The normal distribution is a widely used probability distribution that is bell-shaped and symmetric around the mean. It is characterized by two parameters: the mean (μ) and the standard deviation (σ). The standard normal distribution has a mean of 0 and standard deviation of 1. Any normal distribution can be converted to a standard normal distribution using standardization, allowing probabilities to be easily calculated using standard normal probability tables. These tables provide the probability that a standard normal random variable Z will fall below various z-values.

1. http://www.youtube.com/watch?v=NYd6wz
YkQIM&feature=relmfu
This is the most important and most
widely used probability distribution
1

2. The Normal Distribution
• Properties
– Bell shaped
1
– Area under curve equals 1
– Symmetric around the mean μ
– Mean = median = Mode
– Two tails approach the horizontal axis – never
touch axis
– Empirical rule applies
– Two parameters – μ and σ 2

3. How does the standard deviation affect the shape of the distribution
s=2
s =3
s =4
m = 11
How does the mean affect the location of the distribution
s=2
m = 10 m = 11 m = 12
3

4. The Normal Distribution
Mathematical model expressed as:
1  xm 
2
1  
2 s 
f ( x)  e ,   x  

2s
where   3.14159 and e  2.71828
- notation N(μ ; σ2)
4

5. STANDARDISING THE
RANDOM VARIABLE
• Seen how different means and std dev’s
generate different normal distributions
• This means a very large number of
probability tables would be needed to
provide all possible probabilities
• We therefore standardise the random
variable x so that only one set of tables is
needed
5

6. STANDARDISING THE
RANDOM VARIABLE
• A normal random variable x can be
converted to a standard normal variable
(denoted Z) by using the following
standardisation formula:-
xm
z , x any value of the random variable X
s
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7. The Standard Normal Distribution
• Different values of μ and σ generate
different normal distributions
• The random variable X can be
standardised
– mean = μ = 0
– standard deviation = σ = 1
xm
z , x any value of the random variable X
s 7

8. The Standard Normal Distribution
μ=0
s=1
z
-3 -2 -1 0 1 2 3
z-values on the horizontal axis
• distance between the mean and the point
represented by z in terms of standard deviation 8

9. Finding Normal Probabilities
• Example
– Marks for a semester test is normally
distributed, with a mean of 60
and a standard deviation of 8
– X ~ N(60 , 82)
– If we need to determine the
probability that the mark will x
50 60 65
be between 50 and 65,
we need to determine the size of the shaded area
– Before calculating the probabilities the x-values
need to be transformed to z-values 9

10. P(-∞ < Z < z)
Standard normal probabilities have been
calculated and are provided in a table
The tabulated probabilities
correspond to the area between
0 z
Z = -∞ and some z > 0
z 0.00 0.01 → 0.05 0.06 → 0.09
0.0 0.5000 0.5040 0.5199 0.5239 0.5359
0.1 0.5398 0.5438 0.5596 0.5636 0.5753
↓
1.0 0.8413 0.8438 0.8531 0.8554 0.8621
1.1 0.8643 0.8665 0.8749 0.8770 0.8830
1.2 0.8849 0.8869 0.8944 0.8962 0.9015
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↓

11. • Example continue
– If X denotes the test mark, we seek the
probability
– P(50 < X < 65)
– Transform the X to the standard normal
variable Z
Every normal variable X m Therefore, once
with some m and s, Z probabilities for Z are
can be transformed s calculated, probabilities
into this Z E(Z) V(Z) of any normal variable
11
μ=0 σ2 = 1 can be found

12. • Example continue Mean = μ = 60 minutes
Standard deviation = σ = 8 minutes
X m
50 -- 60 < Z < 65 -- 60 )
X m
P(50 < X < 65) = P(
8s 8s
= P(-1.25 < Z < 0.63) X-m
Z=
s
To complete the calculation we need to compute
the probability under the standard normal distribution12

13. • How to use the z-table to calculate
probabilities Example
Determine the following probability: P(Z > 1.05) = ?
1 - P(Z < 1.05)
= P(Z > 1.05)
0 1.05
P(Z > 1.05) = 1 – 0.8531 = 0.1469
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14. • How to use the z-table to calculate probabilities
Example
Determine the following probability: P(-2.12 < Z < 1.32) = ?
0.9066
P(-∞ < Z < 1.32) = 0.9066
0,5
1.32
P(-2.12 < Z < +∞) = 0.9830
0.9830
0,5
-2.12
+ -2.12 0 1.32
P(-2.12 < Z < 1.32) = (0.9066 + 0.9830) - 1 = 0.8896 14

15. • Example
– Determine the following probabilities:
P(0.73 < Z < 1.40) = ?
P(-∞ < Z < 1.40) = 0.9192
0.9192
1.40
P(-∞ < Z < 0.73) = 0.7673
0.7673 0 0.73 1.40
0.73
P(0.73 < Z < 1.40) = 0.9192 – 0.7673 = 0.1519 15

16. The symmetry of the normal distribution
makes it possible to calculate probabilities
for negative values of Z using the table as
follows:
P(-z < Z < +∞) = P(-∞ < Z < z) 16

17. • Example student marks - continued
P(50 < X < 65) = P(-1.25 < Z < 0.63) 0.7357
0.8944
= 0.8944 + 0.7357 – 1 0.8944
0.8944
0.8944
0.8944 0.8944
= 0.6301 z = 0.63
In this example z = -1.25 , because of symmetry read 1.25
z 0.00 0.01 → 0.05 0.06 → 0.09
0.0 0.5000 0.5040 0.5199 0.5239 0.5359
0.1 0.5398 0.5438 0.5596 0.5636 0.5753
↓
1.0 0.8413 0.8438 0.8531 0.8554 0.8621
1.1 0.8643 0.8665 0.8749 0.8770 0.8830
1.2 0.8849 0.8869 0.8944 0.8962 0.9015 17