Lecture 4 - Statistics & Probability (1) (2) PDF

Lecture (4) Chapter (2) “Describing Data with Numerical Measures” ‫وصف البيانات باستخدام مقاييس عددية‬ Describing data Measures...

Lecture (4) Chapter (2) “Describing Data with Numerical Measures” ‫وصف البيانات باستخدام مقاييس عددية‬ Describing data Measures of Measures of center variability Standard Mean Median Mode Range Variance deviation 2.3 Measures of variability ‫مقاييس التشتت‬ 1) Range ‫المدى‬ 2) Variance ‫التباين‬ 3) Standard deviation ‫اإلنحراف المعياري‬  Range: Range = max – min Example: Given the following data: 2, 5, 10, 1, 4 and 3. Find the range. Range = max – min = 10 – 1 = 9  Variance and Standard Deviation: Population variance (𝜎 2 ): Let x1 , x2 , … , xN be observations in a population then the Population variance is: ∑𝑁 𝑖=1(𝑥𝑖 −𝜇) 2 ∑𝑁 𝑖=1 𝑥𝑖 𝜎2 = where 𝜇 = 𝑁 𝑁 o The population standard deviation (𝜎) is √𝜎 2.  Sample variance (𝑆 2 ): Let x1 , x2 , … , xn be observations in a sample then the sample variance is: ∑𝑛 𝑖=1(𝑥𝑖 −𝑥̅ ) 2 ∑𝑛 𝑖=1 𝑥𝑖 𝑆2 = where 𝑥̅ = 𝑛−1 𝑛 or ∑𝑛𝑖=1 𝑥𝑖2 − 𝑛𝑥̅ 2 𝑆2 = 𝑛−1 Note that: ∑𝑛𝑖=1 𝑥𝑖2 ≠ (∑𝑛𝑖=1 𝑥𝑖 )2 o The sample standard deviation (𝑆) is √𝑆 2. Example: Given the following data: 1, 14, 15, 9, 4, 3 ∑𝑛𝑖=1 𝑥𝑖 𝑥̅ = 1) Calculate the sample variance and standard deviation. 𝑛 ∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2 1 + 14 + 15 + 9 + 4 + 3 2 𝑆 = = 𝑛−1 6 = 7.67 (1 − 7.67)2 + ⋯ + (3 − 7.67)2 = = 35.07 5 𝑆 = √𝑆2 = √35.07 = 5.92 2 ∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2 𝑆 = 𝑛−1 2) Calculate the sample standard deviation using the second formula. ∑𝑛𝑖=1 𝑥𝑖2 − 𝑛𝑥̅ 2 528 − 6 × (7.67)2 𝑆2 = = 𝑛−1 5 𝑛 = 35.07 ∑ 𝑥𝑖2 = 12 + 142 + 152 𝑖=1 𝑆 = √𝑆2 = √35.07 = 5.92 + 92 + 42 + 32 = 528 Example: If the deviation of five observation around their mean: −2, 2, 4, 3, −7. Calculate the sample variance. ∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2 (−2)2 + 22 + 42 + 32 + (−7)2 𝑆2 = = = 20.5 𝑛−1 5−1 Example: Given the following data. ∑𝑛𝑖=1 𝑥𝑖 𝑥̅ = 9, 9, 9, 9, 9, 9, 9, 9, 9, 9. 𝑛 9 + ⋯ + 9 90 Calculate the sample variance. = = 10 10 =9 2 ∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2 (9 − 9)2 + ⋯ + (9 − 9)2 𝑆 = = =0 𝑛−1 10 − 1 Note that if all the observation have the same value the the sample variance (𝑆2 ) is always zero (= 0).  The largest 𝑆 2  the greater the variability..‫كلما كانت التباين كبيرة كلما كانت االنحرافات أكثر‬ Example: Given the following table: Sample 1 Sample 2 𝑥̅ = 18 𝑥̅ = 18 𝑆2 = 9 𝑆2 = 25 Which sample is the best? Sample 1.‫في حال تساوي الوسط فإن قيمة التباين األقل هي األفضل‬ Example: Given the following informatios: 2 𝑛 2 ∑50 𝑖=1 𝑥𝑖 = 300 and the sample variance 𝑆 = 3. Calculate ∑𝑖=1 𝑥𝑖. ∑50 𝑖=1 𝑥𝑖 𝑥̅ = ∑𝑛𝑖=1 𝑥𝑖2 − 𝑛𝑥̅ 2 𝑛 𝑆2 = 𝑛−1 300 = =6 50 ∑𝑛𝑖=1 𝑥𝑖2 − 50 × (6)2 3= (50 − 1) 𝑛 ∑ 𝑥𝑖2 = 1947 𝑖=1 Homework: Q1: You are given n = 10 measurements: 3, 5, 4, 6, 10, 5, 6, 9, 2, 8. a. Calculate the sample mean. b. Find median. c. Find the mode. Q2: You are given n = 8 measurements: 4, 1, 3, 1, 3, 1, 2, 2. a. Find the range. b. Calculate 𝑥̅. c. Calculate 𝑆 2 and 𝑆. ‫ لذلك‬،‫ االنحراف المعياري يأخذ جميع خصائص التباين ماعدا الوحدة فوحدة االنحراف المعياري هي نفس وحدة قياس البيانات‬:‫مالحظة‬.