Week_6_Lab

# Week_6_Lab - z-Score -3.50 -3.25 -3.00 -2.75 -2.50 -2.25...

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z-Score -3.50 0.00023 0.02% -3.25 0.00058 0.06% -3.00 0.00135 0.13% -2.75 0.00298 0.30% -2.50 0.00621 0.62% -2.25 0.01222 1.22% -2.00 0.02275 2.28% -1.75 0.04006 4.01% -1.50 0.06681 6.68% -1.25 0.10565 10.56% -1.00 0.15866 15.87% -0.75 0.22663 22.66% -0.50 0.30854 30.85% -0.25 0.40129 40.13% 0.00 0.50000 50.00% 0.25 0.59871 59.87% 0.50 0.69146 69.15% 0.75 0.77337 77.34% 1.00 0.84134 84.13% 1.25 0.89435 89.44% 1.50 0.93319 93.32% 1.75 0.95994 95.99% 2.00 0.97725 97.72% 2.25 0.98778 98.78% 2.50 0.99379 99.38% 2.75 0.99702 99.70% 3.00 0.99865 99.87% 3.25 0.99942 99.94% 3.50 0.99977 99.98% Area to the Left Area to the Left This table is similar to the table used in our textbook. Select any cell under Column B to see the Excel formula used. See the worksheet Normal Ex for an application example.

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=NORMDIST(17,20,5,TRUE) =NORMDIST(22,20,5,TRUE)-NORMDIST(17,20,5,TRUE) =1-NORMDIST(22,20,5,TRUE) y
Minutes 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 Solution Method 1. ............................... See Solution Method 2 at RIGHT The time per workout an athlete uses a stairclimber is normally distributed, with a mean of 20 minutes and a standard deviation of 5 minutes. An athlete is randomly selected. a. Find the probability that the athlete uses a stairclimber for less than 17 minutes. b. Find the probability that the athlete uses a stairclimber between 17 and 22 minutes. c. Find the probabiliity that the athlete uses a stairclimber for more than 22 minutes. Solution: First, we recall that a normal distribution extends approximately three standard deviations above and below the mean. Thus, the time on the stairclimber will probably be within 3 standard deviations, or 3*5 = 15 minutes. Thus, the possible values of time will extend from 20 - 15 = 5 to 20 + 15 = 35. We create a column of Minutes ranging from 5 to 35, in one minute increments. (Use Edit --> Fill --> Series to create the list.) This column is shown at right. Next, use the NORMDIST function to generate a column of Area to the Left. This function has the format NORMDIST(x,mean,stddev,true). In this case, the formula for cell B12 will look like: =NORMDIST(A12,20,5,true) We then add another column to covert to percentage. Part A: The answer is the area to the left of 17. This is given in the Area to the Left column. Thus, simply look up x = 17. That is, P(x<17) = 0.2743 = 27.43%. Part B: The answer is the area between 17 and 22. Since the table gives Area to the Left, we look up 22 and 17, then subtract the area. (Answer will always be positive, so be sure to subtract smallest from largest.) That is, P(17<x<22) = 0.6554 - 0.2743 = 0.3811 = 38.11% Part C: The answer is the area above 22. Since the table gives Area to the Left (or Area Below), we look up 22 and subtract the area from 1. That is, P(x>22) = 1 - P(x<22) = 1 - 0.6554 = 0.3446 = 34.46%

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0.0013 0.13% 0.0026 0.26% 0.0047 0.47% 0.0082 0.82% 0.0139 1.39% 0.0228 2.28% 0.0359 3.59% 0.0548 5.48% 0.0808 8.08% 0.1151 11.51%
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## This note was uploaded on 08/16/2010 for the course ACC Math 221 taught by Professor Barnhart during the Spring '10 term at DeVry Ft. Washington.

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Week_6_Lab - z-Score -3.50 -3.25 -3.00 -2.75 -2.50 -2.25...

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