The amount of Revenue a company brings in has a normal dist. μ R = $86,200 and σ R = $600. Its Expenses has a μ E = $12,020 and σ E = $800. The company has a good month and doubles its revenue. Let P be the formula for Profit where P = 2R – E Find the probability that the company will make more than $162,500 in Profit. (Hint find the Mean and Standard Dev. of Profit) μ 2R - E = 2∙μ R – μ E = 2(86200) – = $160,380 σ 2 2R - E = 2 2 ∙σ 2 R + σ 2 E = 4(600 2 ) = $ σ 2R - E = $ P(μ 2R - E > 162,500) = Normalcdf(1.47, E99) Z = 1.47 = WARM - UP

A company is interviewing applicants for managerial positions. They plan to hire two people. They have already rejected most candidates and are left with a group of 11 applicants of whom 7 are women. Unable to differentiate further between the applicants, they choose two people at random from this group of 11. Let the random variable X be the number of men that are chosen. Find the probability model for X and find µ & σ. Number of Men P(# of Men) 012 WARM - UP μ x = σ x = 0.645

Chapter 17 - The Bernoulli Trials The Bernoulli Trials: 1.Each observation must fall into one of just Two Categories (“Success” or “Failure”). 2.All observations must be Independent. 3.The probability of success, p i, must be the same for each observation. Bernoulli or NOT Bernoulli? You are rolling 5 dice and need to get at least two 6’s to win the game We record the eye colors found in a group of 500 people. A city council of 11 Dem and 8 Rep. picks a committee of 4 at random. You are finding the Probability of getting 4 Democrats.

The Geometric Distribution The Bernoulli Trials : 1.Each observation must fall into one of just Two Categories (“Success” or “Failure”). 2.All observation must be Independent. 3.The probability of success, p i, must be the same for each observation. 4G. The variable of interest, ‘X’, is the number of trials required to obtain the first success.

The Probability Distribution Function: The Probability that it will take k attempts to obtain the first success. P(x = k) = geometpdf (p, k) The Cumulative Distribution Function: The Probability that it will take at most k attempts to obtain the first success. P(x ≤ k) = P(x = 1) + P(x = 2) + … + P(x = k) P(x ≤ k) = geometcdf (p, k) EXAMPLE: NASA has a 0.20 probability of successfully launching a satellite in orbit. What is the probability that the first successful launch occurs on the 4 th attempt? geometpdf(.20, 4) = = (1 – 0.20) 3 (0.20)

Geometric Mean Expected Value = μ = 1/p Ch The Geometric Distribution (Cont.) EXAMPLE: A baseball player has a probability 0.15 of making a random homerun. Find the average number of at bats he will need to make his first homerun. 1/0.15 = 6.667

EXAMPLE: Let X be the number of basket attempts needed for a college basketball player to make his first free throw. The player has probability 0.82 of making a random free throw. Find the probability that: a.) He makes his first basket on the first attempt. b.) He makes his 1 st basket on the 5 th attempt. c.) It takes 4 or fewer attempts to make a basket. d.) At least 3 attempts are required to make a basket. e.) What is the probability that the player will make 6 baskets in a row before he misses one? P(x = 1) = geometpdf(0.82, 1) = 0.82 P(x = 5) = geometpdf(0.82, 5) = P(x ≤ 4) = geometcdf(0.82, 4) = P(x ≥ 3) = 1 – geometcdf(0.82, 2) = = 1 – P(x ≤ 2) (0.82) 6 ·(0.18) =

EXAMPLE #1: A couple plans to have children until they have their first boy. Let X = the number of children this takes. Each child has a prob. 0.5 of being a boy. a.) Is this a Geometric Setting? What are the parameter(s)? b.)List the Geometric Distribution. X 12345… P( x = X ) YES! G(0.5) On average, how many children should it take to have a boy? Mean = μ = 1/p = 1/0.5 = 2

Geometric Formula P(x = k) = (1 – p) (k – 1) ·p = geometpdf (p, k) Geometric Mean A Geometric Distribution, G(p,k) has mean: Expected Value = μ = 1/p Ch The Geometric Distribution (Cont.) EXAMPLE: A baseball player has a probability 0.15 of making a random homerun. Find the average number of at bats he will need to make his first homerun. 1/0.15 = 6.667