Download Answers - Yale School of Medicine TBL

10/13/2016
Normal distribution
Statistics in medicine
• Properties:
– Bell shape
– Symmetric about the mean
– Mean=median=mode
– The area under the curve =
1
• 68% within mean + 1SD
• 95% within mead + 2SD
• 99% within mean+ 3SD
Workshop 2: Describing variation, and probabilities
October 6, 2016
10:00 AM to 10:50 AM
Hope 110
Fatma Shebl, MD, MS, MPH, PhD
Assistant Professor
Chronic Disease Epidemiology Department
Yale School of Public Health
[email protected]
SLIDE 0
SLIDE 1
Readiness assessment questions
Readiness assessment questions
Q1: A study of serum triglycerides levels among patients
with history of myocardial infarction revealed that serum
triglycerides is normally distributed with a mean of 200
mg/dL and standard deviation of 10 mg/dL. Within the
study population 68% of the serum triglycerides values will
lie within the limits of?
Q1: A study of serum triglycerides levels among
patients with history of myocardial infarction
revealed that serum triglycerides is normally
distributed with a mean of 200 mg/dL and
standard deviation of 10 mg/dL. Within the study
population 68% of the serum triglycerides values
will lie within the limits of?
A.
B.
C.
D.
180 mg/dL to 210 mg/dL
170 mg/dL to 230 mg/dL
180 mg/dL to 220 mg/dL
190 mg/dL to 210 mg/dL
Since 68% lie within mean + 1SD then the limits will be
200 + 10 = 190 mg/dL to 210 mg/dL
SLIDE 2
Probabilities rules
SLIDE 3
Probabilities rules
• Probability
– The probability (P) of an event A is written as P(A)
–P(A)= Number of times A occurs / Number of
times A can occur
• Product(multiplication) rule
– Used to estimate probability of two or more events being
true
– Depends on the assumption of independence or not
• One probability is not influenced by the outcome of
another probability i.e. P(A+|B+)= P(A+|B-)
• Independence assumed
– P(A and B)= P(A) x P(B)
• Independence NOT assumed (used in Bayes theorem)
– P(A and B)= P(A) x P(B|A) = P(B) x P(A|B)
SLIDE 4
SLIDE 5
1
10/13/2016
Probabilities rules
Probabilities rules
• Conditional probability
–P(B|A) = P(A and B)/ P(A) x P(B)
• Addition rule
– All possible different probabilities in situation must
add to one
– Mutually exclusive events
• P(A or B)= P(A) + P(B)
– Mutually inclusive events (non-mutually exclusive)
• Modified addition rule
• P(A or B or both)= P(A) + P(B) – P(A and B)
SLIDE 6
Readiness assessment questions
SLIDE 7
Readiness assessment questions
Q2-3 : An investigator measured body mass
index(BMI) of fitness club members. He reported
that 15% were under-weight, 30% had healthy
weight, and 15% are overweight, 25% are obese,
and 15% extremely obese. The members were
mostly men (75%).
Q2-3 : An investigator measured body mass index(BMI)
of fitness club members. He reported that 15% were
under-weight, 30% had healthy weight, and 15% are
overweight, 25% are obese, and 15% extremely obese.
The members were mostly men (75%).
Q2: What is the probability that a randomly selected member of
the club members will be healthy or overweight?
Explanation: Because are being healthy and being overweight
are mutually exclusive events therefore we can use the addition
rule for mutually exclusive events
P(A or B)= P(A) + P(B)= 30%+15%=45%
Q2: What is the probability that a randomly
selected member of the club members will be
healthy or overweight?
A. 20%
B. 25%
C. 45%
D. 50%
SLIDE 8
Readiness assessment questions
SLIDE 9
Readiness assessment questions
Q2-3 : An investigator measured body mass
index(BMI) of fitness club members. He reported
that 15% were under-weight, 30% had healthy
weight, and 15% are overweight, 25% are obese,
and 15% extremely obese. The members were
mostly men (75%).
Q2-3 : An investigator measured body mass index(BMI)
of fitness club members. He reported that 15% were
under-weight, 30% had healthy weight, and 15% are
overweight, 25% are obese, and 15% extremely obese.
The members were mostly men (75%).
Q3: The probability that a randomly selected member will be male
and of healthy weight?
Q3: The probability that a randomly selected
member will be male and of healthy weight?
A. .0009
B. 009
C. .09
D. . 225
Explanation: Because the probability of being male is independent
of the weight, therefore we can use the multiplication rule to
calculate the probability of the two independent events.
P(A, and B)= P(A) x P(B)= .75 x .3=.225
S L I D E 10
S L I D E 11
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10/13/2016
Variable: is any characteristic of a subject that can
be measured.
Scales
1. Nominal: Used for characteristics that
have no numerical values, no
measurement scales and no rank order.
