1. NORMAL DISTRIBUTION Normal distribution- continuous distribution.
Normal density:
bell shaped,
unimodal- single peak at the center, symmetric.
Completely described by its center of symmetry - mean μ and spread - standard deviation σ. μ
Random variable with normal distribution – normal random variable with mean μ and st. dev. σ: X~N(μ, σ)
Standard normal random variable: mean 0 and st. dev. 1: Z~N(0, 1)

2. NORMAL DISTRIBUTION-CHANGING LOCATION AND SCALE
CHANGING SCALE σ μ1 μ2
“peaky” density
“flatter” density
1=σ1 < σ2 = 2
0=μ1 < μ2 =1
Changes in mean/location cause shifts in the density curve along the x-axis.
Changes in spread/standard deviation cause changes in the shape of the density curve.

3. Why Bother with Normal Distributions?
Normal distributions are great descriptions-models-approximations for many data sets such as weights, heights, exam scores, experimental errors, etc.
Great descriptions of results-outcomes of many chance driven experiments.
Statistical inference based on normal distribution works well for many (approximately) symmetric distributions.
HOWEVER, remember that not everything or everybody is normal!

4. THE EMPIRICAL RULE - THE 68-95-99.7 RULE
Areas under the normal curve are probabilities (as under any density curve).
Special areas under the normal density curve:
Approximately 68% of the observations fall within 1 standard deviation of the mean
Approximately 95% of the observations fall within 2 standard deviations of the mean
Approximately 99.7% of the observations fall within 3 standard deviations of the mean

5. THE EMPIRICAL RULE - contd
The range "within one/two/three standard deviation(s) of the mean" is highlighted in green.
The area under the curve over this range is the rel. frequency of observations in the range.
That is, 0.68/95/99.7 = 68%/95%/99.7% of the observations fall within one/two/three standard deviation(s) of the mean, or, 68%/95%/99.7% of the observations are between (μ – 1/2/3σ) and (μ + 1/2/3σ).

6. AREAS UNDER THE NORMAL CURVE
Normal probabilities = areas under the normal curve are tabulated for the standard normal distribution (front cover of your text book).
In looking for probabilities keep in mind:
Symmetry of the normal curve and P(Z=a)=0 for any a.
FIND:
P(Z < 0.01) = 0.504
P(Z ≤ ) = 0.496
P(Z < 0) = 0.5
P( Z < 2.92)=
P(Z>2.92)= = or, by symmetry =P(Z< )=0.0018
P(-1.32< Z <1.2)= – =0.7915
SUMMARY OF RULES we used above: P(Z>a)=1-P(Z< -a)
P(a < Z < b) = P(Z < b)- P( Z < a)

7. GENERAL NORMAL DISTRIBUTION
IF X~N(μ, σ) then Z = ~N(0, 1) standard normal.
standardization
Example. Suppose that the weight of people in NV follows normal distribution with mean 150 and standard deviation 20 lb. Find the probability that a randomly selected Nevadan weighs
at most 160 lb; b) over 160 lb.
Solution. Let X= weight of a randomly selected Nevadan. X~N(150, 20).
P(X ≤ 160) =
P(X>160)= 1 - P(X ≤ 160) =1 – =

8. NORMAL PERCENTILES
Given that P(Z < p)=0.95 find p. Here p is called 95th percentile of Z.
Inside the table I looked for 0.95.
Found and
Used z-value corresponding to
the midpoint (0.95) between the
two available probabilities
p=1.645
If an available probability is closer
to the one we need, use the
z-value corresponding to
that probability.
0.95
p=?

9. NORMAL PERCENTILES, CONTD.
EXAMPLE. Suppose scores X on a test follow a normal distribution with mean 430 and standard deviation 100. Find 90th percentile of the scores, that is find score x such that P(X ≤ x)=0.9.
Solution. Since we start with a normal but NOT STANDARD normal distribution, we have to standardize at some point:
0.9 = P(X ≤ x) =
get equation:
x =128
x = 558
90% of students scored 558 or less.
0.90
z =1.28

10. EXAMPLE
Height of women follows normal distribution with mean 64.5 and standard deviation of 2.5 inches. Find
a) The probability that a woman is shorter than 70 in.
b) The probability that a woman is between 60 and 70 in tall.
c) What is the height 10% of women are shorter than, i.e. what is the 10th percentile of women heights?
SOLUTION. X= women height; X~N(64.5, 2.5).
a) P(X <70)=P(Z< ( )/2.5)=P(Z<2.2)=0.9861
b) P( 60 < X < 70) = P( ( )/2.5) < Z < ( )/2.5)=P(-1.8< Z < 2.2)= P( Z <-2.2) – P( Z < -1.8) = – =
c) 10th percentile of X =?
0.1=P( X< x) = P( Z< (x-65.5)/2.5), so -2.33=(x-65.5)/2.5; x=
10% of women are shorter than in.