# Blood test "solt

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Write your t**Lecture 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**.

**Column**s are
Things which are being measured on the cases. These are called **variables**.

**Examples:**

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 25^{th} percentile (bottom) and
The third quartile or 75^{th} 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 = 25^{th} percentile (bottom of the box) – 1.5*IQR

Upper
Fence = 75^{th} 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**

- For an event A,
**0 £ Prob(A) £ 1 .** - 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) | Positive | Negative | Total |

Yes | 154 | 225 | 379 |

No | 362 | 23,362 | 23,724 |

Total | 516 | 23,587 | 24,103 |

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(T ^{c }|C^{c})**

**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) | Positive | Negative | Total | Positive | Negative | Total | ||

Yes | 45,000 | 5000 | 50,000 | 9,000 | 1,000 | 10,000 | ||

No | 5,000 | 45,000 | 50,000 | 9,000 | 81,000 | 90,000 | ||

Total | 50,000 | 50,000 | 100,000 | 18,000 | 82,000 | 100,000 |

**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 | 0 | 1 | 2 | 3 | 4 | Total |

Prob | 0.008 | 0.076 | 0.265 | 0.411 | 0.240 | 1.00 |

**LECTURE
3**

**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 **s ^{2}** (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**

**Assumptions:**

**Binary Setting****n = 5 Patients. We assume that these 5 patients are independent (as far as Having AE is concerned).****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 | 0 | 1 | 2 | 3 | 4 | Total |

Prob | 0.008 | 0.076 | 0.265 | 0.411 | 0.240 | 1.00 |

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**

Example:

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 75^{th} 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**

** HW#2**

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).

**20/(20+95)=0.173**

**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) = % 1^{st} 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) = % 1 ^{st} 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) = % 1 ^{st} 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.**

**HW#4**

**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 80 ^{th} 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**

ext here!