Probability Distributions in Math 211: Discrete & Continuous with Binomial & Normal Models, Exams of Mathematics

Download Probability Distributions in Math 211: Discrete & Continuous with Binomial & Normal Models and more Exams Mathematics in PDF only on Docsity! Math 211 Distribution Models Simple Statistical Distributions on the TI-83 Math 211 – ASU In this course we look at basic statistical distributions for both the discrete and continuous cases. We start with a random variable X for some scenario. Discrete Case If for just a finite set of x values, then we would say this is a discrete situation. For example, if we flip a coin ten times and let X = number of times heads appears, then the possible values for X are the integers between 0 and 10 inclusive. For these values, 0)( ≥= xXp )( xXp = will be positive, and we can actually calculate these probabilities using the Binomial Probability Model, which we’ll discuss in a moment. It wouldn’t make sense to let X = 4.5 (i.e. we got 4.5 heads in ten flips). From a probability standpoint, we’d say . We could summarize this scenario by saying that: 0)5.4( ==Xp xotherallxXp xxXp ,0)( 10,9,8,7,6,5,4,3,2,1,0,0)( == =≥= Let’s perform this example. We will use the Binomial Probability Model since this scenario meets the following conditions: it is independent from trial to trial, the probability of success does not change, and we are performing multiple trials. The formula for a binomial probability is , where p = probability of success, k = number of trials and n = number of successes.. However, we will use a built-in feature on the TI-83 (and higher) models called “binompdf” to perform these calculations. Hit 2 nkn nk ppCknpB −−⋅⋅= )1()(),,( nd-DISTR on your calculator to find this choice. Please note there is also a choice for binomcdf which we’ll discuss in a moment. The call sequence for binompdf is binompdf(k,p,n), where the variables are as described in the preceding paragraph. The ‘pdf’ stands for probability density function. For our purposes it calculates the probability at a value n. Say we flip our coin 10 times. The probability of heads is p = 0.5. Say we want 5 heads to appear. The call sequence would be binompdf(10,0.5,5), which gives 0.246. Hence we have about a 24.6% likelihood of getting 5 heads in 10 flips. Suppose we want to know the likelihood of getting at most 5 heads. Now we use the cdf feature, which stands for cumulative density function. This calculates the probability of the event occurring for all values from 0 up to and including n. We’d type binomcdf(10,0.5,5) = 0.623, which says there’s a 62.3% likelihood of us getting up to and including 5 heads in 10 flips. binompdf = probability at a value for n binomcdf = probability for the interval 0 up to n. Math 211 Distribution Models We can set up each probability and sketch a distribution model using the binompdf feature. For illustrative purposes, the following table also shows the cumulative (binomcdf) values as n progresses from 0 to 10. n = number of heads Probability of n heads Cumulative probability 0 binompdf(10,0.5,0) = 0.000977 binomcdf(10,0.5,0) = 0.000977 1 binompdf(10,0.5,1) = 0.009766 binomcdf(10,0.5,1) = 0.010742 2 binompdf(10,0.5,2) = 0.043945 binomcdf(10,0.5,2) = 0.054688 3 binompdf(10,0.5,3) = 0.117188 binomcdf(10,0.5,3) = 0.171875 4 binompdf(10,0.5,4) = 0.205078 binomcdf(10,0.5,4) = 0.376953 5 binompdf(10,0.5,5) = 0.246094 binomcdf(10,0.5,5) = 0.623047 6 binompdf(10,0.5,6) = 0.205078 binomcdf(10,0.5,6) = 0.828125 7 binompdf(10,0.5,7) = 0.117188 binomcdf(10,0.5,7) = 0.945313 8 binompdf(10,0.5,8) = 0.043945 binomcdf(10,0.5,8) = 0.989258 9 binompdf(10,0.5,9) = 0.009766 binomcdf(10,0.5,9) = 0.999023 10 binompdf(10,0.5,10) = 0.000977 binomcdf(10,0.5,10) = 1.000000 Notice the cumulative values in the third column increase and eventually sum to 1. This should not be a surprise: if you flip a coin 10 times, it is certain (p = 1) that you’ll get at most 10 heads. Cumulative probabilities always increase in value and top out at 1. A histogram graph of the individual probabilities is given below. Suppose we want to know the probability of getting at least 5 shots, or . The binomcdf works by accumulating probabilities from the left (from zero), so we need to reword this problem