COMSATS University Islamabad Sahiwal Campus (Department of Computer Science) Course Title: Statistics and Probability Theory Course Instructor: Nosheen Ramzan Semester: 4th Batch: Section: Time Allowed: 60 minutes Student’s Name: Important Instructions / Guidelines: Course Code: MTH 262 Credit Hours: 3-0 Program Name: BSCS A Date: 01-04-2020 30 Maximum Marks: Reg. No. CUI/ /SWL Read the question paper carefully and answer the questions according to their statements. Mobile phones are not allowed. Calculators must not have any data/equations etc. in their memory. (1) 1st Sessional SP21` Q.1) For each of these questions, choose the option (a, b, C or d) that is TRUE. Directions: Answer all Questions and please give me a justification on why you chose that answer. Thank you. 1. If in a discrete series 85% values are less than 30, then it describes what value a. π30 = 85 b. π1 = 30 c. π·85 = 75 d. π·ππ = ππ 2. If the coefficient of variation for prices is 5.483% and the mean price is Rs 15.5, then the variance price π π 2 is equal to a. 0.85 b. 84.9865 c. 28.269 d. 0.7225 π. π· πΆ. π = × 100 ππΈπ΄π π. π· 5.483% = × 100 15.5 0.05483 × 15.5 = π. π· ππππππππ = 0.84982 = 0.7225 3. Through visualization not mathematically determine the location of mean, median and mode classes from the given Histogram of Prices of 300 cars. Page 1 of 21 a. ππππ = 1.449 − 1.995, ππππππ = 1.995 − 2.495, ππππ = 2.495 − 2.995 b. ππππ = 1.449 − 1.995, ππππππ = 2.495 − 2.995, ππππ = 1.995 − 2.495 c. π΄πππ = π. πππ − π. πππ, πππ πππ = π. πππ − π. πππ, πππ π = π. πππ − π. πππ d. ππππ = 1.995 − 2.495, ππππππ = 1.995 − 2.495, ππππ = 1.995 − 2.495 Q.2 4. The mean marks got by 68 students in the subject of Statistics are 65. The mean of the top 10 of them was found to be 80 and the mean of the last 10 was known to be 50. The mean of the remaining 48 students is determined by a. 60 b. 65 c. 0 d. 48 (10 × 80) + (10 × 50) + (48 × πΜ 3 ) 65 = 10 + 10 + 48 5. Which of the following statement is not a property of the standard deviation? a. It is always negative number b. It is affected by extreme values in a data set c. It is based on all the values in the data set d. It is the most widely used measure of dispersion Attempt all short questions a. Two data sets of weights (Pounds) of 469 and 478 applicants for admission in a cadet college at 2012 and 2013 respectively are given below 1. Find the missing values in the table below. Class Mean Variance Median Q1 Q3 Minimum Maximum 2012 219.7741 26.5871 220 200 240 160 300 Page 2 of 21 (5) 2013 219.3646 26.2187 220 200 240 160 300 2. Indicate the skewness or symmetry for both classes. Which class contains outlier(S). identify it? (2+1) ππππ < ππππππ Indicate that distribution for is 2013 less negatively skewed than 2012 In year 2012 upper outlier is 308 and lower outlier is 315 In year 2013 only upper outlier is 308 3. Indicate in which year the cadets have more homogenous weights? (1) √26.5871 πΆ. π2012 = × 100 = 2.3462 219.7741 √26.2187 πΆ. π2013 = × 100 = 2.3342 219.3646 Year 2013 is more homogenous in weights than 2012. b. The histogram and excel outcome of descriptive statistics of One Year return for funds of 479 retired police officers for year 2016 are given below 7.217870564 0.082967258 7.31 8.89 Bin / upper class limits 0.3 1.71 3.12 4.53 1.81582711 5.94 82 22.55% 3.297228092 1.448602426 0.458405475 14.1 -1.1 13 3457.36 479 7.35 8.76 138 147 51.36% 82.05% 10.17 65 95.62% 11.58 12.99 14.4 More 17 3 1 0 99.16% 99.79% 100.00% 100.00% Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count Frequency Cumulative % 1 4 5 16 0.21% 1.04% 2.09% 5.43% 1. Complete a frequency distribution and draw a histogram for a given frequency distribution (5) Lower class limit - 1.1 0.31 