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Variance and Standard Deviation Explained
- 1. UNIVERSITY OF NORTHEASTERN PHILIPPINES Iriga City School of Graduate Studies A/Y 2022-2023 INFERENTIAL STATISTICS VARIANCE AND STANDARD DEVIATION KEN PAUL V. BALCUEVA REPORTER
- 2. VARIANCE • RONALD FISHER • 1981 • THE CORRELATION BETWEEN RELATIVES ON THE SUPPOSITION ON MENDELIAN INHERITANCE
- 3. VARIANCE • RONALD FISHER STANDARD DEVIATION • KARL PEARSON • 1893 • “ROOT MEAN SQAURE ERROR”
- 4. OBJECTIVES • To find the VARIANCE of a data set • To find the STANDARD DEVIATION of a data set
- 6. VARIANCE •Variance is the average squared deviation from the mean of a set of data. •It is used to find the standard deviation.
- 7. VARIANCE •Find the MEAN of the data. • MEAN is the average so add up the values and divide by the number of items. •Subtract the mean from each value – the result is called the deviation from the mean. •Square each deviation of the mean. •Find the sum of the squares. •Divide the total by the number of items.
- 8. VARIANCE FORMULA • The variance formula includes the Sigma notation, 𝛴, which represents the sum of all the items to the right of Sigma. • MEAN is represented by 𝜇 and N is the number of items.
- 10. STANDARD DEVIATION • Standard Deviation shows the variation in data. • If the data is close together, the standard deviation will be small. • If the data is spread out, the standard deviation will be large. • Standard Deviation is often denoted by the lowercase Greek letter Sigma, 𝝈
- 11. STANDARD DEVIATION • Find the Variance • Find the Mean of the data. • Subtract the mean from each value. • Square each deviation of the mean • Find the sum of the squares • Divide the total by the number of items. • Take the square root of the variance.
- 12. STANDARD DEVIATION FORMULA • The standard deviation formula can be represented using Sigma notation. • The standard deviation formula is the square root of the variance.
- 14. Find the Variance and Standard Deviation The math test scores of five students are: 92, 88, 80, 68, and 52
- 15. 1) Find the Mean: (92+88+80+68+52) 5 = 76
- 16. 1) Find the Mean: (92+88+80+68+52) 5 = 76 2) Find the Deviation from the Mean: 92-76= 16 88-76= 12 80-76= 4 68-76= -8 52-76= -24
- 17. 3) Square the Deviation from the Mean: (16)²= 256 (12)²= 144 (4)² = 16 (-8)²= 64 (-24)²= 576
- 18. 4) Find the sum of the squares of the deviation from the mean: 256+144+16+64+576= 1056 5) Divide by the number of data items to find the variance: 1056 5 = 211.2
- 19. 6) Find the square root of the Variance: 211.2 = 14.53 Thus, the standard deviation of the test scores is 14.53
- 20. CONCLUSION • As we have seen, standard deviation measures the dispersion of data. • The greater the value of the standard deviation, the further the data tend to be dispersed from the mean.
- 21. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION
- 22. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION FORMULA MEAN VARIANCE STANDARD DEVIATION E(x)=𝛴xP(x) 𝜎2= 𝛴[(x-𝜇) ² P(x)] 𝜎= 𝜎2 = 𝛴[(x−𝜇) ² P(x)]
