1. Continuous Random Variable Normal Distribution
Unimodal
Symmetrical
Asymptotic to X-Axis X
Curve is Uniquely Determined by two Parameters: µ and σ

2. Normal Curves for Different Means and Standard Deviations20 30 40 50 60 70 80 90
100
110
120

3. Standardized Normal Distribution
A normal distribution with
a mean of zero, and
a standard deviation of one
Z Formula
standardizes any normal distribution
Z Score
the number of standard deviations which a value is away from the mean s = 1

4. We can find the equivalent point on the Standardized Normal Curve
for a point on any other normal Curve. s = ? s = 1 X Z

5. Z Table Second Decimal Place in Z

6. P( 0 ≤ Z ≤ 1 ) =
P( -1 ≤ Z ≤ 1 ) =
P( 0 ≤ Z ≤ 2 ) =
P( -2 ≤ Z ≤ 2 ) =

7. P( Z ≥ 1 ) =
P( Z ≤ ) =
P( 1.25 ≤ Z ≤ 2.50) =
P( ≤ Z ≤ 2.66) =
Find C such that Probability is:
P( Z ≤ C ) = .80

8. s = 10 Areas for a Normal Curve with µ = 80 σ = 10 Z P( X ≥ 90 ) =

9. P( 65 ≤ X ≤ 90 ) =
Find C such that P( X ≥ C ) = .15

10. Ex µ = $951 σ = $96
P( X ≥ $1000 ) =
P( 900 ≤ X ≤ 1100 ) =

11. P( 825 ≤ X ≤ 925 ) =
P( X < 700 ) =

12. Ex 6.12 µ = 200 σ = 47 Find C given this Probability
P( X > C ) = .60
P( X < C ) = .83

13. P( X < C ) = .22
P( X < C ) = .55

14. Normal Approximation to Binomial Distribution Probabilities
Necessary Condition: n ≥ 100 and n•p ≥ 5
We use a Normal Curve which has
the same Mean and Std Dev as
the Binomial Distribution.
Continuity Correction – We will assign all real values for a unit
Interval to the single discrete binomial outcome.
i.e. X = 40  ≤ X ≤ 40.5
Ex: 100 Coin Tosses – n = 100 p= .5 q = .5

15. P100(X=40) ≈ P( 39.5 ≤ X ≤ 40.5)
P100(40≤X≤60) ≈ P( 39.5 ≤ X ≤ 60.5)

16. Ex: 180 Die Rolls – S = 1 Spot n = 180 p = 1/6 q = 5/6
P180(X=25) ≈ P( 24.5 ≤ X ≤ 25.5)
P180(X≤25) ≈ P(X ≤ 25.5)

17. Ex: n = 150 p = .75
P150(X < 105) ≈
P150(110≤X≤120) ≈

18. P150(X > 95) ≈