Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
Source data Americans’ heights The mean, The standard deviation, Men 69.0 inches 2.8 inches Women 64.0 inches 2.8 inches (You will find various different values for these measurements at various internet sites. These are typical.) Excel Graph of this Normal Distribution This was not easy to do. See xzBoth.xlsx American men's heights -4 -3 -2 -1 z-score 0 1 2 3 4 80.2 77.4 74.6 71.8 69 66.2 63.4 60.6 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 57.8 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 x value = Height (inches) Notice that there are TWO x-axes, TWO horizontal axes: x-to-z and z-to-x conversions If you know the x-value (height in inches) and you want to convert it to a z-score (“standard score”) Subtract: the value, minus the mean, . Then divide by the standard deviation, . If you know the z-score and you want to convert it to an x-value of height, in inches: Multiply times the standard deviation, . Add the mean, . Notes.docx 6/16/2011 2:36 PM D.R.S. Notation – Funny ways they write things What they What it TI-84 Excel write means ( ) “z of 0.05”, the value for which the area to the right of this value is exactly 0.05, or 5% of the total area. (If you happen to use 0.05, you get the negative of the desired .) ( ) ( The area to the left of z=1.11 ) ( The area to the right of z=1.11 ) The area between and Notes.docx 6/16/2011 2:36 PM D.R.S. Solution Methods If using the printed tables of Areas of the Standard Normal Distribution Convert all values to values. Work the problem as a problem. If using TI-84 or Excel Special functions are available to work in terms of the values of the mean and standard deviation. TI-84 tools The menu with these functions isUse 2ND DISTR (on the VARS key. Use Function to use normalcdf(low x, high x, mean, stdev) (2ND Comma for EE, which means What it tells you Area between the two x-values ) for and for . Similar to z-problem function =normalcdf(low z, high z) Excel functions These functions for working with area (probability) in terms of x-values are similar to the ones we’ve used for area (probability) in terms of z-values. Function to use What it tells you =NORM.DIST( x-value, mean, stdev, TRUE) Area to the left of that x-value =NORM.INV(probability area, mean, stdev) What x-value has that area to its left. (For Excel 2007 and earlier, omit the periods from the function names.) Similar to z-problem function =NORM.S.DIST(z-value) =NORM.S.INV(probability area) Examples TI-84 solution Problem “What percent of American men are shorter than 5’ 6’’ (66 inches)?” Excel solution =NORM.DIST(66, 69, 2.8, TRUE) gives area to the left of x = 66. Find the area under the curve to the left of x = 66. Answer: 0.1420, or 14.20% Notes.docx 6/16/2011 2:36 PM D.R.S. TI-84 solution Problem “What percent of American men are 6’4’’ (76 inches) or taller?” Excel solution = 1-NORMDIST(76, 69, 2.8, TRUE) Find the area under the curve to the right of x = 76. Subtract 1 minus the area to the LEFT of x = 76 to get the area to the RIGHT of x = 76. = NORMDIST(70, 69, 2.8, TRUE) - NORMDIST(68, 69, 2.8, TRUE) Answer: .0062, or 0.62%, less than 1%. “What percent of American men are within an inch of the mean height, that is, between 68 and 70 inches tall?” Find the area under the curve between x = 68 and x = 70. Answer: 0.2790, or 27.90% “What is the range of heights of the middle 25% of American men?” =NORM.INV(0.3750, 69, 2.8) and =NORM.INV(0.6250, 69, 2.8) 1 - 0.2500 area in the middle = 0.7500 area in the tails, 0.7500 div by 2 = 0.3750 in each tail The low endpoint has 0.3750 area to its left. The right endpoint has the 0.3750 left endpoint + the 0.2500 width = 0.6250 area to its left. Answer: 68.1 to 69.9 inches (5’ 8.1’’ to 5’ 9.9’’) “What is the height that separates the top 10% from everybody else?” =NORM.INV(0.9000, 69, 2.8) Answer: 72.6 inches, or 6’ 0.6’’, about 6 ft + ½ inch tall If the bottom quartile of Midstate League players are under 6’ 2’’ =NORM.S.INV(0.2500) (74 inches) tall, and the mean height is 6’ 4’’ (76 inches), what is =(74-76)/NORM.S.INV(0.2500) the standard deviation of all the players in the league? Or What z-score separates the bottom 25% from the top 75%? It’s 0.6745. Then plug in to or and solve for unknown . (74 – 76) / -0.6745 = 2.97 or 3.0 inches Notes.docx 6/16/2011 2:36 PM D.R.S.