- 23. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION PROBLEM Determine the mean, variance, and standard deviation of the following Probability mass function: x P(x) 1 0.5 2 0.30 3 0.15 4 0.10 5 0.20
- 24. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION x P(x) xP(x) x-𝜇 (x-𝜇)² (x-𝜇)²P(x) 1 0.15 2 0.30 3 0.15 4 0.10 5 0.20
- 25. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION x P(x) xP(x) x-𝜇 (x-𝜇)² (x-𝜇)²P(x) 1 0.15 0.15 2 0.30 0.6 3 0.15 0.45 4 0.10 0.40 5 0.20 1 MEAN a)𝜇=𝛴xP(x) 𝜇 = 1.95 1.95
- 26. x-𝜇 1 – 1.95 = -1.95 2 – 1.95 = 0. 05 3 – 1.95 = 1.05 4 – 1.95 = 2.05 5 – 1.95 = 3.05
- 27. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION x P(x) xP(x) x-𝜇 (x-𝜇)² (x-𝜇)²P(x) 1 0.15 0.15 -1.95 2 0.30 0.6 0.05 3 0.15 0.45 1.05 4 0.10 0.40 2.05 5 0.20 1 3.05 MEAN a)𝜇=𝛴xP(x) 𝜇 = 1.95 1.95
- 28. (x-𝜇)² -1.95 x -1.95 = 3.8025 0. 05 x 0. 05 = 0.0025 1.05 x 1.05 = 1.1025 2.05 x 2.05 = 4.2025 3.05 x 3.05 = 9.3025
- 29. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION x P(x) xP(x) x-𝜇 (x-𝜇)² (x-𝜇)²P(x) 1 0.15 0.15 -1.95 3.8025 2 0.30 0.6 0.05 0.0025 3 0.15 0.45 1.05 1.1025 4 0.10 0.40 2.05 4.2025 5 0.20 1 3.05 9.3025 MEAN a)𝜇=𝛴xP(x) 𝜇 = 1.95 1.95
- 30. VARIANCE (x-𝜇)²P(x) 3.8025 x 0.15 = 0.5704 0.0025 x 0.30 = 0.0008 1.1025 x 0.15 = 0.1654 4.2025 x 0.10 = 0.4203 9.3025 x 0.20 = 1.8605
- 31. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION x P(x) xP(x) x-𝜇 (x-𝜇)² (x-𝜇)²P(x) 1 0.15 0.15 -1.95 3.8025 0.5704 2 0.30 0.6 0.05 0.0025 0.0008 3 0.15 0.45 1.05 1.1025 0.1654 4 0.10 0.40 2.05 4.2025 0.4203 5 0.20 1 3.05 9.3025 1.8605 MEAN a) 𝜇=𝛴xP(x) 𝜇 = 1.95 VARIANCE a) 𝜎2 = 𝛴[(x-𝜇) ² P(x)] 𝜎2 = 3.02 1.95 3.0174
- 33. DISCRETE RANDOM VARIABLE: MEAN, VARIANCE AND STANDARD DEVIATION x P(x) xP(x) x-𝜇 (x-𝜇)² (x-𝜇)²P(x) 1 0.15 0.15 -1.95 3.8025 0.5704 2 0.30 0.6 0.05 0.0025 0.0008 3 0.15 0.45 1.05 1.1025 0.1654 4 0.10 0.40 2.05 4.2025 0.4203 5 0.20 1 3.05 9.3025 1.8605 MEAN a) 𝜇=𝛴xP(x) 𝜇 = 1.95 VARIANCE a) 𝜎2 = 𝛴[(x-𝜇) ² P(x)] 𝜎2 = 3.02 STANDARD DEVIATION a) 𝜎= 𝜎2 𝜎=1.74 1.95 3.0174
- 34. SAMPLE MEAN, VARIANCE, AND STANDARD DEVIATION FOR UNGROUPED DATA FORMULA SAMPLE MEAN SAMPLE VARIANCE STANDARD DEVIATION 𝑥 = 𝑥 𝑛 𝑆2 = 𝑥 − 𝑥 2 𝑛 − 1 𝑆 = 𝑥 − 𝑥 2 𝑛 − 1
- 35. EXAMPLE PROBLEM The following are the scores of 8 randomly selected students in Grade 7. 7,8,12,15,10,11,9 and 14 Compute the following: a. Sample Mean b. Sample Variance c. Sample Standard Deviation
- 36. 1) 7,8,12,15,10,11,9 and 14 𝑥 𝑥 − 𝑥 𝑥 − 𝑥 2 7 8 12 15 10 11 9 14
- 39. 1) 7,8,12,15,10,11,9 and 14 𝑥 𝑥 − 𝑥 𝑥 − 𝑥 2 7 -3.75 14.0625 8 -2.75 7.3625 12 1.25 1.5625 15 4.25 18.0625 10 -0.75 0.5625 11 0.25 0.0625 9 -1.75 3.0625 14 3.25 10.5625 86 55.5 𝒙 = 𝒙 𝒏 = 86 8 𝒙 = 10.75
- 40. 1) 7,8,12,15,10,11,9 and 14 𝑥 𝑥 − 𝑥 𝑥 − 𝑥 2 7 -3.75 14.0625 8 -2.75 7.3625 12 1.25 1.5625 15 4.25 18.0625 10 -0.75 0.5625 11 0.25 0.0625 9 -1.75 3.0625 14 3.25 10.5625 86 55.5 𝒙 = 𝒙 𝒏 = 86 8 𝒙 = 10.75 𝑺𝟐 = 𝒙 − 𝒙 𝟐 𝒏 − 𝟏 = 𝟓𝟓.𝟓 𝟖−𝟏 𝒔𝟐 = 𝟕. 𝟗𝟑
- 41. 1) 7,8,12,15,10,11,9 and 14 𝑥 𝑥 − 𝑥 𝑥 − 𝑥 2 7 -3.75 14.0625 8 -2.75 7.3625 12 1.25 1.5625 15 4.25 18.0625 10 -0.75 0.5625 11 0.25 0.0625 9 -1.75 3.0625 14 3.25 10.5625 86 55.5 𝒙 = 𝒙 𝒏 = 86 8 𝒙 = 10.75 𝑺𝟐 = 𝒙 − 𝒙 𝟐 𝒏 − 𝟏 = 𝟓𝟓.𝟓 𝟖−𝟏 𝒔𝟐 = 𝟕. 𝟗𝟑 𝑺 = 𝒙 − 𝒙 𝟐 𝒏 − 𝟏 = 7.93 𝑺 = 2.82
- 42. Thank you!!!
