1. 08/05/2025 1
Arba-Minch University
College of Medicine and Health sciences
School of Public Health
PROBABILITY AND PROBABILITY DISTRIBUTIONS
By: Etenesh K. (BSc, MPH( Epidemiology & Biostatistics))
2. Learning objectives
At the end of this chapter, the student will be able to:
• Define Probability and related terms
• Categories of Probability
• Properties of probability and Basic Probability Rules
• Discuss Probability distribution of different variables
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3. Why Probability in Statistics?
Nothing in life is certain(uncertainty is high) i.e. we gauge the chances
of successful outcomes.
A probability provides a quantitative description of the chances or
likelihoods associated with various outcomes.
Probability theory was developed out of attempting to solve problems
related to games of chance such as tossing a coin, rolling a die, etc.
i.e. trying to quantify personal beliefs regarding degrees of uncertainty.
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4. Why Probability in Medicine?
Because medicine is an inexact science, physicians seldom
predict an outcome with absolute certainty.
E.g., to formulate a diagnosis, a physician must rely on available
diagnostic information about a patient
History and physical examination
Laboratory investigation, X-ray findings, ECG, etc
4
5. cont..
Although no test result is absolutely accurate, it does affect
the probability of the presence or absence of a disease.
Sensitivity and specificity
An understanding of probability is fundamental for
quantifying the uncertainty that is inherent in the decision-
making process.
5
6. Cont..
• Probability theory is a foundation for statistical inference
Probability theory allows us to draw conclusions about a
population of patients based on known information of sample
patients drawn from that population.
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7. Probability
The chance that an uncertain event will occur/ chance of
occurrence.
Likelihood of an event
Assumes a random process i.e. the outcome is not predetermined
- there is an element of chance
e.g. If a patient taking a certain drug, what is the chance of cure?
Probability can also be defined as the number of times in which
that event occurs in a very large number of trials.
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8. Cont..
Outcome
A specific result of a single trial of a probability experiment
Experiment
any process with an uncertain outcome
• When an experiment is performed, one and only one
outcome is obtained
• An experiment is a trial and all possible outcomes are
events.
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9. Cont..
Event
Something that may happen or not when the experiment is
performed.
• An event is any set of outcomes of interest
Sample Space
is the collection of all possible outcomes
e.g.All 6 faces of a die
P(S) = 1
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10. Cont..
Two Categories of Probability
1.Objective Probability
A. Classical probability
B. Relative frequency probability.
2. Subjective Probabilities
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12. Con..
Example: If we toss a die, what is the probability of coming up (4)?
– E = 1 (which is 4) and S = 6
– The probability of 4 coming up is 1/6
• There are 2 possible outcomes {H, T} in the set of all possible trials
of Tossing of coin
P(H) = 0.5
P(T) = 0.5
SUM = 1.0
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13. Relative Frequency Probability
• Relative frequency probability: The probability that
something occurs is the proportion of times it occurs
when exactly the same experiment is repeated a very
large (preferably infinite!) number of times in
independent trials.
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15. Cont..
• E.g. Suppose that of 158 people who attended a dinner party
99 were ill due to food poisoning.
– The probability of illness for a person selected at random is
Pr(illness) = 99/158 = 0.63 or 63%.
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16. Subjective probability
• Subjective probability: measures the confidence or a wish
that a particular individual has in the truth of a particular
proposition.
• It also known as educated guess
E.g. If some one says that he is 95% certain that a cure for
AIDS will be discovered within 5 years, then he means that
Pr(discovery of cure of AIDS within 5 years) = 95%.
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17. Unions and Intersections
• Intersections of Two Events
• “If A and B are events, then the intersection of A and B,
denoted by AnB , represents the event composed of all basic
outcomes in A and B.”
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18. Cont..
• Unions of Two Events
• The union of A or B, A U B, is the event that either A
happens or B happens or they both happen simultaneously
– P ( A or B ) = P ( A U B )
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19. Mutually exclusive
• Two events A and B are mutually exclusive if they have no
elements in common.
• If A and B are outcomes of an experiment they cannot both
happen at the same time.
• Thus, if A and B are mutually exclusive events, Pr(A or B) = Pr
(A) + Pr(B).
