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Biomathematics 2

Probability, random variables.

Continuous random variable. Normal, standard normal

distribution.

Dr. Beáta Bugyi

associate professor

University of Pécs, Medical School

Department of Biophysics

2020

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CONTINUOUS RANDOM VARIABLE continuous: uncountable, infinite number of values, arises from measurement

Probability – discrete/continuous random variables

Let’s consider that a statistical experiment has an outcome corresponding to

A) a discrete random variable and X = 0 – 10 (finite number of outcomes: 10)

Give the probability that the outcome is 6.

𝑃(𝑋 = 6) =1

10= 0.1

B) a continuous random variable and X = 0 – 10 (infinite number of outcomes)

Give the probability that the outcome is 6. Exactly 6, not 6.1, 6.01, …, 6.00000000001

𝑃(𝑋 = 6) =1

∞= 0

NORMAL DISTRIBUTION

𝑁(𝜇, 𝜎), 𝜇 = 𝑚𝑒𝑎𝑛, 𝜎 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛

Probability density function (PDF)

𝑓(𝑥) =1

√2𝜋𝜎2exp (−

(𝑥 − 𝜇)2

2𝜎2 )

Cumulative density function (CDF)

𝐹(𝑥) = ∫1

√2𝜋𝜎2exp (−

(𝑥 − 𝜇)2

2𝜎2 )𝑥

−∞

Graphical representation of the PDF and CDF of normal distributions.

The normal distribution is defined by its mean (𝜇) and standard deviation (𝜎).

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%. 𝑵(𝝁, 𝝈) = 𝑵(𝟔𝟎, 𝟏𝟎). Represent graphically the following

probabilities.

Q1.1: What is the probability that a student scores 60%? 𝑃(𝑋 = 𝑥 = 60) = ?

Q1.2: What is the probability that a student scores less than 60%? 𝑃(𝑋 < 𝑥 = 60) =?

Q1.3: What is the probability that a student scores more than 60%? 𝑃(𝑋 > 𝑥 = 60) = ?

Q1.4: What is the probability that a student scores less than 80%? 𝑃(𝑋 < 𝑥 = 80) = ?

Q1.5: What is the probability that a student scores between 60% and 80%? 𝑃(𝑥 = 60 < 𝑋 < 𝑥 =

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%. 𝑵(𝝁, 𝝈) = 𝑵(𝟔𝟐, 𝟖).

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; ∞ normal distribution 1 standardized distribution

(standard normal distribution)

How to standardize normal distributions?

𝑁(𝜇, 𝜎)

z score: 𝒛 =𝒙−𝝁

𝝈

z score: how many standard deviations (𝜎) is a given value (𝑥) from the mean (𝜇)

STANDARD NORMAL DISTRIBUTION

𝑆𝑁(0, 1), 𝜇 = 1, 𝜎 = 0

Probability density function (PDF)

𝑓(𝑥) =1

√2𝜋𝜎2exp (−

(𝑥−𝜇)2

2𝜎2 ) , 𝑤ℎ𝑒𝑟𝑒 𝜇 = 0 𝑎𝑛𝑑 𝜎 = 1: 𝑓(𝑥) =1

√2𝜋exp (−

𝑥2

2),

Cumulative density function (CDF)

𝐹(𝑥) = ∫1

√2𝜋exp (−

𝑥2

2)

𝑥

−∞

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%. 𝑵(𝝁, 𝝈) = 𝑵(𝟔𝟎, 𝟏𝟎). Standardize the normal distribution. Give the

probabilities by using the Z table.

Q1.1: What is the probability that a student scores 60%? 𝑃(𝑋 = 𝑥 = 60) = ?

𝑃(𝑋 = 𝑥 = 60) = 0

Q1.2: What is the probability that a student scores less than 60%? 𝑃(𝑋 < 𝑥 = 60) =?

𝑧 =𝑥 − 𝜇

𝜎=

60 − 60

10= 0.00

𝑃(𝑋 < 𝑥 = 60) = 0.5 → 50 %

Q1.3: What is the probability that a student scores more than 60%? 𝑃(𝑋 > 𝑥 = 60) = ?

𝑃(𝑋 > 𝑥 = 60) + 𝑃(𝑋 < 𝑥 = 60) = 1

𝑃(𝑋 > 𝑥 = 60) = 1 − 𝑃(𝑋 < 𝑥 = 60) = 1 − 0.5 = 0.5 → 50 %

Q1.4: What is the probability that a student scores less than 80%? 𝑃(𝑋 < 𝑥 = 80) = ?

𝑧 =𝑥 − 𝜇

𝜎=

80 − 60

10= 2.00

𝑃(𝑋 < 𝑥 = 80) = 0.9772 → 97.72 %

Q1.5: What is the probability that a student scores between 60% and 80%? 𝑃(𝑥 = 60 < 𝑋 < 𝑥 =

80) = ?

𝑃(𝑋 < 80) − 𝑃(𝑋 < 60) = 0.9772 − 0.5 = 0.4772 → 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%. 𝑵(𝝁, 𝝈) = 𝑵(𝟔𝟐, 𝟖). Give the probabilities by using the Z table.

Q2.1: What is the probability that a student scores less than 65%? 𝑃(𝑋 < 𝑥 = 65) =?

