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Biomathematics 2
Probability, random variables.
Continuous random variable. Normal, standard normal
distribution.
Dr. BeΓ‘ta Bugyi
associate professor
University of PΓ©cs, Medical School
Department of Biophysics
2020
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CONTINUOUS RANDOM VARIABLE continuous: uncountable, infinite number of values, arises from measurement
Probability β discrete/continuous random variables
Letβs consider that a statistical experiment has an outcome corresponding to
A) a discrete random variable and X = 0 β 10 (finite number of outcomes: 10)
Give the probability that the outcome is 6.
π(π = 6) =1
10= 0.1
B) a continuous random variable and X = 0 β 10 (infinite number of outcomes)
Give the probability that the outcome is 6. Exactly 6, not 6.1, 6.01, β¦, 6.00000000001
π(π = 6) =1
β= 0
NORMAL DISTRIBUTION
π(π, π), π = ππππ, π = π π‘ππππππ πππ£πππ‘πππ
Probability density function (PDF)
π(π₯) =1
β2ππ2exp (β
(π₯ β π)2
2π2 )
Cumulative density function (CDF)
πΉ(π₯) = β«1
β2ππ2exp (β
(π₯ β π)2
2π2 )π₯
ββ
Graphical representation of the PDF and CDF of normal distributions.
The normal distribution is defined by its mean (π) and standard deviation (π).
The PDF has a characteristic bell shape.
The PDF is symmetric to the mean of the distribution.
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The inflection point of the PDF corresponds to the standard deviation of the distribution.
The width (width at half-maximum) of the PDF is proportional to the standard deviation; the
larger the width the larger the standard deviation.
Probability is given by the area under the PDF (see examples below).
Example 1
The test result of students from Subject 1 follows a normal distribution with a mean of 60% and
standard deviation of 10%. π΅(π, π) = π΅(ππ, ππ). Represent graphically the following
probabilities.
Q1.1: What is the probability that a student scores 60%? π(π = π₯ = 60) = ?
Q1.2: What is the probability that a student scores less than 60%? π(π < π₯ = 60) =?
Q1.3: What is the probability that a student scores more than 60%? π(π > π₯ = 60) = ?
Q1.4: What is the probability that a student scores less than 80%? π(π < π₯ = 80) = ?
Q1.5: What is the probability that a student scores between 60% and 80%? π(π₯ = 60 < π < π₯ =
80) = ?
Example 2
The test result of students from Subject 2 follows a normal distribution with a mean of 62% and
standard deviation of 8%. π΅(π, π) = π΅(ππ, π).
Question:
How can we work with different normal distributions? Do we need the PDF of each and every normal
distribution?
Answer:
Normal distributions can be standardized; β normal distribution 1 standardized distribution
(standard normal distribution)
How to standardize normal distributions?
π(π, π)
z score: π =πβπ
π
z score: how many standard deviations (π) is a given value (π₯) from the mean (π)
STANDARD NORMAL DISTRIBUTION
ππ(0, 1), π = 1, π = 0
Probability density function (PDF)
π(π₯) =1
β2ππ2exp (β
(π₯βπ)2
2π2 ) , π€βπππ π = 0 πππ π = 1: π(π₯) =1
β2πexp (β
π₯2
2),
Cumulative density function (CDF)
πΉ(π₯) = β«1
β2πexp (β
π₯2
2)
π₯
ββ
Graphical representation of the PDF and CDF of the standard normal distribution.
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Z table
summarizes the CDF of the standard normal distribution
Example 1
The test result of students from Subject 1 follows a normal distribution with a mean of 60% and
standard deviation of 10%. π΅(π, π) = π΅(ππ, ππ). Standardize the normal distribution. Give the
probabilities by using the Z table.
Q1.1: What is the probability that a student scores 60%? π(π = π₯ = 60) = ?
π(π = π₯ = 60) = 0
Q1.2: What is the probability that a student scores less than 60%? π(π < π₯ = 60) =?
π§ =π₯ β π
π=
60 β 60
10= 0.00
π(π < π₯ = 60) = 0.5 β 50 %
Q1.3: What is the probability that a student scores more than 60%? π(π > π₯ = 60) = ?
π(π > π₯ = 60) + π(π < π₯ = 60) = 1
π(π > π₯ = 60) = 1 β π(π < π₯ = 60) = 1 β 0.5 = 0.5 β 50 %
Q1.4: What is the probability that a student scores less than 80%? π(π < π₯ = 80) = ?
π§ =π₯ β π
π=
80 β 60
10= 2.00
π(π < π₯ = 80) = 0.9772 β 97.72 %
Q1.5: What is the probability that a student scores between 60% and 80%? π(π₯ = 60 < π < π₯ =
80) = ?
π(π < 80) β π(π < 60) = 0.9772 β 0.5 = 0.4772 β 47.72%
Example 2
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The test result of students from Subject 2 follows a normal distribution with a mean of 62% and
standard deviation of 8%. π΅(π, π) = π΅(ππ, π). Give the probabilities by using the Z table.
Q2.1: What is the probability that a student scores less than 65%? π(π < π₯ = 65) =?
π§ =π₯ β π
π=
65 β 62
8= + 0.375
If a value is not listed in the table, use the following approximation:
+ 0.375 =0.37 + 0.38
2
π(π < π₯ = 65) =0.6443 + 0.6480
2= 0.6462 β 64.62 %
Q2.2: What is the probability that a student scores less than 45%? π(π < π₯ = 45) =?
