Chap 7.2 - West Ada

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Transcript Chap 7.2 - West Ada 

Let’s define the random variable
𝑋 = 𝑛𝑒𝑑 π‘”π‘Žπ‘–π‘› π‘“π‘Ÿπ‘œπ‘š π‘Ž 𝑠𝑖𝑛𝑔𝑙𝑒 $1 𝑏𝑒𝑑 π‘œπ‘› π‘Ÿπ‘’π‘‘.
1. Is this example a discrete or continuous random variable?
2. What are the possible values of 𝑋?
3. Create a probability distribution.
4. What is the player’s average gain?
Chap 7.2
Means and Variance from Random Variables
Mean of a Random Variable

An average of the possible values of β€œπ‘‹β€, but with an
essential change to take into account the fact that not
all outcomes are equally likely

Also called the β€œExpected Value”
Mean of a Discrete Random
Variable
(Expected Value)

To find the mean of β€œπ‘‹β€, multiply each possible value
by its probability, then add all the products

πœ‡π‘‹ = π‘₯1 𝑝1 + π‘₯2 𝑝2 + β‹― + π‘₯π‘˜ π‘π‘˜ =

Back to the warm up…..

π‘₯𝑖 𝑝𝑖
What is the player’s average gain?

Example 7.5
Most states and Canadian provinces have governmentsponsored lotteries. Here is a simple lottery wager, from
the Tri-State Pick 3 game that New Hampshire shares with
Maine and Vermont. You choose a three-digit number; the
state chooses a three-digit winning number at random and
pays you $500 if your number is chosen.

How many three-digit combinations are there to chose
from?

What is the probability of winning?

Let X be the amount your ticket pays you, draw a
probability distribution.

What is your average pay-off from many tickets?
Law of Large Numbers

Draw Independent observations at random from any population with
finite mean µ. Decide how accurately you would like to estimate µ.
As the number of observations drawn increases, the same mean π‘₯ of
the observed values eventually approaches the mean µ of the
population as closely as you specified and then stays that close.

Nutshell: (in the long run)


Proportion of outcomes get closer to the probability of that value

Average outcome gets closer to the distribution mean
β€œLaw of small numbers” is a myth….

Heads and tails….
Rules for Means
1.
If 𝑋 is a random variable and π‘Ž and 𝑏 are fixed
numbers then

2.
πœ‡π‘Ž+𝑏𝑋 = π‘Ž + π‘πœ‡π‘‹
If π‘₯ and 𝑦 are random variables, then

πœ‡π‘₯+𝑦 = πœ‡π‘₯ + πœ‡π‘¦
Variance of a Random
Variable

Variance of a Discrete Random Variable

Example 7.7 pg. 485

The Standard Deviation (𝜎π‘₯ ) of π‘₯ is the square root of
the variance

𝜎π‘₯2 = ( π‘₯1 βˆ’ πœ‡π‘‹ )2 𝑝1 + ( π‘₯2 βˆ’ πœ‡π‘‹ )2 𝑝2 +
β‹― + π‘₯π‘˜ βˆ’ πœ‡π‘˜ 2 =
( π‘₯𝑖 βˆ’ πœ‡π‘– )2 𝑝𝑖
You can not add the standard deviations, but you can add the
variances and then evaluate for the standard deviations.
Rules for Variance
1.
If 𝑋 is a random variable and π‘Ž and 𝑏 are fixed
numbers then

2.
2
πœŽπ‘Ž+𝑏𝑋
= 𝑏2 πœŽπ‘‹2
If 𝑋 and π‘Œ are independent random variable then
2
 πœŽπ‘‹+π‘Œ
= πœŽπ‘‹2 + πœŽπ‘Œ2
2
 πœŽπ‘‹βˆ’π‘Œ
= πœŽπ‘‹2 + πœŽπ‘Œ2
and
Assignment: pg.486 23-25, 32, 34, 38-40
Random trivia: Some people have all the luck…
In June 2005, Donna Goeppert won a million dollar
jackpot playing the Pennsylvania Lottery. The odds of
winning a lottery like the one in Pennsylvania are 1.44
million to 1. But what is extraordinary is that Ms.
Goeppert had previously won $1 million playing the
lottery earlier in the year. The odds of winning twice
vary, depending on how many tickets are scratched. A
university professor in Pennsylvania estimated that if you
played 100 tickets, the odds of winning the lottery twice
are about 419 million to 1.