Probability
distributions
By
Dr. Ameer kadhim Hussein.
M.B.Ch.B. FICMS (Community Medicine).
Probability distributions
1.Discrete probability distributions are the binomial
distribution and Poisson distribution.
2. A continuous probability distribution is a probability
density function. The area under the smooth curve is
equal to 1 and the frequency of occurrence of values
between any two points equals the total area under the
curve between the two points and the x-axis .
The normal distribution
The normal distribution is the most important
distribution in biostatistics. It is frequently
called the Gaussian distribution. It is used for
continuous variables .
The two parameters of the normal distribution
are the mean (µ) and the standard deviation
(σ). The graph has a familiar bell-shaped curve.
The normal distribution
Graph of a normal distribution
(Gaussian distribution)
1. It is symmetrical around the mean .
2. The mean, median and mode are all equal.
3. The total area under the curve above the x-axis is 1
square unit. Therefore 50% is to the right of mean and
50% is to the left of mean.
4. Perpendiculars of:
±1 (σ) contain about 68%;
±2 (σ) contain about 95%;
±3 (σ) contain about 99.7%
of the area under the curve.
Relationship between the normal curve and the standard
deviation:
frequency
All normal curves share this property: the SD cuts off a
constant proportion of the distribution of scores:-
68%
95%
99.7%
-3
-2
-1
mean
+1
+2
+3
Number of standard deviations either side of mean
The standard normal distribution
A normal distribution is determined by µ and
σ. The normal distribution creates a family of
distributions depending on whatever the values
of µ and σ are. The most important member
of that family is the standard normal distribution
which has µ =0 and σ =1.
Standard z score
The standard z score is obtained by creating a
variable z whose value is
Given the values of µ and σ we can convert a
value of x to a value of z and find its probability
using the table of normal curve areas.
Finding probabilities
a. What is the probability that z < -1.96?
(1) Sketch a normal curve
(2) Draw a line for z = -1.96
(3) Find the area in the table
(4) The answer is the area to the left of the line P(z < 1.96) = .0250
Finding probabilities
b) What is the probability that z > 1.96?
(1) Sketch a normal curve
(2) Draw a line for z = 1.96
(3) Find the area in the table
(4) The answer is the area to the right of the line;
found by subtracting table value from 1.0000; P(z >
1.96) =1.0000 - .9750 = .0250
What is the probability that (-1.96 <z < 1.96)?
P(-1.96 < z <1.96) = 0.9750 – 0.0250 = 0.95
Example :
If Z is a standard normal distribution, then
P( Z < 2) = 0.9772
is the area to the left to 2 and it equals 0.9772.
2
19
Example:
P(-2.55 < Z < 2.55) is the area
Between -2.55 and 2.55, Then it
Equals P(-2.55 < Z < 2.55)
=0.9946 – 0.0054
= 0.9892.
Example:
P(-2.74 < Z < 1.53) is the area
Between -2.74 and 1.53.
P(-2.74 < Z < 1.53)
=0.9370 – 0.0031
= 0.9339.
-2.55
-2.74
0
2.55
1.53
20
Example :
P(Z > 2.71) is the area to the
Right to 2.71.
So, P(Z > 2.71) =1 – 0.9966 = 0.0034.
2.71
21
Given the following probabilities. Find Z1:
1. P(z ≤ Z1)= 0.0055
2. P(z1≤ z ≤ 2.98)= 0.1117
0.9986-0.1117= 0.8869
z1= -2.54
Z1= 1.21
How to transform normal distribution (X) to
standard normal distribution (Z)?
This is done by the following formula:
z
x

Example:
If X is normally distributed with µ = 3, σ = 2. Find the
value of standard normal Z, If X= 6?
Answer:
x 63
z

 1.5

2
Example
Suppose that systolic blood pressure among teachers is
approximately normally distributed with mean of 140 and
standard deviation of 50. Find the probability that a
teacher picked at random will have a systolic blood
pressure less than 100. We follow the steps to find the
solution.
(1) Write the given information
µ = 140
σ = 50
x = 100
(3) Convert x to a z score
P(X<100) = P(Z<100-140/50) = P(Z< -0.8)
(4) P (z < -0.8) = 0.2119.
(5) Complete the answer:
the probability that a teacher picked at random will
have a systolic blood pressure less than 100 is 0.2119.
Example:
In a study of children ages 8 to 15 years. The
researchers found that the amount of time
children spend in the upright position
followed a normal distribution with mean of
5.4 hours and standard deviation of 1.3.
If a child selected at random ,then
1-The probability that the child spend less than 3
hours in the upright position 24-hour period
P( X < 3) = P(Z <
3  5.4
1.3
) = P(Z < -1.85) = 0.0322
------------------------------------------------------------------------2-The probability that the child spend more than 5
hours in the upright position 24-hour period
P( X > 5) = P(Z >
5  5.4
1.3
) = P(Z > -0.31)
= 1- 0.3783= 0.6217
-----------------------------------------------------------------------
4-The probability that the child spend from 4.5 to
7.3 hours in the upright position 24-hour period
P( 4.5 < X < 7.3) = P(
4.5  5.4
1.3
= P( -0.69 < Z < 1.46 )
= 0.9279 – 0.2451 = 0.6828
< Z<
7.3  5.4
)
1.3
Skewed Data
Data may have a positive skewness (long tail to
the right, or a negative skewness (long tail to the
left).
Kurtosis
Kurtosis indicates data that are bunched together
or spread out.
Data that are bunched together give a tall, thin
distribution which is not normal. This is called
leptokurtic.
Data that are spread out give a low, flat
distribution which is not normal. This is called
platykurtic.
Kurtosis
Thank you