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21. Basic Probability Rules
1. Addition rule
If events A and B are mutually exclusive:
P(A or B) = P(A) + P(B)
P(A and B) = 0
More generally:
P(A or B) = P(A) + P(B) - P(A and B)
P (event A or event B occurs or they both occur)
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23. Con…
Example: One die is rolled. Sample space = S = (1,2,3,4,5,6)
• Let A = the event an odd number turns up,A = (1,3,5)
• Let B = the event 1,2 or 3 turns up; B = (1,2,3 )
• Let C = the event 2 turns up, C= (2)
Find 1. Pr (A), Pr (B) and Pr (C)
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24. Cont..
Answer
Pr(A) = Pr(1) + Pr(3) + Pr(5) = 1/6+1/6+ 1/6 = 3/6 =
1/2
Pr(B) = Pr(1) + pr(2) + Pr(3) = 1/6+1/6+1/6 = 3/6 = ½
Pr (C) = Pr(2) = 1/6
2.AreA and B;A and C; B and C mutually exclusive?
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25. Cont..
A and B are not mutually exclusive. Because they have the
elements 1 and 3 in common.
Similarly, B and C are not mutually exclusive.They have the
element 2 in common.
A and C are mutually exclusive.They don’t have any
element in common
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26. Example: The probabilities below represent years of
schooling completed by mothers of newborn infants
26
27. cont..
What is the probability that a mother has
completed < 12 years of schooling?
P( 8 years) = 0.056 and
P(9-11 years) = 0.159
Since these two events are mutually exclusive,
P( 8 or 9-11) = P( 8 U 9-11)
= P( 8) + P(9-11)
= 0.056+0.159
= 0.215
27
28. cont..
What is the probability that a mother has completed
12 or more years of schooling?
P(12) = P(12 or 13-15 or 16)
= P(12 U 13-15 U 16)
= P(12)+P(13-15)+P(16)
= 0.321+0.218+0.230
= 0.769
28
29. Con…
2. Multiplication rule
– If A and B are independent events,
– Then P(A ∩ B) = P(A) × P(B)
• More generally (both independent & dependent),
– P(A ∩ B) = P(A) P(B|A) = P(B) P(A|B)
– P (A and B) denotes the probability that A and B both occur
at the same time.
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29
30. Example
• In tossing two coins, what is the probability that a head will
occur both on the first coin and the second coin?
– Since there is independence between events
– Then P(A ∩ B) = P(A) × P(B)
= ½ x ½ = ¼
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31. Independent Events
• Two events A and B are independent events if the fact thatA
occurs does not affect the probability of B occurring or
• The outcome of one event has no effect on the occurrence or
non-occurrence of the other
• Thus, if events A & B are independent, Pr(B/A) =
P(B) and Pr(A/B) = P(A).
• Then, P(A B) = P(A) x P(B)
∩
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32. Cont..
Example: n tosses of a coin and the chances that on each
toss it lands heads.
These are independent events.
The chance of heads on any one toss is independent of the
number of previous heads.
No matter how many heads have already been observed,
the chance of heads on the next toss is ½.
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33. Dependent Events
• When the outcome or occurrence of the first event affects the
outcome or occurrence of the second event in such a way that
the probability is changed, the events are said to be dependent
events.
For example:
– E1 = Rain forecasted on the news
– E2 =Take umbrella to work
• Probability of the second event is affected by the occurrence of
the first event.
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34. Con…
When two events are dependent, the probability of both
occurring is
– P(A and B) = P(A). P(B/A). From this
• P(B/A)= P(A and B)
P(A) this is called conditional probability
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35. Conditional Probability
• Refers to the probability of an event, given that another event is
known to have occurred.
– “What happened first is assumed”
• The conditional probability that event B has occurred given that event
A has already occurred is denoted P(B|A) and is defined
• Provided that P(A) > 0
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36. Example:
• A study investigating the effect of prolonged exposure to bright light
on retina damage in premature infants:
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Light exposure
Retinopathy
Total
Yes No
Bright light 18 3 21
Reduced light 21 18 39
Total 39 21 60
37. Con…
• The probability of developing retinopathy is:
P (Retinopathy) = No. of infants with retinopathy
Total No. of infants
= (18+21)/(21+39)
= 0.65
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38. Con…
• The conditional probability of retinopathy, given exposure to bright
light, is:
• P(Retinopathy/exposure to bright light) =
No. of infants with retinopathy exposed to bright light
No. of infants exposed to bright light
= 18/21 = 0.86 OR
• P (R/BL) = P( R and BL) = 18/60 = 0.86
P(BL) 21/60
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39. Con…
• P(Retinopathy/exposure to reduced light) =
# of infants with retinopathy exposed to reduced light
No. of infants exposed to reduced light
= 21/39 = 0.54
• The conditional probabilities suggest that premature infants exposed to
bright light have a higher risk of retinopathy than premature infants
exposed to reduced light.
