1. Continuous Random Variable
So how do we determine 𝑓(𝑥), the probability density function of 𝑋?
Choose pdf from a class of continuous distributions
Selection depends on the nature of population of interest
Certain types of data have known distributions
Otherwise, make an assumption or educated guess in choosing the most appropriate one
In this class, it should always be clear what the distribution of something is.

2. Normal Distribution The mean, median, and mode are all located
Most prevalent and arguably the most important class of distributions
Bell curve
Symmetric about 𝜇
PDF: 𝑓 𝑥 = 1 𝜎 2𝜋 𝑒 − 𝑥−𝜇 2 /(2 𝜎 2 )
𝑥= value of 𝑋
𝜇=𝐸(𝑋) (population mean of 𝑋)
𝜎 2 =𝑉𝑎𝑟(𝑋) (population variance of 𝑋)  𝜎= 𝑉𝑎𝑟(𝑋) (population standard deviation of 𝑋)
𝜋= … (pi)
𝑒= … (natural number)
The mean, median, and mode are all located
at the same place.

3. Normal Distribution
Most prevalent and arguably the most important class of distributions
Bell curve
Symmetric about 𝜇
PDF: 𝑓 𝑥 = 1 𝜎 2𝜋 𝑒 − 𝑥−𝜇 2 /(2 𝜎 2 )
𝑥= value of 𝑋
𝜇=𝐸(𝑋) (population mean of 𝑋)
𝜎 2 =𝑉𝑎𝑟(𝑋) (population variance of 𝑋)  𝜎= 𝑉𝑎𝑟(𝑋) (population standard deviation of 𝑋)
𝜋= … (pi)
𝑒= … (natural number)
𝑃 𝑋<𝑎 = −∞ 𝑎 1 𝜎 2𝜋 𝑒 − 𝑥−𝜇 2 /(2 𝜎 2 ) 𝑑𝑥 = ? a
Draw this on the board.
Interpretation of the standard deviation for normal distribution: rule
𝑃(𝜇−𝜎≤𝑥≤𝜇+𝜎)≈0.6827
𝑃(𝜇−2𝜎≤𝑥≤𝜇+2𝜎)≈0.9545
𝑃(𝜇−3𝜎≤𝑥≤𝜇+3𝜎)≈0.9973
That is, about 68% of observations lie within one SD of the mean, about 95% of observations lie within two SDs of the mean, and about 99.7% of observations lie within three SDs of the mean.

4. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2

5. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ? c

6. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up! c

7. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
𝑃 𝑍≥1 =0.1587
0.1587
1.00

8. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1
1.00

9. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥1
1.00

10. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥ =1−0.1587=0.8413
0.8413
1.00

11. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥ =1−0.1587=0.8413
𝑃 𝑍≤−1 =𝑃 𝑍≥1
Because the standard normal distribution is symmetric (about 0), the following
is identity is true: 𝑃 𝑍≥𝑐 =𝑃(𝑍≤−𝑐).
-1.00

12. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥ =1−0.1587=0.8413
𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587
0.1587
Because the standard normal distribution is symmetric (about 0), the following
is identity is true: 𝑃 𝑍≥𝑐 =𝑃(𝑍≤−𝑐).
-1.00

13. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
Key idea is to state the probability as 𝑃 𝑍≥𝑐
for some value positive 𝑐 so that we can use the table.
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥ =1−0.1587=0.8413
𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587
𝑃 −1≤𝑍≤0.50
Remember that: 𝑃 𝑎≤𝑋≤𝑏 =𝑃 𝑋≥𝑎 −𝑃(𝑋≥𝑏)
-1.00
0.50

14. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
Key idea is to state the probability as 𝑃 𝑍≥𝑐
for some value positive 𝑐 so that we can use the table.
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥ =1−0.1587=0.8413
𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587
𝑃 −1≤𝑍≤0.50 =𝑃 𝑍≥−1 −𝑃 𝑍≥0.50
Remember that: 𝑃 𝑎≤𝑋≤𝑏 =𝑃 𝑋≥𝑎 −𝑃(𝑋≥𝑏)
-1.00
0.50