so that the cdf features can be used. The complement of making at least 5 is to make at most 4. You would add up the bars from 0 to 4, but leave 5 open. We get binomcdf(10,0.5,4) = 0.376953, from above. Hence, the probability of getting at least 5 heads is 1 – 0.377 (rounded) = 0.623, or about 62.3%. )5( ≥Xp Lastly, suppose we want to know the probability of getting between 3 and 5 heads, or )53( ≤≤ Xp . We can simply add them up: 568.0)5()4()3( ==+=+= XpXpXp (rounded). We can also recognize that this would be the same as )2()5( ≤−≤ XpXp = binomcdf(10,0,5,5) – binomcdf(10,0.5,2) = 0.623047 – 0.054688 = 0.568 (rounded, same as before). In the second method, Math 211 Distribution Models where x is the original value in question and z is its equivalent on the normalized normal curve (called the z-score). So, in this case, z = (105 – 100)/10 = 0.5. We just need to determine the area below 0.5. See the picture for assistance: The left half of the region has an area of 0.5. We just need to find the area between 0 and 0.5. Your calculator has a nifty feature that will calculate this. Go to 2nd-DISTR (above the VARS key) and select “normalcdf” and hit enter. You have to tell it to run from 0 to 0.5, so your command will look like this: normalcdf(0,0.5) This gives 0.19, which added to 0.5 gives 0.69. Hence, the probability a randomly chosen person has an IQ below 105 is . A common trick that works well with the calculator for open- ended intervals is to use -5 and 5 as default left and right bounds when needed (in place of negative and positive infinity). In the above example, we could have typed normalcdf(-5,0.5) and have arrived at the same result. Please note: there is nothing magical with -5 and 5 other than they’re far enough away from the mean to act reasonably well as ‘open-ended’ endpoints. The accuracy is still well within what we want. 69.0)105( =≤Xp Consider a scenario where we want to work backwards. Consider a club that only accepts those who score in the top 2% of IQ. What would be the minimum IQ to get into this club? We treat this as an area problem: we want the z-score that gives an area of 0.02 to the right (see figure). Your calculator has a nifty feature called InvNorm (2nd-DISTR, then choice #3). InvNorm works by returning the z-score for a given area from the left, so in this case we want the z-score that gives an area of 0.98 to the left, which will be the same as the z-score corresponding to the area of 0.02 to the right. InvNorm(0.98) gives us z = 2.05. Now that we have z = 2.05, we can use the renormalization formula to determine x: 2.05 = (x – 100)/10, which with a little algebra, gives x = 120.5. Math 211 Distribution Models Practice problems (Discrete Cases) 1. A salesman has a 40% likelihood of making a sale. One day 25 people visited his shop. Find the probability he… a. Made exactly 12 sales. b. Made at most 12 sales. c. Made at least 13 sales. d. Made between 7 and 12 sales. 2. Jeff target shoots. He has a probability of 0.85 of making his shot. He makes 30 attempts. Find the following probabilities: a. )20( =Xp b. )22( ≤Xp c. )24( ≥Xp d. )2620( ≤≤ Xp Practice problems (Continuous Cases) 1. Using your calculator, determine these probabilities of a normal distribution in the scenario where the mean µ = 0 and the standard deviation σ = 1. This distribution has already been normalized. Use symmetry and complements if needed. =≤≤− )11( Xp =≤≤− )01( Xp =≤≤ )10( Xp Notice anything consistent about the above three? =≤≤− )22( Xp =≤≤− )33( Xp =≤≤ )31( Xp =≤≤ )9.15.0( Xp =−≤ )2.1(Xp =−≥ )5.1(Xp =≥ )75.0(Xp Math 211 Distribution Models 2. In this scenario the mean µ = 12 and the standard deviation σ = 2. Renormalize and find the following probabilities: =≤≤ )139( Xp =≤ )14(Xp =≤≤ )5.107( Xp =≥ )5.8(Xp 3. Suppose the average man’s height is 70 inches (5-10) and the standard deviation is 2.5 inches. a) What is the probability a randomly chosen man stands between 5-10 and 6 feet? b) What is the probability a randomly chosen man stands less than 5-7? c) What is the probability a randomly chosen man stands greater than 5-9? d) What is the probability a man stands above 6-3? e) The tall man’s club admits members in the top 1% of heights. What is the minimum height requirement to get into this club?