Download 7. Biostat Lecture 4a

Bio-Statistic
KUEU 3146 & KBEB 3153
Bio-Statistic
Estimation:
Binomial, Poisson, Normal & Student’s t Distribution
Binomial Distribution
• Two outcomes i.e Yes/No, Head/Tail,
Live/Die, success/fail etc
• Independent event
• Defined by 2 parameters: n,p
• Mean, μ=np
• Standard deviation σ=√(np(1-p)).
• Binomial Probability
P(X=x)=
()
n
x
px(1-p)n-x
Binomial Distribution
According to the Vital Statistics of United States,
there is roughly 80% chance that a person aged 20
will be alive at age 65. Suppose that three people
aged at 20 are selected at random. Find the
probability that the number alive at age 65 will be
•Exacly two person alive. (Ans: P=0.384)
•At least one person alive. (Ans: P=0.104)
•At most one person alive. (Ans: P=0.992)
Poisson Distribution
•
•
•
•
Defined by a parameters: λ
Mean, μ=λ
Standard deviation σ=√ λ.
Binomial Probability
P(X=x)= e-λ
λx
x!
Where e = 2.718
and λ is a positive
real number
Poisson Distribution
Some reports indicate that the number of patients
arriving at the hospital between 6.00pm and 7.00pm
has a poisson distribution with a parameter λ = 6.9.
Determine the probability that, on a given day the
number of patients who arrive at the hospital
between 6.00pm and 7.00 pm will be
•Exacly two patients. (Ans: P=0.095)
•Between 4 and 10 patients. (Ans: P=0.821)
•At most two patients. (Ans: P=0.032)
Poisson Approximation to Binomial
• Determine n (the number of trails) and p
(the success probability)
• Determine whether n ≥ 100 and np ≤ 10. If
they are not, do not use the poisson
approximation.
• Use binomial Probability
Where e = 2.718
P(X=x)=
e-np
(np)x
x!
and λ is a positive
real number
Poisson Approximation to Binomial
Statistical Abstract of the United State reported that
the infant mortality rate (IMR) in Sweden is 3.5 per
1000 live birth. Estimate the probability that 500
randomly selected live birth, there are
•No infant deaths. (Ans: P=0.174)
•At most three infant deaths. (Ans: P=0.899)
Normal Distribution
(Gaussian Distribution)
Properties of Normal Distribution
• It has symmetrical bell-shape curve.
• Defined by 2 parameters: mean (μ) and
standard deviation (σ).
• Has a particular internal distribution for the
area under curve. Example, 68.26% of the
area is contained within μ±1σ, 95.45%
within μ±2σ and 99.74% within μ±3σ.
Normal Distribution
(Gaussian Distribution)
Properties of Normal Distribution
• It has symmetrical bell-shape curve.
• Defined by 2 parameters: mean (μ) and
standard deviation (σ).
• Has a particular internal distribution for the
area under curve. Example, 68.26% of the
area is contained within μ±1σ, 95.45%
within μ±2σ and 99.74% within μ±3σ.
Normal Distriburion Curve
Symmetrical Frequency Distribution
99.74%
95.45%
68.26%
μ-3σ
μ-2σ
μ-1σ
μ
μ+1σ
μ+2σ
μ+3σ
Area Under Normal Curve
x-μ
Z=
σ
34.13%
13.13%
2.15%
-3
-2
-1
0
1
2
3
Area Under Normal Curve
Z Score
x-μ
Z=
σ
0.4332
0.5
-3
-2
-1
0
1
1.5 2
What is the proportion of students having SAT
Maths score between 500 and 650? Mean SAT
Maths score is 500 with standard deviation of
100.
3
Area Under Normal Curve
Z Score
x-μ
Z=
σ
0.3849 + 0.4332 = 0.8181
0.4332
0.3849
-3
-2 -1.2 -1
0
1
1.5 2
What is the proportion of students having SAT
Maths score between 380 and 650? Mean SAT
Maths score is 500 with standard deviation of
100.
3
Area Under Normal Curve
Z Score
Determine the Z score having an area of 0.025
and 0.04 under the standard normal curve as
shown below.
0.05
0.025
Z=?
0
Z=?
Student’s t Distribution
Distribution of sample means
x-μ
t=
s/√n
z=
x-μ
σ/√n
df = n - 1
Z score formula for
population distribution is
different from this one