1.72 3.13 4.54 5.95 7.36 8.77 10.18 11.59 13.00 Bin / upper class limits 0.30 1.71 3.12 4.53 5.94 7.35 8.76 10.17 11.58 12.99 14.4 Frequency Cumulative % 1 4 5 16 82 138 147 65 17 3 1 0.21% 1.04% 2.09% 5.43% 22.55% 51.36% 82.05% 95.62% 99.16% 99.79% 100.00% Page 3 of 21 More 0 100.00% Chart Title 160 138 140 147 120 100 82 80 65 60 40 20 1 4 5 16 17 3 1 0 2. Detect any outlier (s) in this data set if it exist in your opinion (1) No outlier 3. Find a better summary measure of central tendency whether it is the mean or the median? Explain. (1.5) Distribution is negatively skewed and mean = 7.22 and median = 7.31 ππππ < πππ πππ Distribution is less negatively skewed so mean is best measure of central tendency. 4. Identify the symmetry and skewness of 10 year returns of 479 officers. (1.5) Distribution is negatively skewed and mean = 7.22 and median = 7.31 ππππ < πππ πππ officers 5. Find the (approximate) value of the 95th percentile. Give a brief interpretation of this percentile. (1.5) From table column of cumulative% indicate that 95.62% values are less than from 10.17and it is known that (8.77+10.17) / 2 = 9.47 95TH Percentile indicate that a value 9.47 contain less than 95% values and less than 5%. 6. Indicate is there any consistency in 10 year returns of 479 officers? (1.5) 1.8158 πΆ. π2012 = × 100 = 25.1572 7.2178 C.V is high so it indicates less consistency in 10 year returns of 479 Page 4 of 21 7. Find the percentages of the given data values that fall within 1.5 standard deviation of the mean. (πΜ ± 1.5 π. π·) (7.2178 ± (1.5 × 1.8158)) = (4.49, 9.94) LCL 3.13 4.54 5.95 7.36 8.77 UCL 4.53 5.94 7.35 8.76 10.17 (πΜ ± 1.5 π. π·) = frequency 16 82 138 147 65 448 × 100 = 93.53% 479 THE END Page 5 of 21 (4) (2) Q.1) For each of these questions, choose the option (a, b, C or d) that is TRUE. Directions: Answer all Questions and please give me a justification on why you chose that answer. Thank you. 1. The following data represent the numbers of minor penalties accrued by each of the 30 National Hockey League franchises during the 2007–08 regular season. 318 336 337 339 362 363 366 369 372 375 378 381 384 385 386 387 390 393 395 403 405 409 417 431 433 434 438 444 461 480 The 8th Decile penalties of League will be determined as a. 431 b. 417 c. 433 d. πππ+πππ π ππ π × ππ = = ππ ππ ππ ππππ πππππ + ππππ πππππ π«π = ππ 2. Geometric mean and harmonic mean for the values 3, -11, 0, 63, -14, 100 are a. 0 and 3 b. 3 and -3 c. 0 and 0 d. Infinite and impossible π«π = 3. Which of the following is the mean of the squared deviations of x values from the mean? a. Mean deviation from mean b. Variance c. Standard error d. Standard deviation 4. The mean of 20 values calculated as 312.6452. which of the following is always true? a. ∑(π − 312.6452) = 20 b. ∑(π − 312.6452) ≥ 20 c. ∑(πΏ − πππ. ππππ) = π d. ∑(π − 312.6452) ≤ 0 5. In a study of distances travelled to a college by commuting students, data from 100 commuters yielded a mean of 8.73 miles. After the mean was calculated, data came in late from three students, with respective distances of 11.5, 7.6, and 10.0 miles. Calculate the mean distance for all 103 students. a. 8.73 b. 9.021 c. 37.83 Page 6 of 21 d. 8.75 ∑π 100 ∑ π = 100 × 8.73 = 873 8.73 = True sum ∑ π = 873 + 11.5 + 7.6 + 10 = 902.1 902.1 πΜ = = 8.75 103 6. Refer to the graph below Frequency Polygone 139 120 83 52 32 1 64 0 67 9 70 3 73 76 79 82 85 88 2 91 A. Mode weight is a. 78.9 b. 82 c. 80.5 d. 80 B. Median weight is a. 78.9 b. 82 c. 80.5 d. 80 ∑ π = 1 + 0 + 9 + 52 + 83 + 120 + 139 + 32 + 3 + 2 = 441 441 ππππππ = = 220.5 πΆπΉ = 1,1,10,62,145,265 2 265 contain 220.5 value so from graph 6th class contains 82 class mark. C. Is this data skewed in any direction? a. Yes, positively skewed