chapter 7 random variables i can find the probability of a discrete random variable. 6.1a discrete...
TRANSCRIPT
![Page 1: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/1.jpg)
Chapter 7Random Variables
I can find the probability of a discrete random variable.
6.1aDiscrete and Continuous Random
Variables and Expected Valueh.w: pg pg 353: 1 – 13 odd
![Page 2: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/2.jpg)
Discrete and Continuous Random Variables
A random variable is a quantity whose value changes.
![Page 3: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/3.jpg)
Discrete Random Variable
A discrete random variable is a variable whose value is obtained by counting.
number of students present number of red marbles in a jar number of heads when flipping three coins students’ grade level
![Page 4: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/4.jpg)
Probability Distribution The probability distribution of X lists the
values and their probabilities. Value of X: x1, x2, x3, … , xk
Probability: p1, p2 , p3 , … , pk
To find the probability of event pi , add up the probabilities of the xi that make up that event.
![Page 5: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/5.jpg)
Example: Getting Good Grades
A teacher gives the following grades: 15% A’s and D’s, 30% B’s, C’s; 10% F’s on a 4 point scale (A=4). Chose a student at random and find the
probability they get a B or better.
![Page 6: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/6.jpg)
Here is the distribution of X:
Grade: 0 1 2 3 4
Prob: .10 .15 .30 .30 .15 P(get a B or better) s/a P(grade 3 or 4): P(X=3) + P(X=4) = 0.30 + 0.15 = 0.45
![Page 7: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/7.jpg)
Probability Histograms Height of each bar is
the probability Heights add up to 1 Prob. of Benfords Law
vs. random digits
![Page 8: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/8.jpg)
Example: Tossing Coins
a. Find the probability distribution of the discrete random variable X that counts the # heads in 4 tosses of a coin.
Assume: fair coin, independence Determine the # of possible outcomes
![Page 9: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/9.jpg)
X = # heads X = 0, X = 1, X = 2, X = 3, X = 4
![Page 10: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/10.jpg)
b. Find each probabilityP(X=0) = 1/16 = 0.0625
P(X=1) =
P(X=2) =
P(X=3) =
P(X=4) = Do they add up to 1? Yes, so legitimate
distribution.
![Page 11: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/11.jpg)
Make a table of the probability distribution.
Number of heads
0 1 2 3 4
Probability:
0.0625
![Page 12: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/12.jpg)
c. Describe the probability histogram.
It is exactly symmetric. It is the idealization of the relative frequency
distribution of the number of heads after many tosses of four coins.
![Page 13: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/13.jpg)
d. What is the prob. of tossing at least 2 heads?
P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) = 0.375 + 0.25 + 0.0625
= 0.6875
![Page 14: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/14.jpg)
e. What is the prob. of tossing at least 1 head?
P(X ≥ 1): use the complement rule = 1 – P(X=0) = 1 – 0.0625
= 0.9375
![Page 15: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/15.jpg)
Exercise: Roll of the Die
If a carefully made die is rolled once, is it reasonable to assign probability 1/6 to each of the six faces?
a. What is the probability of rolling a number less than 3?
![Page 16: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/16.jpg)
P(X<3) = P(X=1) + P(X=2)
= 1/6 + 1/6 = 2/6 = 1/3
= 0.33
![Page 17: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/17.jpg)
b. Use the TI-83/89 to:
Simulate rolling a die 100 times and assign the values to L1.
MATH:PRB:randInt(1,6,100) store L1
Sort the list in ascending order. STAT:SortA(L1)
![Page 18: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/18.jpg)
Count the outcomes of 1 or 2 and record the relative frequency.
Results will vary but what do we expect the probability to be?
33% P(X=1) = /100 P(X=2) = /100 P(X=1) + P(X=2) =
![Page 19: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/19.jpg)
Exercise: Three Children A couple plans to have three children.
There are 8 possible arrangements of girls and boys.
For example, GGB. All 8 arrangements are approximately equally likely.
![Page 20: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/20.jpg)
a. Write down all 8 arrangements of the sexes of three children.
What is the probability of any one of these arrangements?
BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG
Each has probability of 1/8
![Page 21: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/21.jpg)
b. Let X be the number of girls the couple has.
BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG
What is the probability that X = 2? 3/8 = 0.375
![Page 22: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/22.jpg)
c. Starting from your work in (a), find the distribution of X. That is, what values can X take, and what are
the probabilities of each value? (Hint: make a table.)
Value of X 0 1 2 3
Probability
![Page 23: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/23.jpg)
X is the number of girls the couple has.
Value of X 0 1 2 3
Probability 1/8 3/8 3/8 1/8
![Page 24: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/24.jpg)
Review of Probability The probability of a random variable is an
idealized relative frequency distribution. Histograms and density curves are
pictures of the distributions of data. When describing data, we moved from
graphs to numerical summaries such as means and standard deviations.
![Page 25: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/25.jpg)
The Mean of a Random VariableNow we will make the same move to expand our description of the distribution of random variables.
The mean of a discrete random variable, X, is its weighted average.
Each value of X is weighted by its probability. Not all outcomes need to be equally likely.
![Page 26: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/26.jpg)
Example: Tri-Sate Pick 3 You pick a 3 digit number. If your number is
chosen you win $500. There are 1000, 3 digit numbers. Each pick costs $1.
![Page 27: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/27.jpg)
Taking X to be the amount your ticket pays you, the probability distribution is:
Payoff X: $0 $500 Probability: 0.999 0.001
Find your average Payoff. Normal “average”:
(0 + 500) /2 =$250
![Page 28: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/28.jpg)
Are the outcomes equally likely?
The long run weighted average is: = 500(1/1000) + 0(999/1000)
= $ 0.50
Conclusion: In the long run, the state keeps ½ of what
you wager.
![Page 29: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/29.jpg)
Expected Value (μx)
(The long run average outcome) We do not expect one observation to
be close to its expected value. μx : probabilities add to 1
![Page 30: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/30.jpg)
Mean of a Discrete Random Variable
The mean of a discrete random variable, X, is its weighted average. Each value of X is weighted by its probability.
To find the mean of X, multiply each value of X by its probability, then add all the products.
![Page 31: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/31.jpg)
1 1 2 2X k k
i i
x p x p x p
x p
![Page 32: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/32.jpg)
Example: Benford’s LawRecall: If the digits in a set of data appear “at random,” the nine possible digits 1 to 9 all have the same probability each being 1/9.
The mean of the distribution is:
μx = 1(1/9) + 2(1/9) + … + 9(1/9)= 45(1/9)= 5
![Page 33: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/33.jpg)
But, if the data obey Benford’s law, the distribution of the first digit V is:
Find the μv:
=
V 1 2 3 4 5 6 7 8 9
Prb. .301 .176 .125 .097 .079 .067 .058 .051 .046
![Page 34: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/34.jpg)
= 3.441
The means reflect the greater probability of smaller digits under Benfors’s law.
![Page 35: Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete and Continuous Random Variables and Expected Value h.w:](https://reader035.vdocument.in/reader035/viewer/2022062308/56649e4b5503460f94b3fd5d/html5/thumbnails/35.jpg)
Histograms of μx and μv
We can’t locate the mean of a right skew distribution by eye – calculation is needed.