𝑧 =𝑥 − 𝜇

𝜎=

65 − 62

8= + 0.375

If a value is not listed in the table, use the following approximation:

+ 0.375 =0.37 + 0.38

2

𝑃(𝑋 < 𝑥 = 65) =0.6443 + 0.6480

2= 0.6462 → 64.62 %

Q2.2: What is the probability that a student scores less than 45%? 𝑃(𝑋 < 𝑥 = 45) =?

𝑧 =𝑥 − 𝜇

𝜎=

45 − 62

8= −2.125

If a value is not listed in the table, use the following approximation:

−2.125 =−2.12 + (−2.13)

2

𝑃(𝑋 < 𝑥 = 45) =0.0170 + 0.0166

2= 0.0168 → 1.68 %

Q2.3: What is the probability that a student scores between 45% and 65%? 𝑃(𝑥 = 45 < 𝑋 < 𝑥 = 65) =

?

𝑃(𝑥 = 45 < 𝑋 < 𝑥 = 65) = 𝑃(𝑋 < 𝑥 = 65) − 𝑃(𝑋 < 𝑥 = 45) = 0.6462 − 0.0168 = 0.6294

→ 62.94 %

Q2.4: What is the median of the students’ scores? 𝑃(𝑋 < 𝑥) = 0.5, 𝑥 = ?

𝑃(𝑋 < 𝑥) = 0.5 → 𝑧 = 0.00

𝑧 =𝑥 − 𝜇

𝜎→ 0.00 =

𝑥 − 62

8→ 𝑥 = 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 students’ scores? 𝑃(𝑋 < 𝑥) = 0.25, 𝑥 = ?

𝑃(𝑋 < 𝑥) = 0.25 → 𝑧 = −0.675

𝑧 =𝑥 − 𝜇

𝜎→ −0.675 =

𝑥 − 62

8→ 𝑥 = 56.6

Q2.6: What is the third quartile of the students’ scores? 𝑃(𝑋 < 𝑥) = 0.75, 𝑥 = ?

𝑃(𝑋 < 𝑥) = 0.75 → 𝑧 = 0.675

𝑧 =𝑥 − 𝜇

𝜎→ 0.675 =

𝑥 − 62

8→ 𝑥 = 67.4

Q2.7: Find what percentage of data is between mean ± 1×standard deviation, mean ± 2×standard

deviation, mean ± 3×standard 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.

𝑋 = 𝑙𝑖𝑓𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑎𝑛𝑐𝑦 𝑜𝑓 𝑎 𝑝𝑒𝑟𝑠𝑜𝑛 𝑖𝑛 𝑎 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 (𝑦𝑒𝑎𝑟𝑠)

𝑋 = 𝑥𝑝𝑒𝑟𝑠𝑜𝑛1, 𝑥𝑝𝑒𝑟𝑠𝑜𝑛2, …

We choose samples from the population, each of the samples consists of n persons and by

finding the average lifetime in each sample (�̅�, sample mean) we obtain the distribution of �̅�.

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.

�̅� = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑖𝑓𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑎𝑛𝑐𝑦 𝑜𝑓 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 𝑖𝑛 𝑎 𝑠𝑎𝑚𝑝𝑙𝑒 (𝑦𝑒𝑎𝑟𝑠)

�̅� = �̅�𝑠𝑎𝑚𝑝𝑙𝑒1, �̅�𝑠𝑎𝑚𝑝𝑙𝑒2, …

Properties of the distribution of the sample means

𝜇�̅� = 𝜇𝑋

𝜎�̅� =𝜎𝑋

√𝑛 (standard error of the mean, SEM)

Characteristics of the distribution: Central limit theorem (CLT)

POPULATION SAMPLE

𝑋 = 𝑥

life expectancy of a person in a

population

�̅� = �̅�

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 (𝑛 ≥ 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: 𝑁𝑋(72, 18.2). What is the distribution of �̅� 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

�̅� if n = 10 or n = 40?

n = 10 not known/not normal, n = 40 approximated by normal

Q3.3: What is the mean of �̅� and standard deviation of �̅� (standard error of the mean) if n = 40?

𝜇�̅� = 𝜇𝑋 = 72

𝜎�̅� =𝜎𝑋

√𝑛=

18.2

√40= 2.88

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𝑁�̅�(72, 2.88)

Q3.4: Find 𝑃(𝑋 < 𝑥 = 70) and 𝑃(�̅� < �̅� = 70)?

𝑃(𝑋 < 𝑥 = 70): What is the probability that the life expectancy of a person in the population

is less than 70 years?

𝑁𝑋(72, 18.2)

𝑧 =𝑥 − 𝜇

𝜎=

70 − 72

18.2= −0.109

𝑃(𝑋 < 𝑥 = 70) = 0.4247 → 42.47 %

𝑃(�̅� < �̅� = 70): What is the probability that the average life expectancy of persons in a sample

is less than 70 years?

𝑁�̅�(72, 2.88)

𝑧 =𝑥 − 𝜇

𝜎=

�̅� − 𝜇

𝜎�̅�=

�̅� − 𝜇𝜎𝑋

√𝑛

=70 − 72

2.88= −0.7

𝑃(�̅� < �̅� = 70) = 0.2420 → 24.2 %