c. No, symmetrical b. Yes, negatively skewed d. Not symmetrical nor skewed 7. The Box and Whisker Plots for the weights (Pounds) and height (inches) of 482 applicants for admission in a cadet college at 2014 are given below Page 7 of 21 1. Which of the following is an accurate comparison of the box plots? a. On both plots, the median is exactly half way between the Lower and Upper Quartiles. (Yes or No) b. Weights of cadets has a smaller range & heights of cadets has a larger IQR (Inter Quartile Range = π3 − π1) (Yes or No) Q.2 2. Inter Quartile Range of weights of cadets is equal to 239.25 -200 = 39.25 and Inter Quartile Range of heights of cadets is equal to 82 – 76 =6 Attempt all short questions a. The geometric mean of four numbers were calculated as 107.8482. After it was noted that the third value 111.3 wrongly considered instead of 110.25 value. Find the true geometric mean? (4) Incorrect G.M 1 107.8482 = (π₯1 × π₯2 × 111.3 × π₯3 )4 (107.8482)4 = (π₯1 × π₯2 × 111.3 × π₯3 ) Page 8 of 21 Corrected G.M 1215504.13316 = (π₯1 × π₯2 × π₯3 ) 1215504.13316 = (π₯1 × π₯2 × π₯3 ) 1215504.13316 × 110.25 = (π₯1 × π₯2 × π₯3 × π₯4 ) 1 πΊ. π = (134009330.68129)4 = 107.5929 b. The histogram and excel outcome of descriptive statistics of Assets for funds of 479 retired police officers for year 2016 are given below Column1 Mean 1644.716305 Bin / upper class limits Standard Error 128.1842876 1942.546 371 77.45% Median 547.73 3880.963 57 89.35% Mode 779.93 5819.38 13 92.07% Standard Deviation 2805.450115 7757.797 10 94.15% Sample Variance 7870550.35 9696.214 14 97.08% Kurtosis 9.85459805 11633.631 3 2.968398968 13572.048 5 98.75% Skewness Frequency Cumulative % 97.70% 97th Percentile Range 19384.17 15510.465 5 99.79% Minimum 4.13 17448.882 0 99.79% Maximum 19388.3 19387.29 0 99.79% Sum 787819.11 21325.707 1 100.00% Count 479 More 0 100.00% 1. Complete a frequency distribution and draw a histogram for a given frequency distribution (5) Page 9 of 21 Chart Title 400 371 350 300 250 200 150 100 50 57 13 10 14 3 5 5 0 1 0 0 0 LCL upper class limits Frequency 4.13 1942.546 371 1942.547 3880.963 57 3880.964 5819.38 13 5819.39 7757.797 10 7757.798 9696.214 14 9696.215 11633.631 11633.632 13572.048 5 13572.049 15510.465 5 15510.466 17448.882 0 17448.883 19387.29 0 19387.300 21325.707 1 3 97.70% 2. Detect any outlier (s) in this data set if it exist in your opinion (1) Yes from table and Histogram indicate that outliers exist in upper tail of distribution from 19387.291-21325.707. 3. Find a better summary measure of central tendency whether it is the mean or the median? Explain. (1.5) Median is better summary measure of central tendency. 4. Identify the symmetry and skewness of this data. (1.5) Histogram indicate that distribution is highly positively skewed. Mathematically ππππ > πππ πππ. 5. Find the (approximate) value of the 97th percentile. Give a brief interpretation of this percentile. (1.5) From given frequency distribution table 9696.215 11633.631 Page 10 of 21 97.70% 9696.214 + 11633.631 = 10664.9225 2 A value 10664.9225 contain less than 97% values and 3% greater values. π97 = 6. Indicate is there any homogeneity in weights of 482 cadets? ππππ. ππππ πͺ. π½ = × πππ = π. ππππ ππππ. ππππ (1.5) 7. Find the percentages of the given data values that fall within 1.8 standard deviation of the mean. (πΜ ± 1.8 π. π·) (4) ππππ. ππππ ± (π. π × ππππ. ππππππ) (−ππππ. ππ, ππππ. ππ) upper class limits LCL Frequency 4.13 1942.546 371 1942.547 3880.963 57 3880.964 5819.38 13 5819.381 7757.797 10 451 × 100 = 94.15% 479 THE END (3) Q.1) For each of these questions, choose the option (a, b, C or d) that is TRUE. Directions: Answer all Questions and please give me a justification on why you chose that answer. Thank