Standard Deviation and Frequency Distributions

TUI Frequency Distributions Module 3/Case 10/148/2012 professor Kuleshov Frequency Distributions This assignment is based on Frequency Distributions and will imply the following information 1. The ability to describe the information provided by the type Deviation. 2. The ability to use the Standard Deviation to calculate the percentage of occurrence of a variable either above or below a particular value. 3. The ability to describe a normal distribution as evidenced by a bell shaped curve as well as the ability to prepare a distribution map from a set of data (module 3 Case).Part 1 (1) To suck up the opera hat deal on a CD player, Tom called eight wash room stores and asked the cost of a specific model. The prices he was quoted are listed below $ 298 $ star hundred twenty-five $ 511 $ 157 $ 231 $ 230 $ 304 $ 372 square off the Standard deviation $ 298 + $ 125+ $ 511+ $ 157+ $ 231+ $ 230+ $ 304+ $ 372= 2228/8 = 278. 5(subtract from s) 19,-153, 232, -121, -47, -48, 25, 93 (squ are numbers) 380, 2356, 54056, 14762, 2256, 2352, 650, 8742 = 106(added) (Divide by 7) 15251 (take square root) Standard Deviation = approximately 123. 2) When investigating times involve for drive-through avail, the following results (in seconds) were obtained. Find the float, variance, and standard deviation for severally of the two samples, and thus compare the two sets of results. Wendys 120 123 153 128 124 118 154 110 MacDonalds 115 126 147 156 118 110 cxlv 137 (2) Set 1 Range maximum minimum = 154-110= 44 scrap of cases 8 To note the average, add all of the observations and divide by 8 Mean 125 square deviations (120-125)2 = (-5)2 = 25 (123-125)2 = (-2)2 = 4 (153-125)2 = (28)2 = 784 (128-125)2 = (3)2 = 9 (124-125)2= (-1)2= 1 (118-125)2 = (-7)2 = 49 154-125)2 = (29)2 = 841 (110-125)2 = (-15)2 = 225 Add the shape deviations and divide by 8 Variance = 1938/7 Variance = 276 Standard deviation = mien(variance) = 16 Set 2 Range 156-110 =46 Number of cases 8 To find the mean, add all of the observations and divide by 8 Mean 131 Squared deviations (115-131)2 = (-16)2 = 280 (126-131)2 = (-5. 75)2 = 33 (147-131)2 = (15)2 = 232 (156-131)2 = (24)2 = 588 (118-131)2 = (-13)2 = 189 (110-131)2 = (-21) 2 = 473 (145-131) 2 = (13) 2 = 175 (137-131) 2 = (5) 2 = 27 This is divide by 7 because this is a sample data n-1=7 Add the squared deviations and divide by 7Variance = 1999/7 Variance = 285 Standard deviation = sort (variance) = 16 The standard deviation for restaurant B is slightly littler than that of restaurant A. The range for restaurant A is slightly less the range of B. This shows there is a little more variation in restaurant A with respect to times required for drive through service than in required for drive through service than in B. (3) A company had 80 employees whose salaries are summarized in the frequency distribution below. Find the standard deviation. Find the standard deviation of the data summarized in the granted frequency distribution. Salary Number of Employees ,001 -10,000 14 10,001 15,000 13 15,001 20,000 18 20,001 25,000 18 25,001 30,000 17 The chart gives frequency and salary, traditional formulas cannot be used delinquent to we do not know the actually salary of distributively employee. In order to do these assumptions need to be done with using nerve point. Example (10000-5001) /2 then added to 5001= 7500 5,001- 10,000 =7500 10,001-15000=12500 15001-20000=17500 2001-2500=22500 25001-30,000=27500 wide number of employees = 80 14, 13, 18, 18, 17= 80 Compute the Mean 14 * 7500 = 105000 13* 12500 = 162500 18* 17500 = 315000 18* 22500= 405000 17 * 27500 = 467500 467500 80Add up all frequency Mark values quantity= 1455000 1455000 80 1455000 / 80 = 18187. 5 = 18188 Now standard deviation Total employees 80 Total 1455000 Means= 18188 7500-18188=-10688 12500-18188=-5688 17500-18188=-688 22500-18188=4312 27500-18188=9312 Square the values -10688= 114233344 -5688=32353344 -688=473344 4312=18593344 9312=86713 344 114233344*13=420593472 323553344*13=420593472 Sd2= 3837187520 80-1 = 48571993 (round up) = 48571994 4. The heights of a group of professional basketball players are summarized in the frequency distribution below. Find the standard deviation. Height (in. ) Frequency 70-71 3 72-73 7 74-75 16 76-77 12 78-79 10 0-81 4 82-83 1 To get the standard deviation of these numbers I first calculated the mean by added all the numbers together (3, 7, 16, 12, 10, 4, 1) and divided it by 7. I then took the mean (7. 57143) and calculated the deviance by subtracting the mean from each one of the numbers in the set. Then I squared each of the individual deviations, added those sums together, and divided the number I got from that sum by one less than the data set, which are 6. Then the last step is shrewd the square root, which is the ending result (5. 38074) References Introduction to Frequency Distributions, Retrieved November 12, 2008, http//infinity. os. edu/faculty/woodbury/Stats/tutorial/Data _Freq. htm Slides on frequency distributions, Retrieved November 12, 2008, http//campus. houghton. edu/orgs/psychology/stat3/ Frequency distributions, Retrieved November 12, 2008, http//davidmlane. com/diazoxide/normal_distribution. hypertext markup language Z-Table Calculator, Retrieved November 12, 2008, http//davidmlane. com/hyperstat/z_table. html Z-Table and Standard Normal Distribution, Retrieved November 12, 2008, http//www. oswego. edu/srp/stats/z. htm Example of the normal distribution, Retrieved November 12, 2008, http//www. ms. uky. edu/mai/java/stat/GaltonMachine. html