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U6 - DAY 2 WARM UP! • Given the times required for a group of students to complete the physical fitness obstacle course result in a normal curve, and that the mean time 21 minutes and the standard deviation is 4… Use the Empirical Rule to Answer the FOLLOWING! • • • • • • What percent took longer than 29 minutes? 2.5% What percent took less than 29 minutes? 97.5% What percent took between 13 and 29 minutes? 95% What percent took between 13 and 25 minutes? 81.5% What percent took longer than 17 minutes? 84% What time would a student need to have to be in the fastest 2.5% < 13 minutes Normal Distribution Probabilities with our Calculators!!! If you are looking for values that are not exactly on the standard deviation lines, there are two ways to find the probability – you can use a normal distribution chart OR you can use a calculator. For this course we will ONLY use the calculator!! Normal Distribution Probability Example #1: A Calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores between 82 and 90? DRAW A PICTURE!!!!! TI 83/84 directions: Normal Distribution Probability Example #1 A calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What is the probability that a student has a score between 82 and 90? a. Press [2nd][VARS](DISTR) [2] (normalcdf) b. Press [82] [,] [90] [,] [78] [,] [5] [)][Enter] normalcdf(x1, x2, μ, σ) normalcdf(82,90,78,5) = .2036578048 There is a 20.37% probability that a student scored between 82 and 90 on the Calculus exam. Normal Distribution Probability Example A calculus exam is given to 500 #1: students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. How many students have scores between 82 and 90? Using the probability previously found: 500 * .2037 = 101.85 There are about 102 students who scored between 82 and 90 on the Calculus exam. Normal Distribution Probability Ex#1: A calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores above 60? Hint: Use 1E99 for upper limit (2nd/EE) normalcdf(60,1E99,78,5) = .9998408543 ~ 99.98% Normal Distribution Probability Example #2 A PreCalculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. How many students have scores above 70? TI 84 2nd/EE Normalcdf(70,1E99,78,5) = .9452007106 500*.9452= 472.6 About 473 students have a score above 70 on the Pre-Calculus exam. Normal Distribution Probability Example #3: Find the probability of scoring below a 1400 on the SAT if the scores are normally distributed with a mean of 1500 and a standard deviation of 200. Hint: Use -1E99 for lower limit Normal Distribution Probability Example #3 Find the probability of scoring below a 1400 on the SAT if the scores are normal distributed with a mean of 1500 and a standard deviation of 200. normalcdf(-1E99,1400,1500,200) = .3085375322 There is a 30.85% probability that a student will score below a 1400 on the SAT. Example #4 Kenny and his English Test! • Scores of each of the previous English tests were normally distributed with a mean of 76 and standard deviation of 4.5. Kenny will be taking the test tomorrow. – What is the probability of Kenny getting at least 70 on the test? (He wants to pass!) ~ 90.9% – What is the probability that he makes between a 93 and a 100? (He really wants and A!) ~ .008% – What is the probability that he makes less than a 50? (Oh no…his parents are going to be angry!!) ~ .0000004% Now let’s think Backwards!! • Example #5 Scores of each of the previous calculus tests were normally distributed with a mean of 86 and standard deviation of 2.2. How high must a student’s score be to score in the top 20%? Draw a picture…it will help! – 2nd /vars/ invNorm (area to the left, μ, σ) – invNorm(0.8, 86, 2.2) = ~ 87.9 or 88 PERCENTILES (Use invNorm with percentiles) • A percentile is the score at which a specified percentage of scores in a distribution fall below. • To say a score of 53 is in the 80th percentile is to say that 80% of all scores are less than 53 More Backwards Questions… Ex #6 Heights are generally found to be normally distributed. Assume that the average adult female height is 65.5 inches with a standard deviation of 3 inches. 2nd /vars/ invNorm (area to the left, μ, σ) – What height do 90% of women fall below? 69.3 in. – What height do 60% of women measure above? 64.7 in. – What height is the 75th percentile of women? 67.5 in. – What two heights, symmetric about the mean, contain 80% of all women's heights? 61.7 in. and 69.3 in. HOMEWORK! Day 2 HW