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40. Exercise
• Culture and Gonodectin (GD) test results for 240 Urethral
discharge specimens
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GD test result
Culture result
Total
Gonorrhea No Gonorrhea
Positive 175 9 184
Negative 8 48 56
Total 183 57 240
41. Cont..
A. What is the probability that a man has gonorrhea?
B. What is the probability that a man has a positive GD test and have the disease
Gonorrhea?
C. What is the probability that a man has a negative GD test and does not have
gonorrhea
D. What is the probability that a man has the disease (Gonorrhea) given the test reads
positive
E. What is the probability that a man has not the disease (Gonorrhea) given the test
reads negative
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42. Con…
• What is the probability that a man has gonorrhea?
– P (gonorrhea) = No. of persons with gonorrhea
Total No. of sample persons
= 183/240
= 0.76
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43. Con…
• What is the probability that a man has a positive GD test and have the
disease Gonorrhea?
– P (test positive | they have the disease)
– P (gonorrhea) = No. of persons with true positives for test
Total No. of persons with Disease
Gonorrhea)
= 175/183
= 0.96 * 100% = 96%
– N.B:True positives = Positive test result and the disease
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44. Con…
• What is the probability that a man has a negative GD test and does not have
gonorrhea
– P(test negative | they don’t have the disease)
P (-ve test No gonorrhea) =
∣ No. of persons who are true negatives for test
Total No. of persons without Gonorrhea
= 48/57
= 0.84 = 84%
N.B:True negatives = Negative test result and don’t have disease
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45. Con…
• What is the probability that a man has the disease (Gonorrhea)
given the test reads positive
• P (gonorrhea the test reads positive) =
∣
No. of persons with true positives for test
Total No. of persons with positive test result
= 175/184
= 0.95; 95%
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46. Con…
• What is the probability that a man has not the disease
(Gonorrhea) given the test reads negative
• P (No gonorrhea the test reads negative) =
∣
No. of persons with true negatives for test
Total No. of persons with Negative test result
= 48/56
= 0.86; 86%
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47. Properties of probability
1. Probabilities are real numbers on the interval from 0 to 1.
2. If two events are mutually exclusive ,then Pr(A or B) = Pr(A) + Pr(B).
3. If A and B are two events, not mutually exclusive , then P(AuB) = Pr (A) +Pr
(B) – Pr( A and B).
4. The sum of the probabilities that an event will occur and that it will not occur
is equal to 1; hence, P(A’) = 1 – P(A)
5. If A and B are two independent events, then Pr ( A and B) = Pr (A).Pr(B)
• This means that P(AnB)=P(A).P(B)
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48. Probability distribution
• The term probability distribution: refers to the collection of all
possible outcomes along with their probabilities.
• Every random variable has a corresponding probability distribution.
• A random variable is a variable whose values are determined by
chance.
• Random variables can be categorical, discrete or continuous.
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49. Cont..
• A discrete random variable is able to assume only a finite or
countable number of outcomes.
• A continuous random variable can take on any value in a specified
interval.
• Continuous random variables can assume an infinite number of
values and can be decimal and fractional values.
• With numeric variables, the aim is to determine whether or not
normality may be assumed.
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50. Cont..
• For categorical variables, we obtain the frequency
distribution of each variable.
• A categorical distribution is a discrete probability distribution
that describes the probability that a random variable will take
on a value that belongs to one of K categories, where each
category has a probability associated with it.
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51. Probability distribution of a categorical
variables
• For a distribution to be classified as a categorical distribution,
it must meet the following criteria:
The categories are discrete.
• There are two or more potential categories.
• The probability that the random variable takes on a value in
each category must be between 0 and 1.
• The sum of the probabilities for all categories must sum to 1.
