1. A small software company bids on two contracts
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?
1st Only
2nd Only
Both
Neither
Profit (x)
$50,000
$20,000
$70,000 $0
P(x)

2. Chapter 16 Part 2
Random variables

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

4. Recall that standard deviation = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝜎= 𝑥−𝜇 2 𝑃(𝑥)
variance
Recall that standard deviation = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒

5. From previous example with an expected value of $20, we now calculate the standard deviation:
Policyholder Outcome
Payout x
Probability P(x)
Deviation
(x-μ)
Death
10,000
1/1000
(10,000 – 20) = 9980
Disability
5000
2/1000
(5000 – 20) = 4980
Neither
997/1000
(0 – 20) =
-20
𝜎= − =$𝟑𝟖𝟔.𝟕𝟖

6. 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?
New
P(new and new) = 0.524
10/14
New
P(new and refurb) =
11/15
4/14
Refurbished
New
11/14
P(ref and new) =
4/15
Refurbished
3/14
P(ref and ref) = 0.057
Refurbished

7. Outcome X
P(x)
X-μ
2 refurbs
1000
0.057
1 refurb, 1 new
100
= 0.419
2 new
0.524
1000 – 98.9 = 901.1
100 – 98.9 = 1.1
= -98.9
E(x) = 0(0.524) + 100(0.419) (0.057) = $98.90
𝜎= −
=$𝟐𝟐𝟔.𝟕𝟒

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