Welcome to MM207 - Statistics! Unit 4 Seminar Good Evening Everyone! To resize your pods: Place your mouse here. Left mouse click and hold. Drag to the right to enlarge the pod. To maximize chat, minimize roster by clicking here

Definitions Statistical Experiment: Any process by which we obtain measurements or data. In Unit 3 seminar we discussed roulette. Spinning the roulette wheel is a statistical experiment. Random Variable: A random variable is the outcome of a statistical experiment. We don’t know that this outcome will be before conducting the experiment Discrete random variable: The possible values of the experiment take on a countable number of results. For the roulette case the there were 38 possible results. Continuous random variable: The possible values of the experiment are infinite. For example, measure the weight of 1 year old cows is continuous. The number of possibilities are uncountable. (500.12 pounds, 534.1534 pounds, etc…

Probability Distribution Score, x Probability, P(x) 1 0.16 2 0.22 3 0.28 4 0.20 5 0.14 Probability distribution is the assignment of probabilities to specific values for a random variable or to a range of values for the random variable Plain English: Each random variable (outcome) from a random experiment has a particular probability of occurring. See page 196 Example 2

Probability distribution properties Mean: This is the expected value of a probability distribution. This is the outcome about which the distribution is centered. μ = ∑ x P(x) Standard deviation: This is the spread of the data around the expected value (mean) σ = √ ∑ (x – μ)2 P(x) How do you use the equations? Let’s work an example.

Mean and Standard Deviation: Example 2 (page 199) P(x) x P(x) x - μ (x-μ)2 (x-μ)2 P(x) 1 0.16 1 * 0.16 = 0.16 1 – 2.94 = -1.94 (-1.94)2 = 3.764 3.764 * 0.16 = 0.602 2 0.22 0.44 -0.94 0.884 0.194 3 0.28 0.84 0.06 0.004 0.001 4 0.20 0.80 1.06 1.124 0.225 5 0.14 0.70 2.06 4.244 0.594 ∑ = 2.94 1.616 μ = ∑ x P(x) = 2.94 σ = √(x-μ)2 P(x) = √ 1.616 = 1.271219 ≈ 1.3

Features of the Binomial Experiment Fixed number of trials denoted by n n trials are independent and performed under identical conditions Each trial has only two outcomes: success denoted by S and failure denoted by F For each trial the probability of success is the same and denoted by p. The probability of failure is denote by q and q = 1 - p) The central problem is to determine the probability of x successes out of n trials. P(x) = ?

Example n = 10 p = 0.4 x = 6 Find P(x = 6) Using the binomial table (Table 2 A8-A10) Using the binomial formula Using Excel

Using the Binomial Table p = 0.4 x = 6 Find P(x = 6) Find the block for 10 Find the row for 6 Find the column for 0.4 P(x = 6) = 0.111

Using the Binomial Formula P(x) = nCx px qn-x x = number of successes n = number of trials p = probability of one success q = probability of one failure (1 – p) nCx is the binomial coefficient give by nCx = n! / [x! (n-x)!] Remember 4! = 4*3*2*1 = 24 and is called factorial notation.

Using the Binomial Formula p = 0.4 x = 6 Find P(x = 6) P(x) = nCx px qn-x nCx = 10C6 = 210 px = 0.46 = 0.004096 qn-x = 0.610-6 = 0.64 =0.1296 210* 0.004096 * 0.1296 = 0.111476736 ≈ 0.111

Using Excel n = 10 p = 0.4 x = 6 Find P(x = 6) Click on the cell where you want the answer. Under fx, find BINOMDIST Number_s: Enter 6 Trials: Enter 10 Probability: Enter 0.4 Cumulative: False P(x = 6) = 0.111476736 ≈ 0.111

Finding Cumulative Probabilities x ≤ 6 Find P(x ≤ 6) Find each probability using the binomial table or the formula. P(x ≤ 6) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) = 0.006 + 0.040 + 0.121 + 0.215 + 0.251 + 0.201 + 0.111 = 0.945 Use the complement P(x ≤ 6) = 1 – P(x > 6) = 1 – [P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10)] = 1 – [0.042 + 0.011 + 0.002 + 0.000] = 1 – 0.055 = 0.945

Finding Cumulative Probabilities con’t x ≤ 6 Find P(x ≤ 6) Use Excel Number_s: Enter 6 Trials: Enter 10 Probability: Enter .4 Cumulative: True P(x ≤ 6) = 0.945238118 ≈ 0.945

Mean and Standard Deviation of the binomial probability distribution Mean or expected number of success μ = np Standard deviation σ = √ npq Where: n = number of trials p = probability of success q = probability of failure (q = 1 – p)

Computing the Mean, Standard Deviation, and Variance for a Binomial Distribution n = 10 p = 0.4 Mean μ = np μ = 10 * 0.4 μ = 4 Standard deviation σ = √ npq σ = √ 10 * 0.4 * 0.6 σ = √ 2.4 ≈ 1.549 Variance σ2 = npq = 2.4