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52. • The most obvious example of a categorical distribution is the
distribution of outcomes associated with rolling a dice.There
are K = 6 potential outcomes and the probability for each
outcome is 1/6:E.g.
Value on Face 1 2 3 4 5 6
Probability 1/6 1/6 1/6 1/6 1/6 1/6
• Notice that the total probability is 1.
52
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53. Discrete Probability Distributions
• A discrete random variable is a variable that can assume only
a countable number of values.
• Many possible outcomes:
– No. of patients tested for HIV
– No. of patients attending a health facility per day
• Only two possible outcomes:
– Yes or no responses, positive or negative test result
– Gender: male or female
53
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54. Cont..
• For a discrete random variable, the probability distribution specifies
each of the possible outcomes of the random variable along with the
probability that each will occur
• Examples can be:
– Frequency distribution
– Relative frequency distribution
– Cumulative frequency distribution
54
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55. Example
• The following data shows the number of diagnostic services a
patient receives
• The sum of all the individual probabilities is 1
55
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56. Cont..
• What is the probability that a patient receives exactly 3 diagnostic
services?
P(X=3) = 0.031
• What is the probability that a patient receives at most one diagnostic
service or less than or equals to one?
P (X≤1) = P(X = 0) + P(X = 1)
= 0.671 + 0.229
= 0.900
56
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57. Cont..
• What is the probability that a patient receives at least four
diagnostic services?
P (X≥4) = P(X = 4) + P(X = 5)
= 0.010 + 0.006
= 0.016
57
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58. Cont..
Properties of probability distribution of discrete random variable
0 ≤ P(X = x) ≤ 1
∑ P(X = x) = 1
Examples of probability distributions for discrete random variables:
– Binomial distribution
– Poisson distribution
58
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59. Binomial distribution
It is one of the most widely encountered discrete probability
distributions.
The binomial distribution is the probability distribution that
results from doing a “binomial experiment”.
A process that has only two possible outcomes is called a
binomial process. In statistics the two outcomes are frequently
denoted as success and failure.
The probabilities of a success or a failure are denoted by p and
q, respectively. Note that p + q = 1. 59
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60. Cont..
Binomial assumptions:
1)The same experiment is carried out n times ( n trials are made).
2) Each trial has two possible outcomes ( usually these outcomes are called
“ success” and “ failure”. If P is the probability of success in one trial,
then 1-p is the probability of failure.
3)The result of each trial is independent of the result of any other trial.
4) The mean and variance of the distribution are given by:
= np, 2 = npq, =
μ σ σ
60
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61. • If the binomial assumptions are satisfied and an experiment is
repeated n times and the outcome is independent from one trial to
another, the probability that outcome X occurs exactly x times is:
Pr(X=x) = n! p x
(1- p)n-x
x!(n-x)!
– X! = x (x-1) (x-2) …….. ( 1)
– Note: 0! =1 (by definition)
– n : denotes the number of fixed trials
– x : denotes the number of successes in the n trials
– p : denotes the probability of success
– q : denotes the probability of failure (1- p)
61
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62. Example
1: Suppose that in a certain population 52% of all recorded births
are males.
If we select randomly 10 birth records What is the probability
that exactly.
5 will be males?
Pr(X=5) = 10! 0.52 5
(1- 0.52)10-5
=0.24
5!(10-5)!
3 or more will be males?
Pr(X≥3) = 1- Pr(X<3) = Pr(X=0)+Pr(X=1)+Pr(X=2)
=1-[0.001+0.013+0.111]= 1-0.125=0.875
62
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63. Cont..
2.Suppose we know that 40% of a certain population are
cigarette smokers. If we take a random sample of 10 people
from this population, what is the probability that we will have
exactly 4 smokers in our sample?
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65. Cont..
3. 70% of a certain population has been immunized for polio. If a
sample of size 50 is taken, what is the “expected total number”,
in the sample who have been immunized?
– µ = np = 50(.70) = 35
• This tells us that “on the average” we expect to see 35
immunized subjects in a sample of 50 from this population.