‫ 𝑆 ألن وحدة 𝑆 نفس وحدة البيانات‬2 ‫يفضل استخدام 𝑆 أكثر من‬ 2.4 On the Practical Significance of the Standard Deviation ‫األهمية العلمية لإلنحراف المعياري‬  Tchebysheff’s Theorem )‫كم نسبة البيانات الواقعة ضمن الفترة (تعطينا أقل نسبة للفترة‬ Given the number k any real number greater than or equal 1 and a set of n measurement, 1 at least (1 − 𝑘 2 ) of the measurement will lie within k standard deviation of their mean). 𝑥̅ ± 𝑘𝑆 k ‫ عن ̅𝑥 بمقدار‬S ‫المعنى كم تبعد‬ Example: Let k = 2. 1 3 At least (1 − ) = = 75% will lie in the interval (𝑥̅ − 2𝑆, 𝑥̅ + 2𝑆) 4 4 Let k = 3. 1 8 At least (1 − 9) = 9 = 88.9% will lie in the interval (𝑥̅ − 3𝑆, 𝑥̅ + 3𝑆) Let k = 1. 1 At least (1 − 1) = 0 (Non of the measurement will lie in the interval (𝑥̅ − 𝑆, 𝑥̅ + 𝑆) Example: Find the proportion of the measurement that fall within 2.5 standard deviation of the mean. 1 5.25 At least (1 − 2.52 ) = 6.25 = 84% will lie in the interval (𝑥̅ − 2.5𝑆, 𝑥̅ + 2.5𝑆) Example: Given a set of 40 measurements that have mean 60 and variance 100. 1) Find the proportion of the measurement that lie in the interval [40, 80]. 𝑥̅ − 𝑘𝑆 = 40 60 − 10𝑘 = 40 𝑘=2 1 3 At least (1 − 22 ) = 4 = 75% will lie in the interval [40, 80]. 2) Find the number of measurements that lie in the interval [40, 80]. 3 × 40 = 30 4 3) Find the proportion of the measurement that lie in the interval [30, 90]. 𝑥̅ − 𝑘𝑆 = 30 60 − 10𝑘 = 30 𝑘=3 1 8 At least (1 − 9) = 9 = 88.9% will lie in the interval [30, 90]. Example: If the mean is 75 and the standard deviation is 10. Find the internal that at least 8 of the measurement will lie in it. 9 1 8 1− 2 = 𝑘 9 1 1 = 𝑘2 9 𝑘2 = 9 𝑘 = ±3.‫نهمل القيمة السالبة ألنها طول‬ (𝑥̅ − 2𝑆, 𝑥̅ + 2𝑆) = (75 − 3 × 10, 75 + 3 × 10) = (45, 105)  Empirical Rule Given a distribution of measurement that is approximately mound shape. 1) The interval (𝜇 ± 𝜎) contains approximately 68% of measurements. 2) The interval (𝜇 ± 2𝜎) contains approximately 95% of measurements. 3) The interval (𝜇 ± 3𝜎) contains approximately 99.7% of measurements. Empirical Tchebysheff’s Mound shape Any distribution The value of k is only 1, 2, 3 Use any value within k , 𝑘 ≥ 1 ‫تقدر نسبة البيانات بالظبط‬ ‫تعطينا أقل نسبة للفترة‬.‫ ألنه يعطي قيمة تقريبية دقيقة‬Tchebysheff’s ‫ أدق من‬Empirical Example: Let k = 2. o Empirical: Approximately 95% of measurements will lie in the interval (𝜇 ± 2𝜎) o Tchebysheff’s: 1 3 At least (1 − 4) = 4 = 75% will lie in the interval (𝑥̅ − 2𝑆, 𝑥̅ + 2𝑆) Example: Given that the distribution of a measurement is mound- sahpe with mean and variance 40 and 81, respectively. 1) What is the proportion of measurement that lie within one standard deviation of the mean. k = 1 contains approximately 68% of measurements. 2) What is the proportion of measurement that lie within the interval (22, 58). 𝑥̅ − 𝑘𝑆 = 22 40 − 9𝑘 = 22 𝑘=2 Contains approximately 95% of measurements. 2.5 A Check on the Calculation of S The approximate value of the standard deviation is 𝑅 𝑆= 4 where R = max – min Example: Given the following data: 4, 9, 15, 20, 13, 14, 8, 14, 17, 5. R = max – min 1) Calculate the approximate value of the standard deviation. 𝑅 16 = 20 – 4 𝑆= = =4 4 4 =16 2) Calculate the standard deviation. ∑𝑛𝑖=1 𝑥𝑖 2 − 𝑛𝑥̅ 2 1522 − 10 × (11.9)2 𝑆2 = = = 11.77 𝑛−1 10 − 1 𝑥𝑖 𝑥𝑖 2 4 16 ∑ 𝑥𝑖 9 81 𝑥̅ = 15 255 𝑛 20 400 4 + 9 + 15 + 20 + 13 + 14 + 8 + 14 + 17 + 5 13 169 = 10 14 196 119 8 64 = = 11.9 14 196 10 17 289 5 25 Total 1522 𝑆 = √𝑆 2 = √11.77 = 3.4 3) Calculate the approximate value of the variance. 𝑅 16 𝑆= = =4 4 4 𝑆2 = 42 = 16