Data Unit 6 Test Flashcards
In the previous unit, we looked primarily at the individual outcomes of probability experiments. We will now turn our focus on models that show the distribution for all possible outcomes of an experiment
Creating distribution charts
Random Variable
A function that maps a numerical value to the outcome of an experiment, usually X.
The random variable is used to account for all the possible outcomes of an experiment.
Example:
: If X is the number rolled on a die, there are 6 possible values for x.
: If X is the number rolled on a die, there are 6 possible values for x.
X all possibilities, x each case
Probability Distribution — The set of all possible values of a random variable and the corresponding probabilities.
Example: Let X be the random variable that counts the number of heads in3 flips of a coin.
of Heads, x
P(X=x)
0 - TTT
1/8 = 0.125
1 - TTH, HTT, THT
3/8 = 0.375
2 - HHT, HTH, THH
3/8 = 0.375
3 - HHH
1/8 = 0.125
Note: The sum of the probabilities is always equal to
1`
Example: Let X be the random variable defined by the sum of the top faces of two fair dice. Construct a probability distribution table and probability histogram.
Sum of dice, x
P(X=x)
2
1/36
3
2/36
4
3/36
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36
An expectation or expected value, E(X), is the predicted average of all possible outcomes of a probability experiment.
E(X) = X(P(x)) … for all outcomes
Example: For the dice example,
E(X) = 2(1/36) + 3(2/36) + 4(3/36) + … 11(2/36) + 12(1/36)
E(X) = 7
Example: If you rolled a pair of dice 360 times, how many times would you expect to get a 9? What do you expect you will get when you roll a pair of dice?
P(9) = 4/36 = 1/9
1/9 x 360 = 40 times.
Example: If you flipped a coin 4 times, how many heads would you expect?
of Heads, x
P(X=x)
0
1/16
1
4/16
2
6/16
3
4/16
4
1/16
E(X) = 0(1/16) + 1(4/16) + 2(6/16) + 3(4/16) + 4(1/16)
All tails … all heads
E(X) = 2
Example: A carnival game has the following rules. If you roll a 4 on fair die, you get $3; otherwise, you lose $1. (Pay $1, get $4 back — Quadruple your money!) Is this a fair game?
x
P(X=x)
1
-1
1/6
2
-2
1/6
3
-1
1/6
4
+3
1/6
5
-1
1/6
6
-1
1/6
Here, the outcomes are with respect to the amount of money that can be won or lost.
E(X) = (-1)(5)(1/6) + 3(1/6)
E(X) = -1/3
-$0.33/game
Therefore, not a fair game because in a fair game E(X) = 0.
You need to break even to have an even
Even chance to win or lose
Binomial Distributions
Need to know the difference between binomial and hypergeometric
A binomial distribution looks at the distribution of the outcomes of several trials of a probability experiment in which the:
1) trials are independent
2) only two outcomes for each trial is success or failure
3) The probability of success or failure for each independent trial is unchanged with each trial.
Repeated Trials — A stochastic process in which:
a) experiments are identical
b) experiments are independent
Bernoulli Trials —
repeated trials that have exactly 2 outcomes (Success or failure)
Ex. A bag contains 3 yellow marbles and 4 blue marbles. Three marbles are selected at random from a bag, one at a time with replacement. Construct a probability distribution table for the number of blue marbles in the sample.
Here, there are 3 repeated, independent trials of the same experiment in which there are only 2 outcomes;
success(picking a blue marble) or
failure (picking a yellow one).
Since there is a replacement after each pick, the probability of success is 4/7 and the probability of failure is 3/7, both remaining unchanged. Since all of the trials are independent, we can apply the product rule for independent events.
of blue marbles, x
P(X=x)
0 (all yellow)
(3 C 0) (4/7)^0 (3/7)^3 = 0.079
1
(3 C 1) (4/7)^1 (3/7)^2 = 0.315
2
(3 C 2) (4/7)^2 (3/7)^1 = 0.419
3
(3 C 3) (4/7)^3 (3/7)^0 = 0.187
E(X) = 0(0.079) + 1(0.315) + 2(0.419) + 3(0.187)
E(X) = 1.714, therefore 2 marbles
ROUND depending on which has the highest probability
ROUND UP to 2 marbles because they have the highest probability (0.419 is bigger than 0.315)
in binomial distribution…
- Do Choose, the success bracket, then the failure
- multiply x by probability for each option and add them up
- ROUND depending on which has the highest probability
in binomial, choose when the
Choose when the success trial occurs
robability in a Binomial Distribution formula:
P(X=x) = (n C x)p^xq^n-x
Where p is the probability of success, q is the probability of failure, q = 1 —p
n = # of trials
x = trial we are currently on
The Expectation for a Binomial Distribution of n number of independent trials:
Don’t need the table
E(X) = np
For the marble example above:
E(X) = 3(4/7)
=1.714
round to 2 because its probability is higher
Ex. A test consists of 10 multiple-choice questions each with 5 possible answers, only one of which is correct. If answers are chosen at random, whatis the probability that
a) 6 answers are correct?
P(X=6) = (10 C 6)(1/5)^6(1/5)^4 = 0.0055
x = 6, n = 10, p = 1/5 correct, q = 4/5 failure
6/10 = 60%, pick where you’re right, success of 1/5 - 6 times, fail of 4/5 - 4 times
Ex. A test consists of 10 multiple-choice questions each with 5 possible answers, only one of which is correct. If answers are chosen at random, whatis the probability that:
b) A guesser would get at least 20% (or 2 out of 10)
Indirect
1 - (none correct - 1 correct)
q = 1 - (10 C 0)(1/5)^0(4/5)^10 - (10 C 1)(1/5)^1(4/5)^9
= 0.624
Ex. A basketball player makes 75% of her free throws. What is the expected number of free throws she will make on her next 10 shots?
We know this is a binomial distribution because independent variables (taking one shot does not affect the other -> equal opportunities) have no need for a table, can use E(X) = np
E(X) = np = 10 x 7.5 = 7.5
But she doesn’t shoot 0.5 of a free throw, so is it closer to 7 or 8?
P(X=7) = (10 C 7)(0.75)^7(0.25)^3 = 0.25
P(X=8) = (10 C 8)(0.75)^8(0.25)^2 = 0.282
Therefore, 8 free throws since it has a higher probability.
Geometric Distributions
Subset of Binomial Distribution