Editor's Notes
- Goodmorning mga maam and sir! The topic that was assigned to me is about Variance and Standard Deviation. But before that, let me give you a little trivia about this two.
- Variance was first introduced by Ronald Fisher in his 1981 paper THE CORRELATION BETWEEN RELATIVES ON THE SUPPOSITION ON MENDELIAN INHERITANCE. But his first application of the analysis of variance was published in 1921.
- While Standard Deviation was all due to a historical accident in 1893, when Karl Pearson introduced the term “standard deviation for what had been known as root mean square error. So, dun po nagsimula ang lahat, that even this two great polymath never imagined that their work were still been applicable up to this present. As we proceed, lets identify first the objectives we have this morning.
- (read the obejctives) Lets start by defining first what is variance?
- Question. What is variance used for in real life? Saan po bang particular na situation or event in life na maaari itong iapply? Tayo po bilang mga educator, we can definitely apply this approach or technique to identify and understand our student's progress especially when giving them an assessment every end of the lesson or the quarter. Dito po natin malalaman kung sino po ba ung nakakua o hindi ng binigay nating assessment. Sa tulong na rin po nitong dalawa, eh we can finally provide them the proper diagnosis or remediation to address the learning gap of the students. Same with the investors, they use variance to see how much risk an investment carries and whether will it be profitable or not.
- So, here is the descriptive formula of variance for a much comprehendible solving method. (read)
- Ito naman po ang decriptive formula ng Standard Deviation. (read) Inshort, standard deviation is just the square root of the variance. For us to get the SD, it is prerequisite to find first the Variance. Kung mali ung computation natin sa Variance, mali rin po ung magiging sagot sa SD. Same thing as we apply this statistic to research, we have be very careful in computing, kasi isang maling data or computation may lead to invalidity and unreliability of our studies.
- So here is the sample figure of standard deviation. SD is a measure of how dispersed the data is in relation to the mean. Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out. Ano po ba and ibig ipahiwatig ng SD sa atin? SD is the average amount of variability in your data set. It tells us, on average, how far each score lies from the mean. So how do we interpret the dispersion of SD, Is it better to have a high or low standard deviation? When a high SD shows that the data is widely spread, we could possibly interpret it as less reliable and a low SD shows that the data are clustered closely around the mean, we could feasibly interpret it as more reliable. Lets us not be confused in interpreting the figures. It is not literally how high or low the curves from the mean or the average for us to interpret it as good or bad, but a better SD represents the typical distance, gaano ba kalayo o kalapit and distansiya ng mga data points together from the mean or average point. At hindi po kung gaano kataas o kababa ang curves na nkikita natin from the figures. In lay-mans term, the closer the distance of data points clustered together from the center/ or what we called mean or average, the more reliable and relatively consistent the study. And vice versa to the data points that are spread out, the less the reliability.
- So, I have here a sample problem po.
- (read the Conclusion) Let us now proceed to the MEAN, VARIANCE AND STANDARD DEVIATION OF DISCRETE RANDOM VARIABLE.
