A small software company bids on two contracts. It anticipates

A small software company bids on two contracts. It anticipates a profit of 50,000 if it gets the first (larger) contract and a profit of 20,000 on the second (smaller) contract. The company estimates there’s a 30% chance it will get the larger contract and a 60% chance it will get the smaller contract. Assuming the contracts will be awarded independently, what’s the expected profit? Profit (x) P(x) 1st Only 50,000 2nd Only 20,000 Both 70,000 Neither 0

Chapter 16 Part 2 RANDOM VARIABLES

The payout (x) in a probability model is called a random variable. The value is based on the outcome of a random event.

Standard Deviation 2 𝜎 ( 𝑥 𝜇) 𝑃(𝑥) variance Recall that standard deviation

From previous example with an expected value of 20, we now calculate the standard deviation: Policyholde Payout x Probabili Deviation r Outcome ty P(x) (x-μ) Death Disability Neither 10,000 1/1000 (10,000 – 20) 9980 5000 2/1000 (5000 – 20) 4980 0 997/100 0 (0 – 20) -20 1 2 2 997 2 𝜎 9980 4980 ( 20 ) 𝟑𝟖𝟔. 𝟕𝟖 1000 1000 1000 2 ( ) ( ) ( )

A company that ships computers randomly selects 2 that you’ve requested for a client from the 15 that are in stock. Of those 15, 4 of them were not new but were refurbished. If your client gets 2 new computers, things are fine. If the client gets one new and one refurbished, it will be sent back and replaced at an expense to you of 100. If both are refurbished, the client will cancel the order and you’ll lose 1000. What is the expected value and standard deviation of the company’s loss? 10/14 New New 11/15 4/14 11/14 4/15 Refurbished New P(new and new) 0.524 P(new and refurb) 0.2095 P(ref and new) 0.2095 Refurbished 3/14 Refurbished P(ref and ref) 0.057

Outcome X P(x) 2 refurbs 1000 0.057 1000 – 98.9 901.1 100 0.2095 0.2095 0.419 100 – 98.9 1.1 0 0.524 1 refurb, 1 new 2 new X-μ 0-98.9 -98.9 E(x) 0(0.524) 100(0.419) 1000(0.057) 98.90

Today’s Assignment: Add to HW: p.384 #9-16, 20, 22