Download 3680 Lecture 10

Math 3680
Lecture #10
Normal Random
Variables
Many distributions follow the bell curve rather well. For
example, according to studies, the height of U.S. women has
the following data: m = 63.5 inches and s = 2.5 inches
Accordingly, about 68% of women have heights
between 61 and 66 inches.
However, the bell curve does not model all
data sets well.
Example: What percentage of women has
heights between 60 and 68 inches?
Example: What percentage of women has
heights between 60 and 68 inches?
Is this answer exact or an approximation?
Example: For a certain population of high school
students, the SAT-M scores are normally distributed
with m = 500 and s = 100. A certain engineering
college will accept only high school seniors with
SAT-M scores in the top 5%. What is the minimum
SAT-M score for this program?
Note: Before, the kind of question that was posed
was: “What percentage of students score above 700
on the SAT-M?” Now, the percentage is specified.
Approximating a Binomial(n, p)
distribution with a normal curve
Example: From a heterozygous cross of two pea
plants, 192 seeds are planted. According to the laws of
genetics, there is a 25% chance that any one offspring
will be recessive homozygous, independently of all
other offspring.
Find the probability that between 44 and 55 (inclusive)
of the seeds are recessive homozygous.
Solution #1 (Exact). Let X denote the number of
recessive homozygous offspring. Then
X ~ Binomial(192, 0.25)
Therefore,
P(44  X  55) 
55
 P( X  k )
k  44
192 
44
148
(0.25) (0.75)
 
 44 
192 
55
137
(0.25) (0.75)
   
 55 
 0.6646.
In terms of the probability histogram:
Solution #2 (Approximate). Note that the probability
histogram of the Binomial(192, 0.25) distribution is
approximated well by a bell curve. Since
X ~ Binomial(192, 0.25),
we have
m  (192)(0.25)  48
s  (192)(0.25)(0.75)  6
We convert 43.5 and 55.5 to standard units:
43.5  48
 0.75
6
55.5  48
 1.25
6
Solution #2 (continued)
Therefore,
P(44  X  55)  (1.25)  (0.75)  0.6678
Question: Why did we standardize 43.5 and 55.5?
This subtlety is called continuity correction.
BBinomial(n, 0.5) distribution
BBinomial(n, 0.5) distribution
BBinomial(100, p) distribution
BBinomial(100, p) distribution
BBinomial(n,π) distribution with π small
Limitations On Normal Approximation
The normal approximation is reasonable if n is large
and both p and 1 - p are not small. More precisely,
the approximation is good if
n p  5 and n (1 - p)  5.
• What happens if n is large and p is small?
• What if instead p is close to 1?
Continuity Correction
The primary difficulty with using the continuity correction is
deciding whether to add or subtract 0.5 from the endpoints.
This decision may be facilitated by drawing a rough histogram
and then deciding which rectangles are to be included.
Example: In the previous problem, what should be converted
to standard units to find
• P(24  X  30)
• P(31 < X  38)
• P(X > 25)
• P(X  37)
Continuity Correction
There’s nothing particularly special about the binomial
distribution for this procedure.
The continuity correction can be used whenever a
discrete random variable is being approximated by a
continuous one.
Example: A gambler repeatedly bets on red in
roulette. The chance of winning on one play is p =
9/19. Suppose the gambler plays 100 times. Use the
normal approximation to estimate the probability that
the gambler wins more than 50 times.