Probability Distributions

Transcription

1 Chapter 6 Discrete Probability Distributions Section 6-2 Probability Distributions Definitions Let S be the sample space of a probability experiment. A random variable X is a function from the set S into a target set T. X is a discrete random variable if the target set is (a subset of) the integers. In other words, discrete random variables have values that can be counted. X is a continuous random variable if the target set is (a subset of) the real numbers. Continuous random variables occur in probability experiments whose outcomes are measured rather than counted. Examples of discrete random variables: the number of correct answers on a multiple-choice test, the number of heads when tossing three coins, the number of daily phone calls at a switchboard, the sum of points when rolling two dice. Examples of continuous random variables: the body temperature of rhinoceroses, the thickness of the Greenland Ice Cap, the scores (out of 100) on an exam, the length of a phone call at a switchboard, the value of the Dow Jones Industrial Average. In this chapter we will only be concerned with discrete random variables - whose values can be counted. Example 1 Three coins are tossed. The sample space is {(T,T,T), (T,T,H), (T,H,T), (T,H,H), (H, T, T), (H, T, H), (H, H, T), (H, H, H)}, where T means tails, and H means heads. We consider the random variable X: number of heads when tossing the three coins. We get the following table for the possible values of X, and their probabilities: 33

2 i: Number of heads outcomes (T,T,T) (T,T,H) (H, H, T) (H, H, H) (T,H,T) (H, T, H) (H, T, T) (T,H,H) Probability P (X = i) 1/8 3/8 3/8 1/8 Discrete Probability Distributions The table i: Number of heads Probability P (X = i) 1/8 3/8 3/8 1/8 is an example of a discrete probability distribution. In general, a discrete probability distribution gives the values of a discrete random variable, along with their probabilities. We use the expression P (X = i) to denote the probability of the event that the random variable X has the value i. Properties of Discrete Probability Distributions then: The sum of all probabilities P (X = i) is equal to 1; i.e., P (X = i) =1. If X is a discrete random variable, Each probability P (X = i) must lie between 0 and 1; i.e., 0 P (X = i) 1. Graphs of Discrete Probability Distributions We graph probability distributions by representing the values i of X on the horizontal axis, and the probabilities P (X = i) onthe vertical axis. For the probability distribution in Example 1, we obtain the following graph: P X X 34

3 Example 2 Find the probability distribution for the sum of points when rolling 2 dice. The random variable X denotes the sum of points. i outcomes P (X = i) The graph for this distribution is: Practice Problems for , 6-9, 6-13, 6-15, 6-17, 6-19, 6-23, 6-27 (construct a probability distribution for (a): X=$-value on one draw, (b): X=sum of $-values on two draws (with replacement); also, redo Example 2, where X now denotes the product of points when rolling two dice. 35

4 Section 6-3 Mean, Variance, and Expectation Example 1 Three coins are tossed. Let X denote the number of heads when tossing the three coins. In the previous section, we found the following probability distribution: i: Number of heads Probability P (X = i) 1/8 3/8 3/8 1/8 This means: if the three coins are tossed, say, 800 times, then we expect to have 100 times 0 heads, 300 times 1 head, 300 times 2 heads, and 100 times 3 heads. The expected number of heads is therefore ( )/800 = (1/8) 0+(3/8) 1+(3/8) 2+(1/8) 3=3/8+6/8+3/8 =1.5 Expected Value (Mean) for a Probability Distribution If X is a discrete random variable with possible outcomes i 1,i 2,...,i n, then the expected value E(X) ofx is given by the formula E(X) =P (X = i 1 ) i 1 + P (X = i 2 ) i P (X = i n ) i n E(X) = P (X) X,using Σ-notation Sometimes the expected value E(X) is called the mean of the random variable X. Sometimes we also write µ instead of E(X). Example 2 The profit X for a farmer is governed by the following probabilities. If the season is very dry, the farmer loses $10,000. The probability of a very dry season is 32%. If the season is very wet, the farmer loses $25,000. The probability of a very wet season is 9%. Otherwise, the farmer makes a profit of $50,000. The expected profit is Variance and Standard Deviation for a Probability Distribution If X is a discrete random variable with possible outcomes i 1,i 2,...,i n, then the variance VAR(X) ofx is given by the formula VAR(X) =P (X = i 1 ) i P (X = i 2 ) i P (X = i n ) i 2 n [E(X)] 2 36

5 VAR(X) = P (X) X 2 [E(X)] 2,using Σ-notation Sometimes the variance VAR(X) is denoted by σ 2. We note that VAR(X) =E([X E(X)] 2 ), i.e. it is the expected quadratic deviation from E(X). The standard deviation is given by S(X) = VAR(X). Example 2 (cont d) We found that E(X) = 24, 050. The variance is and the standard deviation is Example 3 A fair die is cast. Find the mean and the standard deviation. We have that P (X = i) = 1 6 for i =1, 2, 3, 4, 5, 6. So, and So, Practice Problems for , 6-39,

6 Section 6-4 The Binomial Distribution Situation A binomial probability experiment consists of n independent repetitions (or trials ) of a basic probability experiment. The basic probability experiment has two possible outcomes or outcomes that can be reduced to two outcomes. One outcome is considered a success, the other a failure. The probability of a success S is denoted by p and the probability of a failure is denoted by q. Then clearly, q =1 p. For a binomial probability experiment with n trials, we consider the discrete random variable X : number of successes Then we say that X is binomially distributed with n trials and probability of success p, orwe write X B(n, p). For a B(n, p)-distributed random variable X, we have that the probability of i successes is: P (X = i) = n! (n i)! i! pi (1 p) n i. Example 1 If a die is cast five times, find the probability of getting exactly four 6 s. The random variable X: number of 6 s is B(5, 1/6)-distributed (5 trials, probability of a six=1/6), and P (X =4)= 5! (5 4)! 4! (1/6)4 (1 (1/6)) 5 4 = (1) ( ) (1/6)4 (5/6) = = We note that the number 5! (5 4)! 4! =5 corresponds to the number of outcomes with exactly four 6 s, which are (S, S, S, S, F ), (S, S, S, F, S), (S, S, F, S, S), (S, F, S, S, S), (F, S, S, S, S). (S denotesa6-a success ; F denotes a 1,2,3,4, or 5 - a failure.) The probability of each outcome is (1/6) 4 (5/6). Remark Note that the expression n! (n i)! i! = n C i. Example 2 Public Opinion reported that 5% of Americans are afraid of being alone in a house at night. If a random sample of 20 Americans is selected, find the following probabilities: 38

7 (1) There are exactly five people in the sample who are afraid of being alone at night. Let X: number of people who are afraid of being alone in a house at night. Then X is B(20, 0.05)-distributed. Here, we need to find P (X =5): (2) There are at most three people in the sample who are afraid of being alone at night. Find P (X 3) = P (X =0)+P (X =1)+P (X =2)+P (X =3): P (X =0)= P (X =1)= P (X =2)= P (X =3)= So, P (X 3) =. (3) There are at least three people in the sample who are afraid of being alone at night. Expected Value and Variance If the random variable X is B(n, p)-distributed, then: the expected value (mean) for X is E(X) =n p, thevarianceforx is VAR(X) =n p (1 p), the standard deviation is S(X) = n p (1 p). 39

8 Example 3 A multiple choice test consists of 10 questions, and each question has of 5 choices. What is the expected number of correct answers if this guy took the test? Practice Problems for , 6-67, 6-69, 6-73, 6-76, 6-79, 6-83,