2. Ordinal: Used for characteristics that
have an underlying order to their values
3. Interval: It is used for characteristics
that can be given numerical values that
are equally spaced and has no true zero
point.
4. Ratio: It is used for characteristics that
can be given numerical values that are
equally spaced and HAS true zero
point.
Types
1. Qualitative
•Nominal
•Ordinal??
2. Quantitative
•Ordinal
•Interval
•Ratio
Readiness assessment questions
Q4: A patient had undergone a cholecystectomy operation.
Post-operatively, he complained of pain. The attending surgeon
asked him to rate his pain on a scale of 0 (no pain) to 10 (the
worst pain). The patient reported that his pain is a 9. After the
administration of 100mg ketoprofen, the patient reported that
the pain is now a 5 on the same scale. After the administration
of morphine sulfate, given as an intravenous push, the pain is 0.
This pain scale is a:
A.
B.
C.
D.
Nominal scale
Continuous scale
Qualitative scale
Ratio scale
S L I D E 12
Readiness assessment questions
S L I D E 13
Application Questions
Q4: The pain scale is subjective and it might be
arguable that a pain score of 6 is not quit as twice as a
score of 3, therefore sometime pain scale is considered
as ordinal scale. However, the pain scale has a true 0,
indicating the absence of pain, therefore it could be
considered as a ratio scale (i.e., for a continuous
variable with a true 0 point). Although the pain scale
is indeed continuous, it is a special and more specific
case of continuous (i.e., ratio). Dichotomous scales are
binary (only two options) and are a special case of
nominal scales. Neither binary nor nominal applies to
the 0-10 pain scale discussed in this question.
Qualitative measures are completely devoid of
objective scales by definition.
Q1- The table below lists the results of blood type evaluation of
50 men and 50 women?
Q1
Blood type
Gender
Male
Female
Totals
O
20
20
40
A
17
18
35
B
8
7
15
AB
5
5
10
50
50
100
Totals
S L I D E 14
S L I D E 15
Application Questions
Application Questions
Q1- What is the probability that an individual selected at
random from the group has type O blood or is a man?
Q1- What is the probability that an individual selected at
random from the group has type O blood or is a man?
A.
B.
C.
D.
Explanation: Sex and blood group are not mutually
exclusive, therefore we will use the general addition rule
0.2
0.35
0.5
0.7
P(A or B)= P(A) + P(B) – P (A and B)
=(40/100) + (50/100) – (20/100)=(70/100)=0.7
S L I D E 16
S L I D E 17
3
10/13/2016
Standard normal distribution (z)
Standard normal distribution (z)
• Properties:
– Bell shape
– Symmetric about the mean
– Mean=median=mode
– Mean=0
– Standard deviation=1
– The area under the curve = 1
• 68% within µ + 1σ
• 95% within µ + 2σ
• 99% within µ + 3σ
• The normal distribution with mean 0
and standard deviation 1
• If the mean#0 and SD#1do z
transformation  allow using the
standard normal table
𝑥−𝜇
– 𝑧=
, where x is the value of the
𝜎
variable, µ is the mean, σ is the SD
• A positive z means the value is above
the mean
• A negative z means the value is below
the mean
• If the z is known you can get the x
– x= µ + zσ
Graph generated by R
Graph generated by R
S L I D E 18
Application Questions
Standard normal distribution (z) tables
Q2- Data on RBC cholinesterase were collected from a
healthy population. The data revealed that RBC
cholinesterase had a mean of 11 µmol/min/ml and SD of 2
µmol/min/ml. What is the probability that an individual
randomly selected from the population will have an RBC
cholinesterase value between 11.95 and 13.95 µmol/min/ml?
Z score
Area under the curve to the left
i.e. below z
Areas under the
standard normal
curve (z scores)
• Could be used to
find proportion
above ,below , or
between any z
scores
• The first column
includes the
stem of the z
value
• The top row
includes the
second and third
digit of the z
value
S L I D E 19
A.
B.
C.
D.
Negative z
.9306
.6844
.2462
.3156
Positive z
Source: http://image.slidesharecdn.com/copyofz-table-130515110049-phpapp02/95/copy-of-ztable-1 -638.jpg?cb=1368615687
S L I D E 20
Application Questions
Application Questions
Q2- It means we want the area between
11.95 and 13.95
RBC cholinesterase
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
S L I D E 21
18 19 20 21 22
S L I D E 22
• Solution:
– Find the z score
equivalent to 11.95
𝑥−𝜇
𝑧=
= (11.95-11)/2=
𝜎
.95/2=.475=.48
– Find the z score
equivalent to 13.95
𝑥−𝜇
𝑧=
= (13.95-11)/2=
𝜎
2.95/2=1.475=1.48
– Find the probability that
z < 0.48  0.6844
Find the probability that
z < 1.48  0.9306
Then the area between
11.95 and 13.95 =
P (.48 < z < 1.48) =
0.9306-0.6844=0.2462
S L I D E 23
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