8-5 Normal Distributions
 continuous distribution found in many application areas
 e.g. SAT scores, people’s heights, leaf lengths
 all are distributed according to the normal distribution
 also called the Gaussian Distribution
 it is the famous "bell-shaped" curve:
 distributions have a mean

and a standard deviation

 the above Normal has

= 430 and

= 100
For any Normal Distribution with mean

and standard deviation

:

68% of all scores lie within

1

of the mean

95% are within

2


99.7% (virtually all) are within

3

8-5 p. 1

Computing areas (probabilities) under normal distributions
Question:
What % of the people that take the SAT make scores between 330 and 530?
Analysis: mean

= 430 s.d.

= 100 picture:
330 =

530 =

% of scores between

and

for any normal distribution?
We know that one; it's 68% !!!
8-5 p. 2

Question:
What % of the people that take the SAT make scores between 430 and 600?
Analysis: mean

= 430 s.d.

= 100
600 = how many

from the mean?
600

430 z = = 1.7

100
We only know percentages for within

1
 
2
 and

3
 of the mean.
Q: Can we find the area under a normal from
 to

+ 1.7

A: Yes, we can look it up in a table.
Here's the picture:
z = 1.7

The table on page 723:
 gives us area ( for any normal distribution !!!)
 from
 to

+ (any number of

) look up 1.7  .4554
Answer : 45.56%
8-5 p. 3

Z-statistics
 by computing the number of

's a score is from the mean
 you are computing a number known as:
 a z-statistic , or
 a standardized score
 the unit of measurement of a z-statistic is always standard deviations
Example: if

= 10 and

= 7, what is the z-statistic or standardized score corresponding to a score of 15? z =
15

7
10
= .714

Now, what % of scores lie between 10 and 15?
Compute the area under a normal between

and

+ .714

Picture:
.71

.2611 Ans: 26% z = .714

8-5 p. 4

Example:
What % of scores lie between 5 and 10?
Picture: z = -1.7

 since the normal curve is symmetric
 the shaded area is the same as for between 10 and 15
 and the answer = 26% as before
Example:
What is the area between scores of 7 and 15?
Picture: z = (7 - 10)/7 = -.43
  z = .714
 shaded area below

(look up .43 in the table) = .1664 shaded area above

(look up .714 in the table) = .2611 total shaded area = .1664 + .2611 = .4275
8-5 p. 5

Example:
What is the area above 15?
Picture: z = .714

 since the total area above the mean is .5
 and the area from 10 - 15 (look up .714) is .2611
 the required area is .5 - .2611 = .2389
Animal Crackers - segue to a new topic
"Animal crackers, and cocoa to drink.
That is the finest of suppers, I think."
The American Express Card is used by 40% of the people buying animal crackers in Austin. 20 animal cracker buyers are surveyed. What is the probability that from 6 to 12 of them use Amex?
 this is a repeated Bernoulli experiment with
 n=20, p=.40, q=.60
 if X = #people using Amex , we want
(using the binomial distribution ) i
12
P(6

X

12) =


6
(.
40 ) i
(.
60 )
20
 i
That's 7 of those nasty terms to compute and add!
What a bore! Is there an easier way? Watch!
8-5 p. 6

Areas under a binomial distribution
 for simplicity, we use an example binomial with:
 n = 5 and p = .4

Let's compute the probability that (X= 1 or X=2):
 p(1) + p(2) =
C
5,1
(.4)(.6) 4 + C
5,2
(.4) 2 (.6) 2 = .2592 + .3456 = .6048
If we graph our binomial, we get:
Now notice: since the bar labeled 1 has width 1 height = (probability) = .2592 the area of bar 1 = 1 x .2592 = .2592 = p(1) the area of bar 2 = 1 x .3456 = .3456 = p(2) the area of bar 1 + the area of bar 2 =
.2592 + .3456 = .6048 = p(1) + p(2)
8-5 p. 7

for discrete distributions :
 areas (under the "curve") represent probabilities
 just as for continuous distributions !
The shaded area is p(1) + p(2)
8-5 p. 8

The Normal approximation to the binomial distribution
We start by overlaying an appropriately-shaped (but rather crudely-drawn) Normal on our binomial distribution:
This makes sense, because of our observation in the last section: one can interpret areas for both the normal and binomial distributions as probabilities
We are creating a normal approximation to the binomial , to be used in the following way:
 instead of computing areas under the binomial
 we will instead compute areas under the normal
 they won't be exactly right . . .
 only an approximation of the "real area" we want
8-5 p. 9

 how good an approximation do we have?
 look:
Areas under the normal approximation will:

"leave out" pieces of the binomial appearing above the normal
 but will "add in" some extra pieces outside the binomial appearing below the normal
 these adjustments do not balance out completely
 but in many cases will provide a good enough approximation!
8-5 p. 10

Using the Normal approximation to compute a binomial probability
We go back to our original problem. We want to compute: p(1) + p(2) = C
5,1
(.4)(.6) 4 + C
5,2
(.4) 2 (.6) 2 which was computed directly to be: .6048
Now we will try to approximate that by computing an area under the normal approximation :
Recall that for the binomial:
Mean:

= np Standard deviation:

= npq
Accordingly, we will use a Normal with:

= 5(.4) = 2

= 5

.
4

.
6 = 1.1
 we want to do p(1) + p(2) BUT . . .
 the limits for the normal will go from .5 to 2.5
( not 1 to 2)
8-5
Why? Look at the picture!! p. 11

So we want to compute this area z = (.5 - 2)/1.1 = -1.36
 z = (2.5 - 2)/1.1 = .45

 shaded area below 2: 1.36 = table = .4131
 shaded area above 2: .45 = table = .1736
 total area: .5867
 remember the exact calculation?: .6048
 not a bad approximation, huh?
8-5 p. 12

Animal Crackers accomplished - no pain!
The American Express Card is used by 40% of the people buying animal crackers in Austin. 20 animal cracker buyers are surveyed. What is the probability that from 6 to 12 of them use Amex?
use a Normal distribution with
Mean:

= np = 20(.40) = 8
Standard deviation:

= npq = 20

.
4

.
6 = 2.19
 compute the area under a Normal
 with mean 8 and standard deviation 2.19
 from 6 - .5 = 5.5
to 12 + .5 = 12.5
 where did the .5's come from?(re-read the previous section!)
Picture: z = (5.5 - 8)/2.19 = -1.14
 z = (12.5 - 8)/2.19 = 2.05

 shaded area below 8: 1.14  table  .3729
 shaded area above 8: 2.05

table

.4798
 total area: .8527
Ans: .8527
8-5 p. 13

When is the Normal approximation appropriate?
The question arises: "When is the Normal Approximation a good enough approximation to the Binomial?"
Here's a rule-of-thumb to use:
Rule-of-Thumb Test
If
0
 and
 n the Normal Approximation is appropriate
 try it on our binomial problem, parameters: n = 5, p = .4
 
= 2,

= 1.1
 remember we got a pretty good approximation?

BUT!

= 2 - (3)(1.1) = -1.3, not

0 it fails the test already!
 so we really shouldn't use the Normal approximation!
 we were just lucky, I guess!
How about Animal Crackers ? n = 20, p = .4,

= 8,

= 2.19

= 1.43

0 OK so far

= 14.57

OK

8-5 p. 14