAL (240, 90, 38)

Answer: 0.00004= 0.004%

Extra problems: In this same context (5.31-5.33) Answer the following:

a) What is the 80th percentile of Blood-glucose levels?

IDF.NORMAL (.8, 90, 38)

Answer: 121.98161

 b) Suppose we consider a random sample Of n= 30 subjects and measure their sample average blood-glucose level. What is The probability that the sample average blood-glucose level will fall in Between 80mg/dL and 100 mg/dL?

SD = 38/sqrt (30) = 6.94

CDF.NORMAL (100, 90, 6.94  ) – CDF.NORMAL (80, 90,  6.94)

Answer: 0.85

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