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Homework for 2.1 Day 1: 41, 43, 45, 47, 49, 51 1) 2) 3) 4) 5) 6) 7) To use the 68-95-99.7 rule to estimate the percent of observations from a Normal Distribution that fall in an interval Use the standard Normal distribution to calculate the proportion of values in a specified interval Use the standard Normal distribution to determine a z-score from a percentile Use Table A to find the percentile of a value from any Normal distribution Make an appropriate graph to determine if a distribution is bell-shaped Use the 68-95-99.7 rule to assess normality of a data set Interpret a normal probability plot A. A specific type of density curve is a Normal Curve B. The distributions they describe are called Normal distributions C. Characteristics: Have the same overall shape: symmetric, single peaked, bell-shaped b) Mean-π and standard deviation is π c) Mean is in the center and so is median (think back to previous section and symmetrical denisity curves) d) π controls the spread a) ο On the board οKeep in mind: these are special properties of normal distributions NOT ALL DENSITY CURVES!!!! ο Example: The mean of MLB batting averages is 0.261 with a standard deviation of 0.034. Suppose that the distribution is normal (this is key to know)with π = 0.261 πππ π = 0.034 a) Sketch a normal density curve for this distribution of batting averages. (Take notice of points that are 1,2,or 3 standard deviations away from the mean) b) What percent of the batting averages are above 0.329? c) What percent of the batting averages are between 0.193 and 0.295? Page 114 ο Z-Scores!! ο Check out the box on page 115 ο Now, letβs look up our Table A in the back of the bookβ¦ ο These are the values of the z scores, the area under the curve to the left of z ο Example: Suppose we wanted to find the proportion of observations in a Normal distribution that were more than 1.53 standard deviations above the mean. ο Ok, so we want to know what proportion of observations in the standard Normal distribution are greater than, z=1.53 ο First, find the area to the left of z=1.53 in table A ο What is that value? ο Z=1.53 ο ο ο ο ο 0.9370 Now, this is the value to the left of 1.53, or you can think of it as β€1.53 If I want to know greater than 1.53 I shouldβ¦ 1-0.9370=0.0630 Why? AREA UNDER THE CURVE ALWAYS = 1 Example: Find the proportion of observations from the standard Normal distribution that are between -0.58 and 1.79 ο Look up the z-scores for those values: ο Z=-0.58β 0.9633 ο Z=1.79 β0.2810 ο Values between: I should subtract them! ο Answer: 0.6823 ο Look at the beige box on page 120 ο Example: In the 2008 Wimbledon tennis tournament, Rafael Nadal averaged 115 MPH on his first serves. Assume that the distribution of his first serve speeds is Normal with a mean of 115 mph, and a standard deviation of 6 mph. About what portion of his speeds would you expect to exceed 120 mph? Conclude: About 20% of Nadalβs first serves will travel more than 120 mph. ο What percent of Rafael Nadalβs first serves are between 100 and 110 mph? Conclude: About 20% of Nadalβs first serves will travel between 100 and 110 mph ο The heights of three-year old females are approximately Normally distributed with a mean of 94.5 cm and a standard deviation of 4 cm. What is the third quartile of this distribution? ο Hint: Look at the z-chart in reverse!! Conclude: The third quartile of three-year old femalesβ heights is 97.18 cm Plot the data- you can use a dotplot, stemplot, or histogram. See if the shape is a bell 2. Check to see if the data follows the 68-95-99.7 Rule A. Find the mean and standard deviation B. Calculate 1, 2, and 3 standard deviations to the right and left C. Find the percent of the data that lies between those standard deviations 1. 12.9, 13.7, 14.1, 14.2, 14.5, 14.5, 14.6, 14.7, 15.1, 15.2, 15.3, 15.3, 15.3, 15.3, 15.5, 15.6, 15.8, 16.0, 16.0, 16.2, 16.2, 16.3, 16.4, 16.5, 16.6, 16.6, 16.6, 16.8, 17.0, 17.0, 17.2, 17.4, 17.4, 17.9, 18.4 No space in the fridge? 4 Frequency 3 2 1 0 12.9 14.1 14.5 14.7 15.2 15.5 15.8 16.2 16.4 16.6 Usable Capacity 17 17.4 18.4 ο Combined with the graph, and the fact that these numbers are extremely close to our rule, we have good evidence to believe this is a Normal Distribution