65
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66. Binomial distribution, generally
X
n
X
n
X
p
p
)
1
(
1-p = probability of
failure
p = probability of
success
X = #
successes out
of n trials
n = number of trials
Note the general pattern emerging if you have only two possible
outcomes (call them 1/0 or yes/no or success/failure) in n independent
trials, then the probability of exactly X “successes”=
66
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67. Poisson distribution
Used to model the number of occurrences of an event that
takes place infrequently in time or space.
Applicable for counts of events over a given interval of
time
Example:
1. number of patients arriving at an emergency department
in a day
2. number of new cases of HIV diagnosed at a clinic in a
month
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68. Cont..
Three assumptions must be met for a Poisson distribution to
apply:
1. The probability that a single event occurs within a given small
subinterval is proportional to the length of the subinterval
2. The rate at which the event occurs is constant over the entire
interval
3. Events occurring in consecutive subintervals are independent
of each other
4. The mean and variance of a Poisson distributed variable are
given by
µ=σ2 = np.
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69. Cont..
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• When the probability of “success” is very small, then px
and
(1 –p)n – x
become too small to calculate exactly by the
binomial distribution.
In such cases, the Poisson distribution becomes useful.
Let λ be the expected number of successes in a process
consisting of n trials,
70. Cont..
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• The probability of observing X successes is
Where:
The symbol e is the constant equal to 2.7183.
λ (Lambda) is the rate at which the event occurs, or
the expected number of events per unit time).
X is a potential outcome of X
71. Example
1.Suppose x is a random variable representing the number of individuals
involved in a road accident each year in USA , which is 2.4 per
10,000 population each year. Given, λ = 2.4
Pr (X=0) = 2.40
e-2.4
= 0.091
0!
2.Pr (X=1) = 2.41
e-2.4
= 0.218
1!
3.Pr (X=2) = 2.42
e-2.4
= 0.262
2!
4.Pr (X>2) = 1- (Pr (X=0) +Pr (X=1) +(X=2)) =0.429
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72. Probability distribution of continuous
variables
• A continuous random variable: any value that can assume
any value in a specified interval or range.
• Under different circumstances, the outcome of a random
variable may not be limited to categories or counts.
• Let us consider the data of grouped frequency distribution
presented in Table under descriptive Statistics and its
histogram.
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73. Cont..
• Smooth curves are used to represent graphically the
distributions of continuous random variables.
• This has some important consequences when we deal with
probability distributions.
• Instead of assigning probabilities to specific outcomes of the
random variable X, probabilities are assigned to ranges of
values.
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76. Cont..
– A continuous probability distribution describes how likely
it is that a continuous random variable takes values
within certain ranges.
• The probability associated with any one particular value is
equal to 0
– Therefore, P(X=x) = 0
– Also, P(X ≥ x) = P(X > x)
– the total area under the curve is equal to one,
76
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77. Cont..
• Probability distributions for a continuous random variable
differ from discrete distributions in several ways:
– An event can take on any value within the range of the
random variable and not just integers
– The probability of any specific value is zero
– Probabilities are expressed in terms of an area under a
curve (probability is measured area under the
curve)
77
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78. Cont..
• Thus, the probability that X will assume some value in the
interval enclosed by two ranges say x1 and x2
• As a continuous variable can take an infinite number of
values, it helps to visualize the probability distribution as a
curve and probabilities as ‘area under the curve’
• It is also called normal distribution
78
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79. Normal Distribution
• The ND is the most important probability distribution in
statistics.
• Frequently called the “Gaussian distribution” or bell-shape
curve.
• A normal distribution is a continuous, symmetric, bell-shaped
distribution of a variable
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81. Cont..
• The ND is vital to statistical work, most estimation
procedures and hypothesis tests underlie ND.
• The concept of “probability of X=x” in the discrete
probability distribution is replaced by the “probability
density function f(x).
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83. Properties of the Normal Distribution
It is a probability distribution of a continuous variable. It extends from
minus infinity( -∞) to plus infinity (+∞).
It is uni-modal, bell-shaped and symmetrical about its mean
The mean, the median and mode are almost equal.
The total area under the curve about the x-axis is 1 square unit
The distribution is completely determined by the parameters and .
μ σ
The curve never touches the x-axis. i.e. It is asymptotic
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84. Cont..
1. The mean μ tells you about location -
– Increase μ - Location shifts right
– Decrease μ – Location shifts left
– Shape is unchanged
2. The standard deviation tells you about narrowness or flatness of the bell
– Increase standard deviation - Bell flattens
• Extreme values are more likely
– Decrease standard deviation - Bell narrows
• Extreme values are less likely
– Location is unchanged
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86. Cont..
