6.2 Homework Questions

Binomial Random Variables Section 6.3 Binomial Random Variables

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

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

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.

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.

Binomial Probabilities Let’s do the children’s gene problem. n=5 , p= 0.25 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)

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”

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

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!

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.

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) 0.23730 0.39551 0.26367 0.08789 0.01465 0.00098

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.

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)

Homework (same as before) Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, 94-105)

Warm Up

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)

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.

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

Homework #3 (again) Pg. 403 (73-75, 77, 79, 80, 82, 84-87, 89-92, 94-105)