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Download Mancosa Analytical techniques Notes and more Exercises Analytical Techniques in PDF only on Docsity! MANAGEMERNT STATISTICS Content Unit Learning Outcomes [ULO] Measures Of Central Location/tendency Mode Median Mean Calculate and Interpret all Measures Of central Location on the basis of both raw and grouped data. The concept of skewness and the interpretation of the reliability of Measures Of Central Location/Tendency Measures Of Dispersion Range IQR QD Variance Standard Deviation Calculate and interpret Measures Of Dispersion for both raw and grouped sets of data Measures Of Central Tendency Raw Data X = The number of faulty ATMs at a bank over a sample of 20 days 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Mode = There is no mode in this distribution. Median = = 33.5 Grouped Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Class/ Freque MidPo CFr. Interva l ncy [ ] int [ ] 10<20 2 15 2 30 225 450 20<30 5 25 7 125 625 3125 30<40 6 35 13 210 1225 7350 40<50 4 45 17 180 2025 8100 50<60 3 55 20 165 3025 9075 Totals 20 710 28100 Mean = ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) = = 35.5 Mode = ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 3.3333333 = 33.3333333 Mode 33 Var[X] = = ∑ ̅ = ( = Var[X] = SD = = S = () SD = Measures Of Dispersion Raw Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Range = = 56 – 17 = 39 = 27.5 = 44.5 IQR = = 44.5 – 27.5 = 17 QD = IQR/2 = = 8.5 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Variance [ ] = ∑( ̅) = ∑ ̅ 34 1156 35 1225 38 1444 41 1681 44 1936 45 2025 46 2116 51 2601 52 2704 56 3136 ∑ 27162 ∑( ̅) ∑( ̅)( ̅) = ∑( ̅ ̅ ) = ∑ ̅ ∑ ∑ ̅ = ∑ ̅ ̅ ∑ ̅ = ∑ ̅ ̅ ̅ ∑ = ∑ ̅ ̅ ∑( ̅) = ∑ ̅ , Therefore: ∑( ̅) = ∑ ̅ Variance [ ] = ∑( ̅) = ∑ ̅ = ( ) = = [ ] = 125.3263158 SD = = SD = = 11.19492366 SD = 11 Totals 20 ∑ = 710 28100 ̅ = ∑ = = 35.5 36 Variance [ ] = ∑ ̅ = ( ) = = Variance [ ] = 114.7368421 SD [ ] = ( ) = 10.71152847 11 25 ̅ 37 Probability Distributions Content Unit Learning Outcomes [ULO] Probability Distributions Binomial Normal Distinguish between the Binomial and Normal Distributions and then calculate the associated probabilities. Measures Of Central P(X = 4) = = Example 2 Two teams the Reds and the Blues play each other across 10 matches in a contest. The probability of the Reds winning a single game is 0.6. Assuming this probability remains constant throughout the 10 games, what is the probability that the Reds will win exactly 3 games? = ( ) = ( ) = ( ) = = 120 P(X = 3) = 120+.6?+0.47 =
MANAGEMERNT STATISTICS Content Unit Learning Outcomes [ULO] Measures Of Central Location/tendency Mode Median Mean Calculate and Interpret all Measures Of central Location on the basis of both raw and grouped data. The concept of skewness and the interpretation of the reliability of Measures Of Central Location/Tendency Measures Of Dispersion Range IQR QD Variance Standard Deviation Calculate and interpret Measures Of Dispersion for both raw and grouped sets of data Alternative Formula for the Mean: Mean = ̅ = ∑ = ∑ Grouped Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Class/ Freque MidPo CFr. Interva l ncy [ ] int [ ] 10<20 2 15 2 30 225 450 20<30 5 25 7 125 625 3125 30<40 6 35 13 210 1225 7350 40<50 4 45 17 180 2025 8100 50<60 3 55 20 165 3025 9075 Totals 20 710 28100 Mean = ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) MANAGEMENT STATISTICS Content Unit Learning Outcomes [ULO] Measures Of Central Location/tendency Mode Median Mean Calculate and Interpret all Measures Of central Location on the basis of both raw and grouped data. The concept of skewness and the interpretation of the reliability of Measures Of Central Location/Tendency Measures Of Dispersion Range IQR QD Variance Standard Deviation Calculate and interpret Measures Of Dispersion for both raw and grouped sets of data Measures Of Central Tendency Raw Data X = The number of faulty ATMs at a bank over a sample of 20 days 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Mode = There is no mode in this distribution. Median = = 33.5 Mean = ∑ = = 35.2 ̅ 35 Interva l ncy [ ] int [ ] 10<20 2 15 2 30 225 450 20<30 5 25 7 125 625 3125 30<40 6 35 13 210 1225 7350 40<50 4 45 17 180 2025 8100 50<60 3 55 20 165 3025 9075 Totals 20 710 28100 Mean = ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) = = 35.5 Mode = ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 3.3333333 = 33.3333333 Mode 33 Median = ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 10⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 5 Median = 35 Looking For the 70th Percentile: = ⟦ ⟧ Measures Of Dispersion Raw Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Range = = 56 – 17 = 39 = 27.5 = 44.5 IQR = = 44.5 – 27.5 = 17 QD = IQR/2 = = 8.5 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Variance [ ] = ∑( ̅) = ∑ ̅ Standard Deviation [= ( ) ] Var[X] = ∑( ̅) = ( ) ( ) ( ) ) ( ) ∑( ̅) ∑( ̅)( ̅) = ∑( ̅ ̅ ) = ∑ ̅ ∑ ∑ ̅ = ∑ ̅ ̅ ∑ ̅ = ∑ ̅ ̅ ̅ ∑ = ∑ ̅ ̅ ∑( ̅) = ∑ ̅ , Therefore: ∑( ̅) = ∑ ̅ Variance [ ] = ∑( ̅) = ∑ ̅ = ( ) = = [ ] = 125.3263158 SD = = SD = = 11.19492366 SD = 11 Considering The Mean = ̅ = 35.2 Mean Daily Break Downs = [35 11] ATMs Range from 24 to 46 ̅ 35 Lower bound of the mean = -11 + 35 = 24 Upper bound of the mean = +11 + 35 = 46 25 ̅ 37 Probability Distributions Content Unit Learning Outcomes [ULO] Probability Distributions Binomial Normal Distinguish between the Binomial and Normal Distributions and then calculate the associated probabilities. Measures Of Central Location/Tendency The Binomial Distribution n = number of trials performed success = p failure = q = 1-p p + q = 1 pmf of the binomial Distribution: P(X = x) = throughout the 10 games, what is the probability that the Reds will win exactly 3 games? P(X = x) = || n = 10; p=0.55; q=0.45 = ( ) = ( ) = ( ) = = 120 P(X = 3) = = .074603106 Example 3 What is the probability of rolling exactly 2 sixes in 6 rolls of a die? P(X = x) = || n = ; p= ; q= X {1, 2, 3, 4, 5, 6} x 1 2 3 4 5 6 P(X = x = ( ) = ( ) = ( ) = = 15 P(X = 4) = ⌊ ⌋ *⌊ ⌋ = .200938786 This brings us to Equation 1, below. This is an extremely important rule in all Probability. This rule states that all relevant Probabilities in every Probabilistic scenario will add up to 1. This rule is shown in Equation [1]. Therefore, in our scenario, all scenario associated Probabilities are: 0 sixes = P(X = 0) 1 six = P(X = 1). 2 sixes = P(X = 2) 3 sixes = P(X = 3) 4 sixes = P(X = 4) 5 sixes = P(X = 5), and 6 sixes = P(X = 6). All these Probabilities should add up to 1, i.e., P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) = 1 … [1] The Probabilities which are directly relevant to the question are the ones that are in BLACK in the Equation [1] above. This means therefore that you will need to calculate all these five individual Probabilities and then add them together. It’s an onerous task, but still doable. I have done the calculation, but I have used what is called a probability Generating function [PGF]. This method is not in your syllabus, but it is simple enough because it is basically based on a Binomial Series expansion. For now I do not advise you to use it. I only used it because it helps me generate or calculate all scenario Probabilities very quickly and in one calculation based on the Binomial Series Expansion. For you, please use the BASIC method we used in class. The BASIC Method we used in class is also part of the Binomial Series expansion. I urge you to calculate all the individual probabilities using the method we employed in the webinar Saturday. So below, I used the Binomial PGF to generate all the probabilities associated with our scenario of rolling a die six times. All the Probabilities from P(X=0) throughout to P(X=6) are generated. This is a smart way of doing things because if a question requires me to calculate 5 Probabilities out of 7 scenario Probability calculation, I might as well calculate the full range to ensure that I can answer any question associated with the whole scenario. Below are the Probabilities: ( ) = = ∑ ( )( ) = ( ) [ ] [ ] +( ) [ ] [ ] +( ) [ ] [ ] +( ) [ ] [ ] +( ) [ ] [ ] + ( ) [ ] [ ] + ( ) [ ] [ ] = 1*1 [ ] + [ ] [ ] + [ ] [ ] + [ ] [ ] + [ ] [ ] [ ] [ ] + [ ] [ ] = [ ] + [ ]+ [ ]+ +[ ] + + = + + + + = P(X 2) = .263224451 The indirect way recognizes the relevance of Equation [1]. It recognizes that the relevant Probabilities are P(X=2), P(X=3), P(X=4), P(X=5) and P(X=6), but they are way too many to calculate individually. It recognizes that if we add these five probabilities to the ones that are not directly relevant to this question, the result will always be 1. But the ones that are not directly relevant are just two, so if we calculate these two [P(X=0) + P(X=1)], and subtract the result from1, we will