2. Learning Outcomes
• Continuous random variable (CRV)
• Examples of random variables
• Probabilities for continuous random variables
– Probability Mass Function and Probability Density Function: PMF vs. PDF
• Expectations and Variances of Continuous Random Variables
• Numerical Example
3. Continuous Random Variable (CRV)
• A continuous random variable is the outcome of a
continuous random experiment
• Theoretically they have infinite values
• They are quantitative and represented by real
(decimal valued) numbers
• Their level of measurement is either interval or
ratio.
6. Probabilities in CRVs
• Let X be a DRV
• E.g. : X={1,2,3,4}
– The events in X are countable and finite
– We can find probabilities of every single event present in X.
– These probabilities of events in X are also called Probability Mass Function (PMF)
of X
• Let X be a CRV
• E.g. 2 ≤ 𝑋 ≤ 4
• Now in actuality there are infinite events present in X.
• Is it possible to find probability of any single event in X ? E.g. P(X=2.67=?)
– The probability of any individual event will be approximately equal to 0.
– Why?
– Not because it is not present in X but rather because we are trying to find
probability of an event that is very small fraction of the total.
• Hence PMF concept cannot be used in CRV.
• But it is important to note that instead of finding probability of just one event if a range
of events’ probability is found then this could be determined.
• This gives the idea and concept of Probability Density Function or PDF.
7. Probability Density Function (PDF)
• Therefore we do not use the idea of finding individual
probability of events in CRV.
• Rather we are more interested in fining the area within
which the desired event can be found.
• Mathematically areas are always determined on
functions
• The function of random variable X that can be used to
find these areas are called Probability Density Function
or PDF
8. Probabilities in CRVs
• PDFs are function of X that has the following
properties
– They show the profile and trend of how the
function varies with changing values of X
– Area of any portion of the PDF will give the
probability density of the range of value of X
– Total area of a PDF follows the law of sum of
probabilities for all mutually exclusive events
from the execution of a random experiment.
9. Probabilities in CRVs
• We find the areas using a mathematical operator called
“integration”. This is identified as follows:
• Let 𝑓 𝑥 is a function of 𝑥 and is the PDF of X. Then integral
of 𝑓 𝑥 provides the PDF of X and is given as;
𝑃𝐷𝐹 =
𝑥=−∞
∞
𝑓 𝑥 𝑑𝑥
Integrations always give us the area under the curve and also
shows the summation as in discrete case given by ∑
Integrations when used as above gives the probability density
or area of X for a certain interval
11. Example 1
• Consider a continuous random variable 𝑋 defined
by the PDF given as:
𝑃𝐷𝐹 = 𝑓 𝑥 =
𝑥 + 1
8
𝑓𝑜𝑟 2 ≤ 𝑋 ≤ 4
0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Determine the following:
a) Prove that 𝑓 𝑥 is actually a PDF
b) 𝑃 𝑋 < 3.5
c) 𝑃 2.7 < 𝑋 < 3.5
d) Predicted outcome of 𝑋
12. Solution
a) In to prove that given function is
actually a PDF , it must satisfy the
following condition
𝑃 2 < 𝑋 < 4 = 1
This is done as follows:
2
4
𝑥 + 1
8
𝑑𝑥 = 1
18. Normally Distributed CRV
Learning Outcomes
• Concept of normally distributed CRV
• Concept of Z
• Finding probabilities using Z
• Finding Z and X using probabilities
20. Normal PDF
𝑀𝑒𝑎𝑛 = 𝜇
Let 𝜇 = 100
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝜎
Let 𝜎 = 20
𝑋
𝑓(𝑥)
21. Normal PDF
𝑋
𝑓(𝑥)
𝜇 = 100
𝜎 = 20
1. What does 𝜎 = 20 mean?
2. How many standard deviations is 𝑋 = 125?
3. How many standard deviations is 𝑋 = 75?
4. How many standard deviations correspond to
75 < 𝑋 = 125?
23. Normal PDF
If I am 125 units away from mean
then in terms of standard deviations what does
it mean?
Solution
100 units on X makes 0 standard deviation
20 units on X from 100 (i.e. 120-100)
makes 1 standard deviation ( on either side of mean)
125 unit on X will make=> 125−100
20
=
1.25 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠
