Transcript Week_4_Lecture_ILS

MATH 221
Integrated Learning System
Week 4 Lecture
The Central Limit Theorem;
Normal Approximations to the
Binomial Distribution
Definition
A sampling distribution is a probability distribution of
sample statistics that is formed when samples of size n
are repeatedly taken from a population. If the sample
statistic is the mean then the distribution is called the
sampling distribution of the sample means.
Properties of Sampling
Distributions
1. The mean of the sampling distribution of the sample
means x is equal to the population mean .
2. The standard deviation of the sampling distribution
of the sample means x is equal to the population
standard deviation  divided by the square foot of
the sample size n. x is called the standard error of
the mean.

x 
n
Example
Four people in a carpool paid the following amounts for
textbooks this semester: $120, $140, $180, and $200.
Find the mean and standard deviation of the above
population. List all possible samples of size two from the
population. Find the mean and standard deviation of the
sampling distribution of the means and compare them to
the population parameters.
Solution
Sample Sample
Mean
120,120 120
120,140 130
120,180 150
120,220 170
140,120 130
140,140 140
140,180 160
140,220 180
Sample Sample
Mean
180,120 150
180,140 160
180,180 180
180,220 200
220,120 170
220,140 180
220,180 200
220,220 220
Using Technology: TI 83
How did we get the values in the previous slides using
the TI 83? You may not be familiar with the lists
feature of the TI 83 that allows you to input data for
analysis. Follow the instructions in the notes below to
see how this feature was used in the previous slides.
The Central Limit Theorem
If a random sample is drawn from any population, the
sampling distribution of the sample means is
approximately normal for a sufficiently large sample size.
The larger the sample size, the more closely the sampling
distribution of the sample means will resemble a normal
distribution. In general, unless the underlying population
has a normal distribution, a sample size of n > 30 is
recommended.
Probabilities and the Central
Limit Theorem
x  x x  x
z 

x
 n
Example
The population mean annual salary for plumbers is  =
$32,500. A random sample of 42 plumbers is drawn from
this population. What is the probability that the mean
salary of this sample is less than $30,000? (Assume that
 = $5,600)
Solution
z 
30,000  32,500
5600 / 42
 2.89
Looking up z = -2.89 in a standard normal table we see that
P(x < $30,000) = 0.0019.
Normal Approximation to
Binomial Distributions
If np  5 and nq  5, then the binomial random variable
x is approximately normally distributed with mean
 = np and   npq
Correction for Continuity
P(x = c)  P(c – 0.5 < x < c + 0.5)
Process
1. Verify that the binomial distribution applies. Specify n,
p, and q.
2. Determine if the normal approximation can be used.
np > 5, and nq > 5.
3. Find the mean and the standard deviation using the
formulas.
4. Apply the continuity correction.
5. Find the z-scores for c – 0.5, and c + 0.5.
6. Find the probability.
Example
Thirty-four percent of Americans have type A+ blood. You
randomly select 32 Americans and ask them if their
blood type is A+.
1. Find the probability that 12 people say that they have
A+ blood.
2. Find the probability that at least 12 people say they
have A+ blood.
3. Find the probability that fewer than 12 people say they
have A+ blood.
Example (Continued)
4. A blood drive would like to get at least 60 people with
type A+ blood. If there are 150 donors, what is the
probability that there will be enough type A+ blood
donors?
Solution
np  0.3432  10.88, nq  0.6632  21.12,   0.340.6632  2.68
So the normal approximation to the binomial distribution
can be used.
Solution (Part 1)
c = 12
z1 = (11.5 – 10.88)/2.68 = 0.231
z2 = (12.5 – 10.88)/2.88 = 0.604
P(11.5 < x < 12.5) = 0.136 [used TI 83]
Solution (Part 2)
P(x  12) = 1- P(x < 11.5) = 1 – 0.591 = 0.409.
[Used TI 83 to calculate P(x < 11.5)]
Solution (Part 3)
This part is easy since we have already calculated the
value we need in Part 2. P(x < 12) = 0.59.
Solution (Part 4)
np  0.34150  51, nq  0.66150  99,   .34.66150  5.80
We want to know the probability of having at least 60 A+
donors. That probability is 1 – P(x < 59.5).
1 – P(x < 59.6) = 1 – 0.929 = 0.071.
Exercise
Fifty-two percent of adults say chocolate chip are their
favorite cookie. You randomly select 40 adults and ask
each if chocolate chip is their favorite cookie.
1. Find the probability that at most 15 people say
chocolate chip is their favorite cookie.
2. Find the probability that at least 15 people say chocolate
chip is their favorite cookie.
3. Find the probability that more than 15 people say
chocolate chip is their favorite cookie.
(Continued on next slide)
Exercise (Continued)
4. A community bake sale has prepared 350 chocolate
chip cookies. If the bake sale attracts 650 customers
and the each buy one cookie, what is the probability
that there will be enough chocolate chip cookies?