Bài giảng môn học Quản trị kinh doanh - Chapter twenty: Two business statistics

BUSINESS STATISTICSChapter Twenty-TwoCopyright © 2014 by The McGraw-Hill Companies, Inc. All rights reserved.McGraw-Hill/IrwinDefine and calculate the mean.Explain and calculate a weighted mean.Define and calculate the median.Define and identify the mode.LU22-1: Mean, Median, and ModeLearning unit objectivesLU 22-2: Frequency Distributions and GraphsPrepare a frequency distribution.Prepare bar, line, and circle graphs.Calculate price relatives and cost comparisons.Explain and calculate the range.Define and calculate the standard deviation.Estimate percentage of data by using standard deviations.LU 22-3: Measures of Dispersion (Optional)Mean – The average used to indicate a single value that represents an entire group of numbers.Median - A measurement that indicates the center of the data (average).TerminologyMode - A measurement that records values; the value that occurs most often.MeanMean =The accountant of Bill’s Sport Shop told Bill, the owner, that the average daily sales for the week were $150.14. The accountant stressed that $150.14 was an average and did not represent specific daily sales. Bill wanted to know how the accountant arrived at $150.14.Sun. Mon. Tues. Wed. Thur. Fri. Sat. $400 $100 $68 $115 $120 $68 $180Mean = $400 + $100 + $68 + $115 + $120 +$68 + $180 = $150.14 7Sum of all valuesNumber of valuesExample:Weighted MeanWeighted meanHow Jill Rivers calculated her GPA to the nearest tenth: Credit Grade PointsCourses Attempted Received (Credits x Grade)Intro to Comp. 4 A 16 (4 x 4)Psychology 3 B 9 (3 x 3)English Comp. 3 B 9 (3 x 3)Business Law 3 C 6 (2 x 3)Business Math 3 B 9 (3 x 3) 16 4949 16= 3.1= Sum of products Sum of frequenciesExample:Finding the Median of a Group of ValuesLook below at the following yearly salaries of the employees of Rusty’s Clothing Shop.Alice Knight $95,000 Jane Wang $67,000Jane Hess $27,000 Bill Joy $40,000Joel Floyd $32,000Find the median value of all employees.$95,000 + $27,000 + $32,000 + $67,000 + $40,000 5= $52,200Finding the Median of a Group of ValuesStep 1. Orderly arrange values from the smallest to the largest.Step 2. Find the middle value.95, 27, 32, 67, 40 27, 32, 40 , 67, 95 32, 40, 67, 95 95, 32, 67, 40 a. Odd number of values: Median is the middle value. Divide the total number of numbers by 2 (5/2 = 2 ½). The next-higher number is the median.b. Even number of values: Median is the average of the two middle values.40 40 + 67 2 = 53.5Mode3, 4, 5, 6, 3, 8, 9, 3, 5, 3Mode is the value that occurs most often.If two or more numbers appear most often, you may have two or more modes.If all the values are different, there is no mode.3 is the mode since it is listed 4 times.Frequency DistributionFrequency Distribution -- A way of collecting and organizing raw data.Price of Tally FrequencyComputer$1,000 llll 5 2,000 l 1 3,000 llll 5 4,000 l 1 5,000 ll 2 6,000 ll 2 7,000 l 1 8,000 l 1 9,000 l 110,000 l 1100070004000100050001000300050002000300030003000800090003000600060001000100001000Computer costsFrequency distribution tableBar GraphFrequency of purchasePrice of Computers200040008000100006000Bar GraphClass Frequency$1000 - $ 3,000.99 11$3001 - 5,000.99 3$5001 - 7,000.99 3$8001 - 9,000.99 2$9001 - 11,000.99 1$3,001-$5,000.99$7,001-$9,000.99Frequency of purchaseCircle Graph12.9%12.9%56.9%17.3%Revenues1st Qtr. $20,4002nd Qtr $27,4003rd Qtr $90,0004th Qtr $20,400Measure of DispersionRange = Highest value – Lowest valueFind the range of the following values:83.6, 77.3, 69.2, 93.1, 85.4, 71.6Range = 93.1 – 69.2 = 23.9Measure of Dispersion – A number that describes how the numbers of a set of data are spread out or dispersed.Range – The difference between the two extreme values (highest and lowest) in a group of values or a set of data.Index NumbersPrice relative = Current price x 100 Base year’s priceA calculator may cost $9 today relative to a cost of $75 some 30 years ago. What is the relative price? $9 $75x 100 = .12 = 12%Example:Consumer Price Index (in percent) Step 1. Find the mean of the set of data.Step 2. Subtract the mean from each piece of data to find each deviation . Step 3. Square each deviation (multiply the deviation by itself).Step 4. Sum all squared deviations.Step 5. Divide the sum of the squared deviations by n - 1, where n equals the number of pieces of data.Standard deviation is intended to measure the spread of data around the mean.Standard DeviationStep 6. Find the square root of the number obtained in Step 5. This is the standard deviation.Step 1 (1 + 2 + 5 + 10 + 12) = 6 (Mean) 5 Step 2 Step 3Data Data -- Mean (Data -- Mean)1 1- 6 = -5 252 2 - 6 = -4 165 5 - 6 = -1 110 10 - 6 = 4 1612 12 - 6 = 6 36 Total 0 94 (Step 4)Step 5 Divide by n-1: 94 = 94 = 23.5 5 - 1 4Step 6 The square root of 23.5 is 4.8.Data Set A x x x x x0 1 2 3 4 5 6 7 8 9 10 11 12 13 Standard DeviationThe standard deviation of data set A is 4.8.Step 1 (4 + 4 + 5 + 8 + 9) = 6 (Mean) 5 Step 2 Step 3Data Data -- Mean (Data -- Mean)1 4- 6 = -2 42 4 - 6 = -2 45 5 - 6 = -1 110 8 - 6 = 2 412 9 - 6 = 3 9 Total 0 22 (Step 4)Step 5 Divide by n-1: 22 = 22 = 5.5 5 - 1 4Step 6 The square root of 5.5 is 2.3. Data Set B x x x x x0 1 2 3 4 5 6 7 8 9 10 11 12 13 Standard DeviationThe standard deviation of data set B is 2.3.