π§ =π₯ β π
π=
45 β 62
8= β2.125
If a value is not listed in the table, use the following approximation:
β2.125 =β2.12 + (β2.13)
2
π(π < π₯ = 45) =0.0170 + 0.0166
2= 0.0168 β 1.68 %
Q2.3: What is the probability that a student scores between 45% and 65%? π(π₯ = 45 < π < π₯ = 65) =
?
π(π₯ = 45 < π < π₯ = 65) = π(π < π₯ = 65) β π(π < π₯ = 45) = 0.6462 β 0.0168 = 0.6294
β 62.94 %
Q2.4: What is the median of the studentsβ scores? π(π < π₯) = 0.5, π₯ = ?
π(π < π₯) = 0.5 β π§ = 0.00
π§ =π₯ β π
πβ 0.00 =
π₯ β 62
8β π₯ = 62
Note: The mean of a data set following normal distribution is equal to its median.
Q2.5: What is the first quartile of the studentsβ scores? π(π < π₯) = 0.25, π₯ = ?
π(π < π₯) = 0.25 β π§ = β0.675
π§ =π₯ β π
πβ β0.675 =
π₯ β 62
8β π₯ = 56.6
Q2.6: What is the third quartile of the studentsβ scores? π(π < π₯) = 0.75, π₯ = ?
π(π < π₯) = 0.75 β π§ = 0.675
π§ =π₯ β π
πβ 0.675 =
π₯ β 62
8β π₯ = 67.4
Q2.7: Find what percentage of data is between mean Β± 1Γstandard deviation, mean Β± 2Γstandard
deviation, mean Β± 3Γstandard deviation.
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IMPORTANCE OF NORMAL DISTRIBUTION
CENTRAL LIMIT THEOREM
Example 3
In a population of persons let X = life expectancy of a person (in years). The distribution of X
has a mean and standard deviation of 72 and 18.2 years, respectively.
π = ππππ ππ₯ππππ‘ππππ¦ ππ π ππππ ππ ππ π ππππ’πππ‘πππ (π¦ππππ )
π = π₯ππππ ππ1, π₯ππππ ππ2, β¦
We choose samples from the population, each of the samples consists of n persons and by
finding the average lifetime in each sample (οΏ½Μ οΏ½, sample mean) we obtain the distribution of οΏ½Μ οΏ½.
Sampling distribution of sample means: a distribution of the sample means calculated from all
possible random samples of a specific size (n) taken from a population.
οΏ½Μ οΏ½ = ππ£πππππ ππππ ππ₯ππππ‘ππππ¦ ππ ππππ πππ ππ π π πππππ (π¦ππππ )
οΏ½Μ οΏ½ = οΏ½Μ οΏ½π πππππ1, οΏ½Μ οΏ½π πππππ2, β¦
Properties of the distribution of the sample means
ποΏ½Μ οΏ½ = ππ
ποΏ½Μ οΏ½ =ππ
βπ (standard error of the mean, SEM)
Characteristics of the distribution: Central limit theorem (CLT)
POPULATION SAMPLE
π = π₯
life expectancy of a person in a
population
οΏ½Μ οΏ½ = οΏ½Μ οΏ½
average life expectancy of persons in a
sample
normal distribution normal distribution for any n
not normal/not known distribution
CLT: if n is large enough (π β₯ 30)
approximated by normal distribution
the larger n, the better the approximation
http://onlinestatbook.com/stat_sim/sampling_dist/index.html
Q3.1: Consider that X has normal distribution: ππ(72, 18.2). What is the distribution of οΏ½Μ οΏ½ if n
= 10 or n = 40?
n = 10 normal, n = 40 normal
Q3.2: Consider that the distribution of X is not known/not normal. What is the distribution of
οΏ½Μ οΏ½ if n = 10 or n = 40?
n = 10 not known/not normal, n = 40 approximated by normal
Q3.3: What is the mean of οΏ½Μ οΏ½ and standard deviation of οΏ½Μ οΏ½ (standard error of the mean) if n = 40?
ποΏ½Μ οΏ½ = ππ = 72
ποΏ½Μ οΏ½ =ππ
βπ=
18.2
β40= 2.88
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ποΏ½Μ οΏ½(72, 2.88)
Q3.4: Find π(π < π₯ = 70) and π(οΏ½Μ οΏ½ < οΏ½Μ οΏ½ = 70)?
π(π < π₯ = 70): What is the probability that the life expectancy of a person in the population
is less than 70 years?
ππ(72, 18.2)
π§ =π₯ β π
π=
70 β 72
18.2= β0.109
π(π < π₯ = 70) = 0.4247 β 42.47 %
π(οΏ½Μ οΏ½ < οΏ½Μ οΏ½ = 70): What is the probability that the average life expectancy of persons in a sample
is less than 70 years?
ποΏ½Μ οΏ½(72, 2.88)
π§ =π₯ β π
π=
οΏ½Μ οΏ½ β π
ποΏ½Μ οΏ½=
οΏ½Μ οΏ½ β πππ
βπ
=70 β 72
2.88= β0.7
π(οΏ½Μ οΏ½ < οΏ½Μ οΏ½ = 70) = 0.2420 β 24.2 %