Chapter 4 DISCRETE RANDOM VARIABLES Introductory statistics Chapter 4 DISCRETE RANDOM VARIABLES PowerPoint Image Slideshow

Sec. 4.2: Probability distribution function for a discrete random variable A Random Variable (RV) is a characteristic of interest in a population being studied. A discrete probability distribution function (PDF) has two characteristics: Each probability is between zero and one, inclusive. The sum of the probabilities is one.

Example: The following table is a valid probability distribution for a random variable X. What must be the value for P(2) to complete the table? X P(X) 0.15 1 0.2 2 3 0.4 What is P(2 or less)? What is P(1 or greater)?

Find the probability of flipping 2 heads. Example: A random variable X is defined as the number of heads observed when a coin is tossed 4 times. The probability distribution for this random variable is shown below. Find the probability of flipping 2 heads. What is the probability of flipping 2 or more heads? What is the probability of flipping 2 or fewer heads? Which is greater, the probability of not tossing 2 heads, or the probability of tossing 2 heads? X P(X) 1 16 1 4 16 2 6 16 3 4

Sec. 4.3: Mean or expected value and standard deviation The expected value is often referred to as the "long-term" average or mean (denoted by µ). This means that over the long term of doing an experiment over and over, you would expect this average. To find the expected value or long term average, μ, multiply each value of the random variable by its probability and add the products.

Example: Using the previous PDF, find the expected value P(X) 1 16 1 4 16 2 6 16 3 4 Expected value: 0∙ 1 16 +1∙ 4 16 +2∙ 6 16 +3∙ 4 16 +4∙ 1 16 =0+ 4 16 + 12 16 + 12 16 + 4 16 = 32 16 =2

Example: Raffle Tickets Raffle tickets sell for $1 each. Imagine 10 tickets were sold for $1 each. The winner of the raffle will take home $5 (half of ten), and everyone else loses. Find the expected value of each ticket. X = value of a ticket (There are two options, you win (net $4), or lose (net -$1)) The probability of winning is 1/10 or 0.1 for winning and 9/10 or 0.9 for losing. The expected value is: 4∙0.1+ −1 ∙0.9=0.4−0.9=−0.5 This means, on average, you will lose $.50 per ticket.

Another raffle example Suppose that in order to raise money for a local seniors citizens home, the town council for Pickering decides to hold a charity lottery: Enter the Charity Lottery One Grand Prize of $20,000 20 additional prizes of $500 Tickets only $10 After reading the fine print, you discover that 10,000 tickets will be sold. Is this a good bet?

Which investment has a better expected value? Below are two alternative investment projects involving and initial investment of $1000. The possible payoffs for each alternative, and the respective probabilities of that payoff are listed below. : Investment 1 Investment 2 Payoff Probability $500 .20 $800 .25 $1000 .40 .35 $1500 $1375 Which investment has a better expected value?

Standard Deviation with PDFs (Create a column that is 𝑥−𝜇 2 𝑃(𝑥), then add that column and take the square root).

Example: Airline Delays A small airline company has only three flights per day. The number of delayed flights per day is regarded as a random variable, calculate the expected value and standard deviation given the probability distribution: # of Delays Probability of Delay 0.5 1 0.3 2 0.1 3 Add (X-mu)P(X) Column

Example: stereo Speakers Some stereo speakers are manufactured with a probability of 20% being defective. Three are selected off a continuous assembly line. Define the random variable X as the number of the defective parts. Find: a) the density function and the expected value for the defective parts b) the expected value for the good parts c) the standard deviation X P(X) 1 2 3 P(0)=(0.8)^3 P(1)=3(0.8)^2 (0.2) P(2)=3(0.8) (0.2)^2 P(3)=(0.2)^3 Create (x-mu)P(X) column.

Sec. 4.4: Binomial Distribution There are three characteristics of a binomial experiment. There are a fixed number of trials, denoted by the letter n. There are only two possible outcomes, called "success" and "failure," for each trial. The letter p denotes the probability of a success on one trial, and q denotes the probability of a failure on one trial. p + q = 1. The n trials are independent and are repeated using identical conditions. Because the n trials are independent, the outcome of one trial does not help in predicting the outcome of another trial. The outcomes of a binomial experiment fit a binomial probability distribution, written as 𝑋~𝐵(𝑛,𝑃). The random variable X = the number of successes obtained in the n independent trials.

