A 1 2 B C D E F G H I J K L M N O P Name: MTH 243 Lab 1 3 4 5 The Normal Distribution We will examine the Normal Distribution and use area under the curve to compute probabilities. 6 Notes & Requirements: Not all unimodal, symmetric curves are normal. 9 For this lab, we will assume everything is normal or can be modeled by a normal distribution. 10 For this lab, complete it in groups of either 1 or 2 or 3 students. Please email one completed spreadsheet to me. Clearly indicate in the email which class you are in and that this is lab 1. 7 8 11 12 13 14 1.) True or False: The area undera probability density curve must sum to 1. Explain your answer: 15 16 Recall z-scores, we use these compute position of a value relative to the mean. 17 The Context: The mean ACT score of all those who took the test in 2010 was 21 and the standard deviation was 4.7. 20 2.) Compute the z-score for a score of 23. Describe what this means for the position of the score. 21 z = 22 Meaning: 18 19 23 To compute the probability that a student gets a certain score, we can use the normal distrubution to make an approximation. We assume that the distribution of score is approximatley normal. 26 To use a continuous probability distribution, that is approximtely normal, we must work with intervals rather than specific numbers. 27 3). Why must we work with intervals of values? 24 25 28 29 30 31 32 33 34 35 36 37 38 39 40 41 In class, when working with normal distritbutions, we relied on the table of values in the z-table (pp. 288-298) . Alternately, we can evaluate probability density functions and add up the area under the curve by using functions in either Excel or Google Docs or on your Ti-83/84 calculator. Excel instructions The function we will use is NORM.DIST, and here is the usage: Google Docs instructions NORM.DIST( x, mu, sigma, logical) NORMDIST( x, mu, sigma, C) where where x = the score (raw data or z score) x = the score (raw data or z score) mu = the mean (0 if using a z-score) mu = the mean (0 if using a z-score) sigma = the standard deviation (1 if using a z-score) sigma = the standard deviation (1 if using a z-score) logical = TRUE if you wan the area under the curve up to (or to the left of x) C = 1 (binary TRUE) if you wan the area under the curve up to (or to the left of x) logical = FALSE if you wan the value of the probability density function (just the y coordinate of the graph) C = 0 (binary FALSE) if you wan the value of the probability density function (just the y coordinate of the graph) (Depending on the version of Excel, you may have to use NORMDIST(x , mu, sigma, logical) - just remove the dot!) 42 Ti-84 instructions 2nd --> VARS(DIST) --> 2:normalcdf( -->enter in: lower bound-comma-upperbound-comma-mean-comma-St Dev 45 Note: use very large values, like 1000, when for bounds that are infinite 43 44 46 Example: Looking at the z table, for a z-score of -0.42 We read that P(z < -0.42) = 0.3372 49 In Excel, we compute this as: "=NORM.DIST(-0.42, 0, 1, TRUE)" 0.33724273 50 On the Ti-84, we compute as: normalcdf(-1000,-.42,0,1) = 0.3372427624 [remember to use the "negative" key not the "subtract" key] 47 48 51 4.) Use Excel or Ti claculator to compute the following probabilities (don't forget the rules of probability and DRAW PICTURES TO HELP YOU!!!!!): a.) P( Z > -2.98 ) = 0.00144124 54 b.) P( -.25 < Z < 1.96 ) = 55 c.) P( Z < -2.56 or Z > 1.39 ) = 56 Hint for part c: Find P(Z<-2.56) + P(Z>1.39) = 52 53 57 58 59 60 61 62 63 64 65 66 We can also reverse this process much like you may have done in your homework. That is, given a probability (area) can you determine the corresponding z value. In Excel, this is accomplished by "=NORM.INV(probability, mu, sigma)" where Google Docs: NORMINV( prob, mu, sigma) probability = the area to the left of the z score, or the probability where mu = mean prob = the area to the left of the z score, or the probability sigma = the standard deviation mu = mean Example, a z-value of -3.4 has a probability of 0.0003 according to our z table. sigma = the standard deviation If we use "=NORM.INV(0.0003, 0, 1)"we get -3.4316144 (Same issue: you may have to remove the dot and use NORMINV(prob, mu, sigma) ) 67 68 69 Ti-84 instructions 2nd --> VARS(DIST) --> 3:invNorm( -->enter in: prob for less than the value-comma-mean-comma-St Dev 70 71 72 Example, a z-value of -3.4 has a probability of 0.0003 according to our z table. If we use "invNorm(0.0003, 0, 1)" we get -3.4316144 73 74 5.) Now you try it. HINT: DRAW PICTURES BEFORE YOU RUN TO TECHNOLOGY!!!! a.) If P( Z < a ) = 0.9933, find a. 77 b.) If P( Z >a ) = 0.7337, find a. 78 c.) If P( -b < Z < b ) = 0.1976, find b 79 d.) If P( 0 < Z < a ) = 0.9595, find a. 75 76 80 81 82 83 84 85 86 6.) So now that we have the skills, let's get back to the problem and use them. a.) ACT scores are not a continuous variable. Why? Even though ACT scores are not continuous, if we assume the distribution is bell-shaped, we can use the normal distribution to approximate the distribution of ACT scores Assume that the distribution is normal and that the number of scores is large enough so our approximation makes sense. Recall: The mean ACT score of all those who took the test in 2010 was 21 and the standard deviation was 4.7. b.) Given the above information about ACT scores (mean and standard deviation), what proportion of WOU students scored a 25 or higher? 17 to 25? 87 88 89 90 91 92 93 94 95 96 97 98 c.) Suppose Western Oregon University requires an ACT score of 19 or higher, what is the probabiility that a senior in high school earns a score high enough to be admitted at WOU