you. 1. Consider the following two data sets. Data Set I 12 25 37 8 41 Data Set II 19 32 44 15 48 Notice that each value of the second data set is obtained by adding 7 to the corresponding value of the first data set. The mean and standard deviation for Data Set I is 24.6 and 14.6389 respectively. Then the mean and variance for Data Set II is equal to a. Mean = 24.6 and Variance = 214.3 b. Mean = 31.6 and Variance = 14.6389 c. Mean = 31.6 and Variance = 214.3 d. Mean = 24.6 and Variance = 14.6389 π =π+7 πΜ = πΜ + 7 πΜ = 24.6 + 7 = 31.3 ππ΄π (π) = ππ΄π (π + 7) Page 11 of 21 ππ΄π (π) = ππ΄π (π) + ππ΄π (7) ππ΄π (π) = ππ΄π (π) ππ΄π (π) = 14.63392 = 214.3 2. If the coefficient of variation for prices is 5.483% and the variance price π π 2 0.7225 then mean is equal to a. 1.3177 b. 15.5 c. 7.5889 d. 0.5483 π. π· πΆ. π = × 100 ππΈπ΄π √0.7225 5.483% = ππΈπ΄π √0.7225 ππΈπ΄π = = 15.5 0.05843 3. Through visualization not mathematically determine the location of mean, median and mode classes from the given Histogram. a. b. c. d. ππππ = 10.3, ππππππ = 10.1, ππππ = 10.1 π΄πππ = ππ. ππ, πππ πππ = ππ. π, πππ π = ππ. π ππππ = 10.0, ππππππ = 10.2, ππππ = 10.1 ππππ = 10.3, ππππππ = 10.2, ππππ = 10.2 Highest frequency = 14 so Mode = 10.1 π 2+6+8+14+12+9+4+4+6 65 Median = 2 = = 2 = 32.5 2 From cumulative frequency frequency 2 6 8 14 12 9 Cumulative 2 8 16 30 42 51 frequency 4 55 4 59 6 65 Class mark = 10.2 = median For group data mean ∑ ππ₯ (9.5 × 2) + (9.9 × 6) + (10 × 8) + β― + (10.6 × 6) 662.7 πΜ = = = ∑π 65 65 = 10.19 Page 12 of 21 4. On a 300-mile auto trip, Lisa covered 52 mph for the first 100 miles, 65 mph for the second 100 miles, and 58 mph for the last 100 miles. The average speed covered by Lisa for the trip equals to a. 59 b. 58.09 c. 58.33 d. 57.85 3 Harmonic mean 1 1 1 = 57.85 ( + + ) 52 65 58 5. A small country bought oil from three different sources in one week, as shown in the following table. Source Barrels Purchased. Price per Barrel ($) Mexico 1000 51 Kuwait 200 64 Spot Market 100 70 The mean price per barrel for all 1300 barrels of oil purchased in that week equals to a. 61.667 b. 433.33 c. 54.46 d. 64 (1000×51)+(200×64)+(100×70) ππππβπ‘ππ ππππ = πΜ π = =54.46 1000+200+300 Q.2 Attempt all short questions a. One Year return for funds of 479 retired police officers for year 2016 collected and this information has been summarised on the box plots below. 1. Describe the five number value summary Minimum value Maximum value Median 1st Quartile 3rd Quartile mean 0.21 158.00 53.00 29.00 81.00 62.04 Page 13 of 21 (5) 2. Is the data set skewed in any direction? If yes, detect it as skewed to the right or to the left? Does this data set contain any outliers? If yes then identify it. (2+2) Yes, Box and Whisker plot shows positive skewness. All outliers lie at upper tail of plot and Lower outlier is 162 and highest outliers is 501. There are 7 outliers in upper tail of plot. b. The histogram and excel outcome of descriptive statistics of hights (inches) of 439 applicants for admission in a cadet college at 2000 are given below 78.93849658 0.18088487 79 81 Bin / upper class limits 65 68 71 74 3.789958918 77 84 32.80% 14.3637886 0.516420999 -0.395616882 80 83 86 117 142 30 Range 28 89 4 Minimum 63 92 2 59.45% 91.80% 98.63% 99th 99.54% Percentile 100.00% Maximum Sum 91 34654 More 0 100.00% Count 439 Heights (Inches) Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Frequency Cumulative % 2 0 8 50 0.46% 0.46% 2.28% 13.67% 1. Complete a frequency distribution and draw a histogram for a given frequency distribution (5) Histogram 160 142 140 117 120 100 84 80 60 50 40 20 30 2 0 63-65 66-68 8 4 2 87-89 90-92 0 Lower class limit 69-71 72-74 Bin / upper class limits 75-77 78-80 Frequency Page 14 of 21 81-83 84-86 63 66 69 72 75 78 81 84 87 65 68 71 74 77 80 83 86 89 90 92 2 0 8 50 84 117 142 30 4 99th Percentile 2 2. Detect any outlier (s) in this data set if it exist in your opinion There are 2 outliers in lower tail of distribution (1) 3. Find a better summary measure of central tendency whether it is the mean or the median? Explain. (1.5) Distribution is negatively skewed then median is better centre of tendency value. 