AP Statistics
AP Statistics
Chapter 7
Random Variables
|7.1 Discrete and Continuous Random |Objective: |
|Variables |-recognize and define discrete random variables, and construct a probability distribution table and a probability |
| |histogram for the random variable. |
| |-recognize and define a continuous random variable, and determine probabilities of events as areas under density curves. |
| |-Given a normal random variable, use the standard normal table or a graphing calculator to find probabilities of events as|
| |areas under the standard normal distribution curve |
|random variable | |
| |A random variable assumes any of several different values as a result of some random event. Random variables are denoted |
| |by a capital letter such as X and the values that the random variable can take on are denoted by the same lowercase |
| |variable ([pic] ). All possible values that the random variable can assume, with their associated probabilities, form the|
| |probability distribution for the random variable. |
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| |A discrete random variable X has a countable number of possible outcomes. |
| |The probability distribution of X lists the values and their probabilities: |
|discrete random variable |Value of X: [pic] |
| |Probability: [pic] |
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| |The probability [pic]must satisfy two requirements: |
| |Every probability[pic]is a number between 0 and 1. |
| |[pic] |
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|Example 7.2 page 394 |[pic] |
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| |P(X = 0) = P(X = 1) = |
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| |P(X = 2) = P(X = 3) = |
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| |P(X = 4) = |
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| |Number of heads: 0 1 2 3 4 |
| |Probability: |
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| |[pic] |
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| |What is the probability of tossing at least two heads? |
| |P(X ( 2) = |
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|Probability histogram | |
| |What is the probability of tossing at least one head? |
| |P( ( 1) = |
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| |a. P(X = 5) = |
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| |b. |
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| |c. P(X ( 3) = |
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| |d. P(X < 3) = |
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| |e. |
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|Practice Problem 7.3 page 396 | |
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| |a. |
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| |b. |
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| |Suppose that we want to choose a number at random between 0 and 1. The sample space is an entire interval of infinitely |
|Practice Problem 7.5 page 396 |many numbers: S = {all numbers x such that 0 ( x ( 1). |
| |How do we assign probabilities to events like 0.3 ( x ( 0.7? Can we assign probabilities to each individual value of x? |
| |[pic] |
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| |P(X ( 0.5) = |
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| |P(X > 0.8 ) = |
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| |P(X ( 0.5 or X >0.8) = |
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| |To solve this problem, we assign probabilities as areas under a density curve (remember those back in chapter 2). Any |
|Continuous Random Variables |density curve has area exactly 1 underneath it, which for us conveniently corresponds to a probability of 1. |
| |[pic] |
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| |A continuous random variable X take all values in a an interval of numbers. |
| |The probability distribution of X is described by a density curve. |
|uniform distribution |The probability of any event is the area under the density curve and above the values of X that make up that event. |
| |The probability of any individual outcome is 0. |
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| |a. |
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| |b. |
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| |d. |
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| |e. |
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| |f. |
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| |The most familiar density curve is the normal curve. Normal distributions are probability distributions. If X has the |
| |N(μ, σ) distribution, the standardized variable |
| |Z = [pic] |
| |is a standard normal random variable having the distribution N(0,1) |
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| |a. P( ) = |
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| |b. P( ) = |
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| |c. P( ) = |
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|Practice Problem 7.6 page 401 | |
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|Practice Problem 7.8 page 402 | |
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| |Objective: |
| |-calculate the mean and variance of a discrete random variable. Find the expected payout in a raffle or similar game of |
| |chance. |
| |-use simulation methods and the law of large numbers to approximate the mean of a distribution. |
| |-use rules for mean and variances to solve problems involving sums, differences, and linear combinations of random |
| |variables |
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| |The mean of a discrete random variable is also called the expected value. Although we cannot predict the outcome of any |
| |single event of a random variable, the expected value of a random variable is its theoretical long run average. It is the|
| |center of its distribution and is denoted by μ or E(X). To remind ourselves we are talking about the mean of a random |
| |variable X, we often use [pic] |
| |Mean: [pic]= E(X) = ( x (P(x) |
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| |Variance: Var(X) = [pic] = ((x - [pic])² ( P(x) |
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| |Standard Dev.: SD(X) = [pic] = [pic] |
| |[pic]=[pic] |
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| |At a carnival a game involves spinning a wheel that is divided into 60 equal sectors. The sectors are marked as follows: |
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| |$20 1 sector |
| |$10 2 sectors |
| |$5 3 sectors |
| |No prize 54 sectors |
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| |The carnival owner wants to know the average expected payoff for this game and the amount of variability (as measured by |
| |the standard deviation) associated with it. |
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| |Define the random variable: |
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| |Make a table: |
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|7.2 Means and Variances of Random | |
|Variables | |
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| |Find the expected value, [pic]: |
| |[pic]= E(X) = ( x (P(x) |
| |[pic]= |
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| |Find the standard deviation: |
| |[pic]=[pic] |
| |[pic]= |
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|Example: | |
| |Steps: |
| |X([pic] |
| |P(X)([pic] |
| |STAT ( CALC ( 1 VARS STAT [pic],[pic]( ENTER |
| |[pic]=_______ |
| |[pic]=_______ |
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| |When you are asked to calculate the expected winnings for a game of chance, don’t forget to take into account the cost to |
| |play the game. |
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| |Suppose the cost to play the game in the previous example is $1. |
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| |Make a table: |
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| |Calculate the [pic]and [pic]: |
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| |The law of large numbers says that the mean outcomes of many trials, (, gets closer to the actual distribution mean μ as |
| |more trials are made. |
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|Find [pic]and [pic]on calculator: |Rule 1. If X is a random variable and a and b are fixed numbers, then |
| |[pic] |
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| |Rule 2. If X and Y are random variables, then |
| |[pic] |
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|Example: | |
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| |b. |
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| |c. |
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| |Rule 1. If X is a random variable and a and b are fixed numbers, then |
|Law of Large Numbers |[pic] |
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|Rules of Means |Rule 2. If X and Y are random variables, then |
| |[pic] |
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| |[pic] |
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| |Rule 3. If X and Y have correlation [pic], then |
| |[pic] |
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| |a. |
|Practice Problems page 425 #7.37 | |
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| |b. |
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|Rules of Variances | |
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|Practice Problems page 426 #7.38 | |
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