15. Standard Normal Distribution
Random variable 𝑍 has a standard normal distribution if
𝑍 is normally distributed
𝐸 𝑍 =𝜇=0
𝑉𝑎𝑟 𝑍 = 𝜎 2 =1 → 𝜎=1
PDF: 𝑓 𝑧 = 1 2𝜋 𝑒 − 𝑧 2 /2
𝑃 𝑍≥𝑐 = 𝑐 ∞ 𝜋 𝑒 − 𝑧 2 /2 𝑑𝑧= ?
Still can’t do this by hand, so you have to use a table to look it up!
Key idea is to state the probability as 𝑃 𝑍≥𝑐
for some value positive 𝑐 so that we can use the table.
𝑃 𝑍≥1 =0.1587
𝑃 𝑍≤1 =𝑃 𝑍<1 =1−𝑃 𝑍≥ =1−0.1587=0.8413
𝑃 𝑍≤−1 =𝑃 𝑍≥1 =0.1587
𝑃 −1≤𝑍≤0.50 =𝑃 𝑍≥−1 −𝑃 𝑍≥0.50
=𝑃 𝑍≤1 −𝑃 𝑍≥0.50 = 1−𝑃 𝑍≥1 −𝑃 𝑍≥0.50 = 1− −0.3085=0.8413−0.3085=0.5328
0.5328
Remember that: 𝑃 𝑎≤𝑋≤𝑏 =𝑃 𝑋≥𝑎 −𝑃(𝑋≥𝑏)
-1.00
0.50

16. Standard Normal Distribution
Suppose that we have the probability that something occurs p, and need to know the value of c beyond which the probability of occurrence is p.
That is, 𝑃 𝑍≥𝑐 =𝑝 → 𝑐= ?
If we assume that the probability distribution of Z follows a standard normal distribution, then this is simple to find using a table! 𝑝 c
Now, let’s say we want to find the value of c, such that the probability of a value greater than c is equal to p.

17. Standard Normal Distribution
Suppose the situation is like that given in the picture.
What do we see?
c is greater than 0.
Then, we find in the table, the value of c that gives 𝑝≈ in the “Area beyond z in tail.”
0.3085
𝑐= 𝑧 𝑝 c
𝑃 𝑍≥𝑐 = → 𝑐= 𝑧 =0.50

18. Standard Normal Distribution
Suppose the situation is like that given in the picture.
What do we see?
c is greater than 0.
Then, we find in the table, the value of c that gives 𝑝≈ in the “Area beyond z in tail.”
0.3085
𝑐= 𝑧 𝑝
c = 0.50
𝑃 𝑍≥𝑐 = → 𝑐= 𝑧 =0.50

19. Standard Normal Distribution
Suppose the situation is like that given in the picture.
What do we see?
c is greater than 0.
We want to find c, such that 𝑃 𝑍≤𝑐 =𝑝
Then, we first convert to finding the value of c in 𝑃 𝑍≥𝑐 =1−𝑝. that gives 1−𝑝≈ in the “Area beyond z in tail.”
0.8413 c
𝑃 𝑍≤𝑐 =1−𝑃 𝑍≥𝑐 = → 𝑃 𝑍≥𝑐 =1−0.8413= → 𝑐= 𝑧 =1.00

20. Standard Normal Distribution
Suppose the situation is like that given in the picture.
What do we see?
c is less than 0.
We want to find c, such that 𝑃 𝑍≤𝑐 =𝑝
Then, we first convert to finding c in 𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 =𝑝 that gives 𝑝≈ in the “Area beyond z in tail.”
0.1587 c
𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 = → −𝑐= 𝑧 =1.00→𝑐=−1.00

21. Standard Normal Distribution
Suppose the situation is like that given in the picture.
What do we see?
c is less than 0.
We want to find c, such that 𝑃 𝑍≤𝑐 =𝑝
Then, we first convert to finding c in 𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 =𝑝 that gives 𝑝≈ in the “Area beyond z in tail.”
0.1587
c = -1.00
𝑃 𝑍≤𝑐 =𝑃 𝑍≥−𝑐 = → −𝑐= 𝑧 =1.00→𝑐=−1.00