4. Identify the symmetry and skewness of this data. (1.5) Negatively skewed distribution mean < median < mode 5. Find the (approximate) value of the 99th percentile. Give a brief interpretation of this percentile. (87+89) π99 = 2 = 88 (1.5) This value contains less than 99% of data set values and 1% greater than it. 6. Indicate is there any homogeneity in hights of 439 cadets? (1.5) 3.7899 C.V = 78.9385 × 100 = 4.18% 7. Find the percentages of the given data values that fall within 1.8 standard deviation of the mean. (πΜ ± 1.8 π. π·) (4) (78.9385 ± (1.8 × 3.7899)) = (72.11, 85.76) 50 + 84 + 117 + 142 + 30 423 = × 100 = 96.33% 439 439 LCL 72 75 78 81 84 UCL Frequency 74 50 77 84 80 117 83 142 86 30 The End Page 15 of 21 (4) Q.1) For each of these questions, choose the option (a, b, C or d) that is TRUE. Directions: Answer all Questions and please give me a justification on why you chose that answer. Thank you. 1. Calculate The following data represent the numbers of minor penalties accrued by each of the 40 National Hockey League franchises during the 2007–08 regular season. 1.6 3.3 4.4 1.9 3.4 4.5 2.2 3.4 4.7 2.5 3.4 4.7 2.6 3.5 2.6 3.5 2.9 3.6 3.0 3.7 3.0 3.7 3.1 3.7 3.1 3.8 3.1 3.8 3.1 3.9 3.2 3.9 3.2 4.1 3.2 4.1 3.3 4.2 3.3 4.3 The 9th Decile penalties of League will be determined as a. 4.3 b. 4.4 π.π+π.π c. π d. 4.2+4.3 2 9 × 40 = 36 10 2. If the three observations are a = 2, b = - 2 and c = 2 then their geometric mean will be: a. 2 b. 0 c. infinite d. – 2 3. The following data give the numbers of times 10 persons used their credit cards during the past 3 months. Mean deviation from mean is equal to π·9 = X π − πΜ |π − πΜ | 9 6 9-10.9 6-10.9 1.9 4.9 a. 0 b. 64.8 c. 6.48 28 2810.9 14 1410.9 17.1 3.1 2 2-10.9 8.9 18 1810.9 7.1 7 3 7-10.9 3-10.9 3.9 7.9 16 1610.9 5.1 6 6-10.9 4.9 d. 10 ∑π = 10.9 π ∑|π − πΜ | 64.8 π. π· = = = 6.48 π 10 4. In a study of speed of a car covered a distance travelled to a college by a student, a harmonic mean yielded 16.7442 miles/ hour. After the mean was calculated that one value of 16 misprinted mistakenly and true value was 10. Then the true Harmonic mean is equal to. a. 10.7442 b. 13 c. 15.7442 d. 13.846 3 16.7442 = 1 1 1 + + π π 16 1 1 1 + + = 0.1792 π π 16 πΜ = Page 16 of 21 3 = 13.846 1 1 (0.1792 − (16) + (10)) π». π = 5. Consider the following two data sets. Data Set I: 4 8 15 9 11 Data Set II 8 16 30 18 22 Notice that each value of the second data set is obtained by multiplying the corresponding value of the first data set by 2. The mean and variance of Data Set I is calculated as 9.4 and 4.03732 respectively. The mean and variance of Data Set II will be equal to a. Mean = 2 × 9.4 and Variance = 4.03732 b. Mean = 2 × 9.4 and Variance = 2 × 4.03732 c. Mean = π × π. π and Variance = π × π. πππππ d. Mean = 4 × 9.4 and Variance = 4 × 4.03732 6. Refer to the graph below Frequency polygone 175 158 62 1 0.395 55 5 0.615 0.835 1.055 1.275 1.495 8 13 1.715 1.935 0 2.155 1 2.375 1 2.595 A. Mode expense rate is a. 0.835 b. 1.055 (Highest frequency 175 class ) c. 1.275 d. 0.945 B. Median expense rate is a. 0.835 b. 1.055 479 ππππππ = 2 = 239.5 Cumulative frequency frequency CF 1 1 5 6 62 68 175 243 158 401 55 456 8 464 13 477 0 477 1 478 c. 1.275 d. 0.945 C. Is this data skewed in any direction? c. Yes, positively skewed