1. 6.2 Homework Questions

2. Binomial Random Variables
Section 6.3
Binomial Random Variables

3. Binomial Setting The four conditions for a binomial setting are:
Success/Failure
Independent Trials
Constant “p” (probability of success)
Set number of trials, n

4. Geometric The four conditions for a geometric setting are:
Success/Failure
Independent Trials
Constant “p” (probability of success)
No set number of trials, n

5. Binomial Random Variable
The count of X successes in a binomial setting is a binomial random variable.
The probability distribution of X is a binomial distribution with parameters n and p.
The possible values of X are the whole numbers from 0 to n.

6. Binomial?
Genetics says that children receive genes from each of their parents independently. Each child of a particular pair of parents has probability 0.25 of having type O blood. Suppose these parents have 5 children. Let X = the number of children with type O blood.
Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y = the number of aces you observe.
Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat this process until you get an ace. Let W = the number of cards required.

7. Binomial Probabilities
Let’s do the children’s gene problem.
n=5 , p= or B(5, 0.25)
P(X=0)
P(none of the children have type O)=
P(X=1)
P(one child has type O)
P(X=2)

8. Building the formula
P(X = k) = = P(exactly k successes in “n” trials)= = (number of arrangements)∙ 𝑝 𝑘 (1−𝑝 ) 𝑛−𝑘 So we need a nice way of finding the “number of arrangements”

9. Number of arrangements: Binomial Coefficient
The number of ways of arranging k successes among n observations is given by the binomial coefficient:
𝑛 𝑘 = 𝑛! 𝑘! 𝑛−𝑘 !
You may know this as nCr
CAUTION : 𝑛 𝑘 is NOT the fraction 𝑛 𝑘

10. Binomial Probability For B(n,p) , that is to say, for any Binomial:
P X=k = 𝑛 𝑘 𝑝 𝑘 (1−𝑝 ) 𝑛−𝑘
This is on the formula sheet!

11. Examples:
Find the probability that exactly 3 children have type O blood.
B(5,0.25)
Should the parents be surprised if more than 3 of their children have type O blood? Justify your answer.

12. Mean and Standard Deviation of a Binomial Distribution
Blood Type Probability Distribution:
Above is the binomial from the blood type problem.
Find the expected value and standard deviation. X 1 2 3 4 5
P(X)

13. Mean and Standard Deviation of Binomial Random Variables
Given B(n,p):
𝜇 𝑥 =𝑛𝑝
𝜎 𝑥 = 𝑛𝑝(1−𝑝)
Remember – these formulas ONLY work for binomial distributions!
Both of these are on the formula sheet as well.

14. Examples Continued: Together, let’s do numbers Page 403: 69-72
Using the first one as an example B(20, 0.85) find:
𝑃(𝑋=17)
𝑃 𝑋≤12
𝑃 𝑋=0 +𝑃 𝑋=1 +…+𝑃(𝑋=12)

15. Homework (same as before)
Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, )

16. Warm Up

17. Normal Approximation for Binomial Distributions
As a rule of thumb, we will use the Normal approximation when n is so large that:
𝑛𝑝≥10
𝑛(1−𝑝)≥10
That is, the expected number of successes and failures are both at least 10.
𝑛≤ 1 10 𝑁 (independence)

18. Example:
Suppose that exactly 60% of all adult US residents would say “agree” if asked if they think shopping is frustrating. A survey asked nationwide sampled 2500 adults.
Let X = the number of people who agree.
Show that X is approximately a binomial random variable.
Check the conditions for using a Normal approximation in this setting.

19. Example Continued:
Use a Normal distribution to estimate the probability that 1520 or more of the sample agree.
Find the mean
Find the standard deviation
Use the Normal curve

20. Homework #3 (again)
Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, )