get the same answer [.263224451 or ] as we got using the direct method: P(X=0) + P(X=1) = + = = .736775549 We subtract the Probabilistic totals from 1 to obtain the Probability of rolling at least 2 sixes. First using Fractions: Remember in our case 1 = ; therefore – - = . This is the same answer we got above and if you convert this to a decimal number, you will get .263224451. Now, using decimals: 1 - .736775549 = .263224451. I think this part of the syllabus has been comprehensively covered. We can therefore use the individual probabilities to calculate any case relevant to the scenario given in the question. Some questions therefore follow: a. What is the probability that at most two sixes will show up in 6 rolls of the die? b. What is the probability that less than 2 sixes will show up in 6 rolls of the die? c. what is the probability that between 3 sixes and 5 sixes will show up in 6 rolls of the die? d. What is the probability that 7 sixes will show up in 6 rolls of the die? e. What is the probability that no sixes at all will show up in 6 rolls of the die? f. What is the probability that 6 sixes will show up in 6 rolls of the die? Since all the Probabilities have been generated, it is now possible to answer any question arising from this scenario and there are many questions that can arise. With all the Probabilities now available, it is just a matter of ingenuity and common sense to get all the correct answers to any question that can arise from this scenario. From the example I have already given above, you should be able now to reason your way to the correct answers to all the questions above. On the next page is another example you can optionally use for PRACTICE. Statistics show that only 25% of South African taxpayers submit their tax return before the deadline date. For a sample of six South African adults, what is the probability that exactly three have submitted their tax return before the deadline date? P(X = x) = || n = ; p=; q= MANAGEMERNT STATISTICS Content Unit Learning Outcomes [ULO] Measures Of Central Location/tendency Mode Median Mean Calculate and Interpret all Measures Of central Location on the basis of both raw and grouped data. The concept of skewness and the interpretation of the reliability of Measures Of Central Location/Tendency Measures Of Dispersion Range IQR QD Variance Standard Deviation Calculate and interpret Measures Of Dispersion for both raw and grouped sets of data Measures Of Central Tendency Raw Data X = The number of faulty ATMs at a bank over a sample of 20 days 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Mode = There is no mode in this distribution. Median = = 33.5 Mean = ∑ = = 35.2 ̅ 35 Interva l ncy [ ] int [ ] 10<20 2 15 2 30 225 450 20<30 5 25 7 125 625 3125 30<40 6 35 13 210 1225 7350 40<50 4 45 17 180 2025 8100 50<60 3 55 20 165 3025 9075 Totals 20 710 28100 Mean = ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) = = 35.5 Mode = ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 3.3333333 = 33.3333333 Mode 33 Median = ⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 10⟦ ⟧ = 30 + ⟦ ⟧ = 30 + 5 Median = 35 Looking For the 70th Percentile: = ⟦ ⟧ Measures Of Dispersion Raw Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Range = = 56 – 17 = 39 = 27.5 = 44.5 IQR = = 44.5 – 27.5 = 17 QD = IQR/2 = = 8.5 MANAGEMENT STATISTICS Content Unit Learning Outcomes [ULO] Measures Of Central Location/tendency Mode Median Mean Calculate and Interpret all Measures Of central Location on the basis of both raw and grouped data. The concept of skewness and the interpretation of the reliability of Measures Of Central Location/Tendency Measures Of Dispersion Range IQR QD Variance Standard Deviation Calculate and interpret Measures Of Dispersion for both raw and grouped sets of data Alternative Formula for the Mean: Mean = ̅ = ∑ = ∑ Grouped Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Class/ Freque MidPo CFr. Interva l ncy [ ] int [ ] 10<20 2 15 2 30 225 450 20<30 5 25 7 125 625 3125 30<40 6 35 13 210 1225 7350 40<50 4 45 17 180 2025 8100 50<60 3 55 20 165 3025 9075 Totals 20 710 28100 Mean = ̅ = ∑ = ( ) ( ) ( ) ( ) ( ) = 40 ⟦ ⟧ = 40 + 10⟦ ⟧ = 40 + ⟦ ⟧ = 40 + = 42.5 Looking For the 3rd Quartile: = ⟦ ⟧ Var[X] = = ∑ ̅ = ( = Var[X] = SD = = S = () SD = Measures Of Dispersion Raw Data 17 19 22 24 27 28 29 31 32 33 34 35 38 41 44 45 46 51 52 56 Range = = 56 – 17 = 39 = 27.5