AP Stats Exam Review: Probability etharrington@wcpss.net TeacherWeb.com

The Normal Distribution Symmetric, mound-shaped Mean = median Area under the curve = 100% 68-95-99.7% rule - sometimes called the Empirical Rule Crazy formula – no need to know it!

The 68-95-99.7% Rule The following shows what the 68-95-99.7 Rule tells us:

For example: 1998 #4 (partial): A company is considering implementing one of two quality control plans – plan P and plan Q – for monitoring the weights of car batteries that it manufactures. If the manufacturing process is working properly, the battery weights are approximately Normally distributed with mean = 2.7 lbs and standard deviation = 0.1 lbs. (a) Quality Control Plan P calls for rejecting a battery as defective if its weight falls more than 1 standard deviation below the mean. If a battery is selected at random, what is the probability that it will be rejected by plan P?

This is the area we’re interested in! Solution to 1998 #4 (a): This is the area we’re interested in! The answer: About 16%

Finding Normal Probabilities When a data value doesn’t fall exactly 1, 2, or 3 standard deviations from the mean, we used to use a Z Table to find the area to the left of the value. Now, we mostly just use our calculators to find areas under a normal curve.

Finding Normal Probabilities (cont.) Table Z is the standard Normal table. We have to convert our values to z-scores before using the table. The figure below shows how to find the area to the left when we have a z-score of 1.80:

Calculator Use Normalcdf – gives area under a normal curve. You can use z-scores or raw data values: normcdf (lower, upper, mean, st dev) invNorm – gives z-score or raw data value associated with a percentile: invNorm(area to left as decimal, mean, st dev) Provide complete communication on FR questions – use drawings and/or identify the values in your calculator syntax.

For example, 2003 #3: #3: Men’s shirt sizes are determined by their neck sizes. Suppose that men’s neck sizes are approximately normally distributed with mean 15.7” and standard deviation 0.7”. A retailer sells men’s shirts in sizes S, M, L, and XL, where the shirt sizes are defined in the table below: Shirt Size Neck Size S 14”-15” (b) Using a sketch of a normal M 15”-16” curve, illustrate and calculate L 16”-17” the proportion of men whose XL 17”-18” shirt size is Medium.

Solution to 2003, #3 (b) (shirt sizes): Normalcdf (15, 16, 15.7, 0.7) = 0.5072

Question: When do we multiply? Answer: when we want to find the probability of a series of consecutive events Example: We just found that 0.5072 of all men wear size Medium shirts. Suppose we select 3 men at random from a huge population. What is the probability that all three of them wear Medium shirts? Answer:

The BINOMIAL Distribution! What if we are counting how many successes will occur in repeated trials? The BINOMIAL Distribution! (1) Fixed number of trials (“n”) (2) Fixed probability of success on each trial (“p”) (3) Two outcomes per trials (success or failure) (4) Trials are independent of each other

Binomial on the Calculator Binomialpdf(n,p,x-value) > only the probability of that specific x-value Binomialcdf (n,p,x-value) > the sum of the probabilities of x=0 up through and including that specific x-value Example: We found that 50.72% of all shirts are mediums. If we select 12 shirts at random, what is the probability that… Exactly 4 of them are Mediums? No more than 4 of them are Mediums? At least 4 of them are Mediums? 0.1139 0.1802 0.9337

More Binomial! For a Binomially-distributed variable, the parameters are… “n” > the number of repeated trials “p” > the success probability in each trials If X is a Binomially-distributed variable, then µ = E(x) = np & σ = sqrt {n·p·(1-p)} ** We found that 0.5072 of men wear Medium shirts. If n=12 are repeatedly sampled, what is the mean and standard deviation of X, the # of Mediums that we find?

Question: When do we add? Answer: when the question can be construed as an “OR” situation Example: We found that 0.5072 of all men wear size Medium shirts. Out of a random sample of 12 men, what is the probability that either 3 or 4 (exactly) are Mediums? Pr(x=3 or x=4) = Pr(x=3) + Pr(x=4) Do each of these as a binomialpdf = 0.049 + 0.1139 = 0.1629

When ADDING, be careful about over-counting! Example: Using the chart on the board, if I select one bird at random, what is the probability that the bird I select is either a chick or is yellow? Pr(chick or yellow) = Pr(chick) + Pr(yellow) minus Pr(both chick & yellow) = 18/25 + 13/25 – 8/25 = 23/25

Note: If there hadn’t been any yellow chicks, then there wouldn’t be anything to subtract, because we wouldn’t have overcounted. The characteristics of “yellow” and “chick” would then be called MUTUALLY EXCLUSIVE.

Combining Random Variables X+Y or X-Y Means will be added or subtracted just like the random variables Variances will be added ONLY IF VARIABLES ARE INDEPENDENT! Remember: Variance = standard deviation squared ADD variances even if subtracting variables

Simulations When describing a simulation, be sure to include details about: How you will assign random digits to represent outcomes How you will move through the table How duplicates will be handled What will be recorded for each trial When will the simulation be stopped

The 2017 A.P. Statistics Exam! Thursday, May 11, at noon