c. No, symmetrical d. Yes, negatively skewed d. Not symmetrical nor skewed Page 17 of 21 1 479 7. The Box and Whisker Plots for one year and three-year return for funds of 479 retired police officers for year 2016 are given below 1. Which of the following is an accurate comparison of the box plots? a. The median for the three-year return suggests better than the oneyear-return on average. (Yes or No) 1-year-return ππππππ = 0.41 3-year-return ππππππ = 8.09 b. Both have outliers in left tail and right tail. (Yes or No) c. On both plots, the median is exactly half way between the Lower and Upper Quartiles. (Yes or No) Q1= -3.59 and Q3=11.34 for 1-year-return ππππππ = 0.41 Q1=6.14 and Q3=9.86 for 3-year-return ππππππ = 8.09 d. Three-year-return has a smaller range & one-year-return has a larger IQR (Inter Quartile Range = π3 − π1) (Yes or No) Range = 15.32 - 0.77 = 14.55 IQR = 2.59- (-3.59) = 6.18 Q.2 2. Inter Quartile Range of one-year-return is equal to 6.18 and Inter Quartile Range of three-year-return is equal to 9.86-6.14=3.72 Attempt all short questions a. How much does the typical American family spend on average to go away on vacation each year? Twenty-five randomly selected households reported the following vacation expenditures (rounded to the nearest hundred dollars) during the past year: 2500 500 800 0 100 0 200 2200 0 200 0 1000 900 321,400 500 500 100 0 8200 900 0 1700 1100 600 3400 a. Find mean, median and mode of given data. (3) Page 18 of 21 Mean Median Mode 13872 500 0 b. Label the best measure of central tendency to answer the original question? (1) ππππ > πππ πππ > πππ π Given distribution is highly positively skewed and median is best measure of central tendency. b. The histogram and excel outcome of descriptive statistics of heights (Inches) of 441 applicants for admission in a cadet college at 2014 are given below Column1 Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 78.95464853 0.180517434 79 81 3.790866107 14.37066584 0.228646046 Bin / upper class limits 65 68 71 74 77 80 83 -0.354535468 28 63 91 More 34819 441 Frequency 86 89 92 Cumulative % 1 0 9 52 83 120 139 91.61% 32 3 2 0 0.23% 0.23% 2.27% 14.06% 32.88% 60.09% 90th Percentile 98.87% 99.55% 100.00% 100.00% 1. Complete a frequency distribution and draw a histogram for a given frequency distribution (5) LCL 63 66 69 72 75 78 81 UCL 65 68 71 74 77 80 83 84 87 90 86 89 92 Frequency 1 0 9 52 83 120 139 90th Percentile 32 3 2 Page 19 of 21 2. Detect any outlier (s) in this data set if it exist in your opinion (1) The given frequency distribution and graph represent one outlier at lower tail of distribution. In class 63-65 contains one outlier. 3. Find a better summary measure of central tendency whether it is the mean or the median? Explain. (1.5) The distribution seems to be negatively skewed but less negatively skewed. ππππ < πππ πππ so median is best measure of central tendency 4. Identify the symmetry and skewness of this data. (1.5) The distribution seems to be negatively skewed but less negatively skewed. ππππ < πππ πππ 5. Find the (approximate) value of the 90th percentile. Give a brief interpretation of this percentile. (1.5) For class 81-83 Cumulative frequency percentage is 91.61%. We can find ππ+ππ 90th Percentile as π = ππ π·ππ = ππ represent that 90% values contain less than 82 value and 10% greater than 82 value. 6. Indicate is there any homogeneity in hights of 441 cadets? (1.5) π. ππππ πͺ. π½ = × πππ = π. ππππ ππ. ππππ 7. Find the percentages of the given data values that fall within 0.5 standard deviation of the mean. (πΜ ± 0.5 π. π·) (4) (78.9546 ± (0.5 × 3.7908)) = (77.05, 80.85) LCL 75 78 81 UCL Frequency 77 83 80 120 83 139 83+120+139 441 The End Page 20 of 21 × 100 = 77.55% Page 21 of 21
