1. Day 46 Agenda:
DG minutes

2. Advanced Placement Statistics
Section 8.1 (Part II): Mean, Variance, and
Standard Deviation for a Binomial
Random Variable
EQ: How do you calculate the mean, variance, and standard deviation for a binomial random variable?
Get out the Ex 8.1 Worksheet from yesterday (President’s Physical Fitness Test).

3. Refer back to the Section 8
Refer back to the Section 8.1 Ex Handout on the President’s Physical Fitness Test to answer #1 - 3.
Calculate the mean, variance and standard deviation, respectively, of this probability distribution.
HINT: Create a probability distribution for the random variable X.
(0)(.6561)+(1)(.2916)+(2)(.0486)+(3)(.0036)+(4)(.0001) = 0.4
REMEMBER: This is a COUNT, not a PERCENTAGE.
0.4
(0-.4)2(.6561)+(1-.4)2(.2916)+(2-.4)2(.0486)+(3-.4)2(.0036)
+(4-.4)2(.0001) = 0.36

4. 2. a. What are the values for the parameters?
n = ______ p = _______ q = _____ 4
0.1
0.9
(4)(.1) = 0.4
b. Find np.
How does it compare to ?
Find npq.
(4)(.1)(.9) = 0.36

5. FORMULAS TO KNOW:
For a binomial random variable,
SEE FORMULA SHEET:

6. RECALL: NORMAL DISTRIBUTION
.4 , .6
b. The Empirical Rule states approximately 95% of the x values lie within ­­­­____ standard deviations 2

7. This occurs from ___ to __ on your density
curve.
-.8
1.6
d. Since X cannot be negative, we would expect (in the long run), a large majority (95%) of the adults (out of groups of four) who pass the fitness test to be between _______ and _____.
1.6

8. # of successes # of failures
Binomial
Distribution
Normal
Approximation
If: n is large enough implies
n and p satisfy np > 10 and n(1 – p) > 10
# of successes
# of failures
Describes distribution
mean
Standard deviation

9. In Class Assignment:
p #22, 23
Get in groups so we can go over these problems together.
Get out a sheet of notebook paper.

10. This distribution can be described as B(509,.116).
p. 529 #22
X = number of homeruns out of 509 randomly selected at bats by Mark McGuire in 1998
Criteria for a Binomial Distribution:
Fixed number of trials; n = 509 at bats
Success  hit a homerun
Failure  did not hit a homerun
Result of one at bat is independent of the result of another at bat.
Probability of hitting a homerun remains the same from trial to trial. p = q = .884
This distribution can be described as B(509,.116).

11. REMEMBER: THIS IS A BINOMIAL DISTRIBUTION
Find the expected number of homeruns we could expect Mark McGuire to have hit in 509 randomly selected at bats.
REMEMBER: THIS IS A BINOMIAL DISTRIBUTION
On average, in the long run, we would expect Mark McGuire to hit homeruns in 509 at bats.
Use a binomial distribution to calculate the probability that Mark McGuire will hit at least 70 homeruns in 509 at bats.
The probability that Mark McGuire will hit at least 70 homeruns in 509 at bats is 7.64%.
1 - binomialcdf (509, 0.116, 69)

12. *Is Normal Approximation Appropriate?
np > 10 ? nq > 10 ?
(509)(.116) > (509)(.884) > 10
> > 10
Criteria met; Normal Approximation Appropriate
RECALL:
What does this allow us to do?
Standardize using z scores

13. Standardize using z scores
The probability that Mark McGuire
will hit at least 70 homeruns in
509 at bats is 6.43%.
How does the probability of Mark McGuire hitting at least 70 homeruns using a normal distribution compare to the probability calculated using a binomial distribution?
The normal distribution gave a much smaller probability of 6.4% compared to 7.6% for the binomial distribution.

14. This distribution can be described as B(400,.92).
p. 530 #23
X = number of people out of a random sample of 400 adults in Richmond who approve of the President’s response
Criteria for a Binomial Distribution:
Fixed number of trials; n = 400 people
Success  approve President’s response
Failure  did not approve President’s response
One person’s opinion is independent of another person’s opinion.
Probability of person approving President’s response remains the same from trial to trial. p = .92 q = .08
This distribution can be described as B(400,.92).

15. Use a binomial distribution to calculate the probability that at most 358 of the 400 adults in the Richmond poll approve of the President’s response.
binomialcdf (400, 0.92, 358)
The probability that at most 358 out of 400 adults in the Richmond poll approve of the President’s response to the WTC attacks is 4.41%.

16. REMEMBER: THIS IS A BINOMIAL DISTRIBUTION
Calculate the expected number of adults in Richmond that approve of the President’s response.
REMEMBER: THIS IS A BINOMIAL DISTRIBUTION
On average, in the long run, we would expect 368 adults in Richmond to approve of the President’s response to the WTC attacks.
Calculate the standard deviation for the number of adults in Richmond that approve of the President’s response.

17. *Is Normal Approximation Appropriate?
np > 10 ? nq > 10 ?
(400)(.92) > (400)(.08) > 10
368 > > 10
Criteria met; Normal Approximation Appropriate
RECALL:

18. Standardize using z scores
The probability that at most 358 adults in Richmond approve of the President’s response to the WTC attacks is 3.29%.
The normal distribution gave a much smaller probability of 3.3% compared to 4.4% for the binomial distribution.
How does the probability calculated using a normal distribution compare to the probability calculated using a binomial distribution?

19. ACCURACY OF NORMAL APPROXIMATION: Increases when p is close to ½.
Less accurate when p is close to 0 or 1.
RULE OF THUMB: BINOMIAL OR NORMAL ?
If n, p, and q satisfy the conditions
np > 10 and nq > 10
you are EXPECTED to use a Normal Approximation instead of a Binomial Distribution.

20. Assignment :
WS Mean, Variance and Standard Deviation for Binomial Random Variable