Every random variable normally distribute with its mean and SD,
Since a normal distribution could be an infinite number of
possible values for its mean and SD, it is impossible to tabulate
the area associated for each and every normal curve.
Instead only a single curve for which μ = 0 and σ = 1 is
tabulated.
The curve is called the standard normal distribution (SND).
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87. Standard normal distribution.
• It is a normal distribution that has a mean equal to 0 and a
SD equal to 1, and is denoted by N(0, 1).
• The main idea is to standardize all the data that is given by
using Z-scores.
• These Z-scores can then be used to find the area (and thus
the probability) under the normal curve.
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89. Z - Transformation
• If a random variable X~N(µ, ) then we can transform it to a SND with
σ
the help of Z transformation
• Translate from x to the standard normal (the “z” distribution) by
subtracting the mean of x and dividing by its standard deviation:
• Z score tells you how many standard deviations away from the
mean an individual value (x) lies
08/05/2025 89
90. Cont..
• This process is known as standardization and gives the
position on a normal curve with µ=0 and δ=1, i.e., the SND, Z
• A Z-score is the number of standard deviations that a given x
value is above or below the mean (µ)
08/05/2025 90
91. Cont..
E.g. If x is normally distributed with mean of 100 and standard deviation
of 50, the z value for x = 250 is:
• This says that x = 250 is three standard deviation (3 increments of 50
units) above the mean of 100
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92. Z-score/Z-value
Z-scores are important because given a Z – value we can find
out the probability of obtaining a score larger or lower than that
Z value. ( look up the value in a z-table).
To look up the probability of obtaining a Z-value larger than a
given value, look up the first two digits of the Z-score in the left
hand column and then read the hundredths place across the top.
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94. Example
• Scores on exam are normally distributed with a mean
of 65 and standard deviation of 9. find the probability
of the scores
• A. less than 54
• B. at least 80
• C. between 70 and 86
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95. Solution
Given
• Mean= 65 SD= 9
A. P(X < 54)=??
= P(Z < (54-65)/9) = P(Z < -1.2222)
• Since the Z table is set up to handle only two decimal place we
round this to -1.22. then, we go to the Z table and look up Z=-
1.22. Look for -1.2 in the first column and 0.02 at the top
• There fore, the probability that X is less than 54 is the
probability that Z is less than -1.22 which gives 0.1112 or
11.12% of the score below 54.
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96. Cont..
B. at least 80 = P(X > 80)=??
= P(Z >(80-65)/9) = P(Z >1.67)
= 0.0475
C. between 70 and 86= p(70<x<86)??
= P(Z>(70-65)/9)= P(z>0.56)=0.2877,
from this, (P(z<0.56)= 1-P(z>0.56)=0.7123
= P(Z<(86-65)/9)= P(z<2.33)=0.0099
Then, 1-(P(z<0.56)+P(z>2.33))=0.2778
• Therefore, the probability that exam score lie between 70 and 86
is equal to 0.2778 Or 27.8% of the score were between 70 and 86
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97. Note that –Empirically proven
• For a normal distribution with mean and standard deviation
• About 68% (most) of the observations is contained within one
SD of the mean.
• About 95% (majority) of the probability is contained within
two SDs
• And 99% (almost all) within three SDs of the mean.
97
101. Exercise
1. Suppose that total carbohydrate intake in 12-14 year old
males is normally distributed with mean 124 g/1000 cal and
SD 20 g/1000 cal.
a) What percent of boys in this age range have carbohydrate
intake above 140g/1000 cal?
b) What percent of boys in this age range have carbohydrate
intake below 90g/1000 cal?
101
![Example
1: Suppose that in a certain population 52% of all recorded births
are males.
If we select randomly 10 birth records What is the probability
that exactly.
5 will be males?
Pr(X=5) = 10! 0.52 5
(1- 0.52)10-5
=0.24
5!(10-5)!
3 or more will be males?
Pr(X≥3) = 1- Pr(X<3) = Pr(X=0)+Pr(X=1)+Pr(X=2)
=1-[0.001+0.013+0.111]= 1-0.125=0.875
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