The outcomes of a binomial experiment fit a binomial probability distribution, written as 𝑋~𝐵(𝑛,𝑃). The random variable X = the number of successes obtained in the n independent trials. The mean, μ, for the binomial probability distribution is μ = np and the standard deviation, σ, is 𝜎= 𝑛𝑝𝑞 .

Example: Notation What is the probability of rolling exactly two sixes in 6 rolls of a die? There are five things you need to do to use binomial distribution. Define Success first. Success must be for a single trial. Success = "Rolling a 6 on a single die" Define the probability of success (p): p = 1/6 Find the probability of failure: q = 5/6 Define the number of trials: n = 6 Define the number of successes out of those trials: x = 2 Notation: Probability of 2 sixes in 6 rolls: P(x=2) Binomial distribution: 𝑋~𝐵(6,1/6).

Using a Calculator Go into 2nd DISTR. The syntax for the instructions are as follows: To calculate P(x = value): binompdf(n, p, number) if "number" is left out, the result is the binomial probability table. Ex. Calculate P(x=2) from last example (enter Binompdf(6,1/6,2)) To calculate P(x ≤ value): binomcdf(n, p, number) if "number" is left out, the result is the cumulative binomial probability table. Ex. Calculate P(x≤2) from last example (enter binomcdf(6,1/6,2)) To calculate P(x > value): 1 - binomcdf(n, p, number) Ex. Calculate P(x>2) from last example (enter 1-binomcdf(6,1/6,2))

Example: Hospital Records Hospital records show that of patients suffering from a certain disease, 75% die of it. Six patients are chosen at random. Define X. What values can X take on? X=number of patients recovering. X=0,1,2,…,6 Give the distribution of X. 𝑋~𝐵(_,_). 𝑋~𝐵(6,.25). How many of the 6 do you expect to survive? 0.25·6 Find the probability 4 of the 6 patients will survive. P(x=4)=binompdf(6,0.25,4) Find the probability at most 4 patients will survive. P(x≤4)=binomcdf(6,0.25,4)

Example: Telephones In the old days, there was a probability of 0.8 of success in any attempt to make a telephone call. (This often depended on the importance of the person making the call, or the operator's curiosity!) Calculate the probability of having 7 successes in 10 attempts. X=number of successful phone calls. X=0,1,2,…,10 𝑋~𝐵(10,0.8). P(x=7) = binompdf(10,0.8,7)

Example: A Blind Marksman A (blindfolded) marksman finds that on the average he hits the target 4 times out of 5. If he fires 4 shots, what is the probability of (a) more than 2 hits? (b) at least 3 misses? X=number of hits, X=0,1,2,3,4 𝑋~𝐵(4,4/5). P(x>2)=1-binomcdf(4,4/5,2) P(x≤1)=binomcdf(4,4/5,1)

Find Mean and Standard Deviation A manufacturer of metal pistons finds that on the average, 12% of his pistons are rejected because they are either oversize or undersize. What is the probability that a batch of 10 pistons will contain (a) no more than 2 rejects? (b) at least 2 rejects? (c) Find the mean number of rejects and standard deviation. Answer: X=number of rejects, 𝑋~𝐵(10,0.12). P(X≤2)=binomcdf(10,0.12,2) P(X≥2)=1-binomcdf(10,0.12,1) μ = np = 10·0.12 = 1.2 and 𝜎= 𝑛𝑝𝑞 = 10∙0.12∙0.88 =1.028

Practice: It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose you randomly survey 11 California residents. We are interested in the number who have adequate earthquake supplies. (a) In words, define the random variable X. (b) List the values that X may take on. (c) Give the distribution of X. X ~ _____(_____,_____) (d) What is the probability that at least eight have adequate earthquake supplies? (e) Is it more likely that none or that all of the residents surveyed will have adequate earthquake supplies? Why? (f) How many residents do you expect will have adequate earthquake supplies?