- BEFORE WE GIVE THE PROBLEM, LET US FIRST GIVE THE FORMULA. Here are the formulas in solving the Mean, Variance, and Standard Deviation of a Discrete Random Variable. First we have Mean, so commonly it is denoted as Capital E of X, or minsan what we call Expected Value, reason to write it in capital letter E as we use it in notating for our Mean. Expected value is equal to summation of X multiply by P of X. So what is X, it is outcome value of the random variable. And as to P, is the Probability of the outcome value of X. Next, is the Variance. Variance is commonly denoted as Sigma squared. So the formula of variance is sigma squared is equal to the summation of open bracket and open parenthesis X minus MU squared multiply by P of X close parenthesis and close bracket. So what is MU in Variance Formula? Ito po ung Mean natin that is notated as E of X Ok po. how about Standard Deviation? Let us remember po that SD is just the Square root of the Variance. So SD, is commonly notated as Sigma is equal to the square root of sigma squared or ung Variance. Or kun gusto po nating kompletong formula ng SD, so Sigma is equal to the square root of summation open bracket and open parenthesis X minus MU squared multiply by P of X close parenthesis and close bracket.
- So we have here, problem number 1. Determine the mean, variance, and standard deviation of the following Probability mass function: So we have the given mass function under the instruction the value of X and the probability of X First we have to find the mean using its formula
- As we can see in this figure, the value of X is in the first column. And in the second column is the Probability of X or P of X. Third is the product of x and P of X. Next column is the X minus MU, Fifth column is the X minus MU squared. And last column is the product of X minus MU squared multiply by the P of X. Ok po. Since we have already the given value of X and P of X manifested from the example, kokopyahin nlang po natin ito SA COLUMN X and P of X. So under column X, we have number 1,2,3,4,5. then the corresponding Probabilities sa column naman po ng P of X are 0.15, down to 0.20. Since we are looking for the Mean, kukunin muna po natin ung product ng X and P of X, which is in our third column. Then kukunin natin ung summation nito for us to get the value of Mean.
- Lets start this by getting the product of the first row which is 1 times 0.15 is equal to 0.15. Second row is, 2 times 0.30 is equal to 0.6. Ok, The same procedure for the third row down to fifth, so kukunin lang po ung product nila. Then after getting the product, iaadd lang po natin ito to get the Mean. So 0.15 + 0.6 + 0.45 + 0.40 + 1 is equal to 1.95. Therefore, our mean is 1.95. Since nakuha na natin ung Mean, makukuha na natin ung data na hinahanap sa next column which is X minus MU. Ang gagawin natin, kukunin lang po natin ung difference ng value ng X at ung summation ng product ng X at P of X.
- So 1 minus 1.95 is equal to -1.95. Next, 2 minus 1.95 is equal to 0.05. Same process down to 5 minus 1.95 is equal to 3.05 So ito na po ung magiging difference natin for this column X minus MU.
- Next, para naman po dito sa column ng X minus MU squared. I mumultiply lang po natin ung Mga difference nato by itself.
- (read the figures) Ok, this result…
- …will be our input for the column X minus MU squared. Since we are done computing the Mean and the given formula, Ngayon naman po is ung Variance. And for us to get the answer for variance. All we need to do is to get the product of X minus Mu squared formula multiply by P of X. Then finally get the summation of the product for us to get the Variance.
- So ganito po ung magiging equation nila. (read the equation) After we get the product of this equation, and for us to get the Variance. We will add them all ang get the summation of the product of X minus MU squared and the P of X.
- So ito na po ung result. The variance of this equation is 3.0174 or if rounded off will be 3.02. 3.02 is our final answer for the Variance. And for us to get the SD, kukunin lang po natin ung square root nito.
- The formula of SD is Sigma is Equal to the square root of Sigma Squared. Therefore, the answer for the equation Sigma is equal to the square root of 3.02 is 1.74. 1.74 is our final answer for Standard Deviation.
- So we have here the scores of 8 selected students from grade 7, 7, 8, 12, 15, 10, 11, 9, and 14 As we can see in the table, the first column are the scores of 8 students under the instruction of X. Second column is X minus Xbar. And the third column is the X minus Xbar squared. Since we already have input for X, ang hahanapin nlang po natin is itong subtrahend na Xbar sa second Column. But before we get there. Kelangan muna po natin hanapin ung sagot sa sample mean by getting the summation of scores acquired by 8 students of grade 7.
- So ganito po un.
- So ganito po un. 7+8+12+15+10+11+9+14 = 86
- So we have here the scores of 8 selected students from grade 7, 7, 8, 12, 15, 10, 11, 9, and 14 As we can see in the table, the first column are the scores of 8 students under the instruction of X. Second column is X minus Xbar. And the third column is the X minus Xbar squared.