Stat Assignment #3 (1)

Assignment #3 Chapter 1 Describing distributions with numbers: pp. 27-49: Homework: #1-6 + 1.65, 1.66, 1.67 Center: Mean ( ࠵?̅ ) — the average (sum of the values divided by the number of values): Median ( M ) — the center value (rank order first): if two middle values, take the average of those two values Spread/variability Variance ( s 2 ) [the sum of squared deviations divided by N -1] Note: x i is each individual value in the distribution ࠵? 2 = ∑(࠵? ࠵? − ࠵?̅) 2 ࠵? − 1 Standard deviation ( s ) [the square root of the variance] Quartiles : 1 st Quartile (biggest) : median of the upper half of the distribution; 3 rd Quartile (smallest) : median of the lower half of the distribution; do NOT include the overall median in these computations Five number summary : (lowest, 3 rd quartile, median, 1 st quartile, highest) IQR: Inter-quartile range : The 1 st quartile (largest) minus the 3 rd quartile (smallest) Note : it is often easy to get “1 st ” and “3 rd ” mixed up. If you want to avoid any error, it is just as acceptable to call them: “u pper quartile ” and “ lower quartile ” . You know the upper quartile is bigger numbers, lower is smaller numbers. Box and whisker plots (“modified boxplot”) : boundaries illustrated below from largest to smallest value 1 . Large outliers (if they occur) shown as dots 2. Line and “whisker”; start at the 3 rd quartile 1.5 x IQR above the line (if no outliers, whisker at largest value) 3. The 1 st quartile (top of the box) 4. The median (line in the box between the 1 st and 3 rd quartiles) 5. The 3 rd quartile (bottom of the box) 6. Line and “whisker”; start at the 1 st quartile 1.5 x IQR below the line (if no outliers, whisker at smallest value) 7. Small outliers (if they occur) shown as dots Resistant measure: Value that does not change central measures or variability a great deal with a few changes of individual values, even if those values are large. Example: medians tend to be more resistant than means Linear transformations (x new = a + bx) #1 . Using the distribution [2 3 5 6 7 9 10] compute the mean ࠵? ̅ #2 . Using the distribution [2 3 5 6 7 9 10] compute the median ( M ) #3 . Using the distribution [2 3 5 6 7 9 10] compute the variance ( s 2 ) DUE 128 24 11 59pm SUNDAY Example of whatit Looks like in Formula upper fer 2 375 6 79410 42 4217 60 213,516,719,10 median D 52 12 67 13 67 15 672 16 612 17617 9 6 2 110 672 52 1 47 13771 4110171117131414122 52 161 04 916 561 8.66676

#4 . Using the distribution [2 3 5 6 7 9 10] compute the standard deviation ( s ) 1.65. (p. 48) The following chart notes the sale of Double stout (beer) in 12 English cities between 1913 (pre-World War 1) and 1925 (post-World War 1). The 12 numbers in the chart are percentage in 1913 beer sales of the sale of beer in 1925 . For instance, “ Bristol 94” means that sale of beer in 19 25 was 94% of the sales in 1913 — hence the sale of beer in Bristol decreased. Th e “ Liverpool 140” means that 19 25 sales were 140% of the sales in 1913 — hence the sale of beer in Liverpool increased. Bristol 94 English O 68 Glasgow 66 Manchester 190 Cardiff 112 English P 46 Liverpool 140 Newcastle-on-Tyne 118 English Agents 78 English R 111 London 428 Scottish 24 a. Mean of these data: (show work) b. Median of these data: (show work) c. Which measure does a better job at identifying a center value? 1.66. (p. 48) Refer to the data in the previous problem a. compute the standard deviation for these data The variance ( s 2 ) The standard deviation ( s ): b. compute the quartiles c. provide a 5-number summary of these data d. which provides a better description of the spread of data? BYE 52.944Mt 24 46 66 68 78 94 111 1 112 118 140 190 428 1475 mean 1475 12 122.9 24,4666 68178 94 111 112 118 140 190 428 9411 102.5 The median is 102.5 The median because it is less influenced by theoutlier 428 52 124 122.972 146 122 9 2 166 122.97 68 122 9 21 78 122.9 t t 52 194 122.917 111 122.977 112 122.912 1118 122.9121140122.972 190 122.912 1428 122.912 12 1 52 11.178 447 552 178.447 105.728 118 140 2 129 upperquartile median 1025 6668 2 67 Lowerquartile 2467 102.5 129 4283 ummary lowest La median UQ highest value The 5 summary based on the median because the mean Standard deviation are shewed by the outlier

#5 : Using the formula for conversion-of-metrics (x new = a + bx), convert the five highest high jumps of 2017 from centimeters to inches: Centimeter values [241, 240, 238, 238, 237] Formula: height (inches) = 0 + .3937(centimeters) 241: 240: 238: 238: 237: #6 : Using the conversion formula change the following Celsius temperatures [23 q , 37 q , 42 q , 15 q , -25 q ] to Fahrenheit. The formula is: temperature (fahrenheit) = 32 + 1.8(celsius) 23 q C 37 q C 42 q C 15 q C -25 q C PICTURE ON PHONE FOR M DER STAND IN 241 3937 94.880 240 3937 94.49 2381.3937 93.700 238 39377 93 70 237 3937 93.310 32 1.8 23 7 32 41.4 73 4 F 32 1.8 37 32 66.6 986 F 32 1.8 4 2 32 75 6 107 6 of 32 1.81 5 32 27 59 O F 32 1.8 25 32 45 13 0 I

Example of interquartile plot: Distribution of values: 2, 5, 8, 7, 4, 9, 10, 12, 15, 23, 36, 14, 18 Compute the median: First rank order from low to high: 2 4 5 7 8 9 10 Æ the median (the middle value) 12 14 15 18 23 36 Now compute the quartiles: [exclude the median (10) when computing quartiles] The first quartile (Q 1 ) is computed with the higher values The third quartile (Q 3 ) is computed with the lower values First quartile (Q 1 ) 12 14 15 Æ the first quartile (Q 1 = 16.5) 18 23 36 Third quartile (Q 3 ) 2 4 5 Æ the third quartile (Q 3 = 6) 7 8 9 Five number summary : {2, 6, 10, 16.5, 36} Interquartile range : the first quartile value minus the third quartile value 16.5 – 6 = 10.5 1.5 x Interquartile range : 1.5 x 10.5 = 15.75 Boxplot: 38 Outlier (36) Inner fence (16.5 + 15.75 = 32.25) Q 1 (first quartile) = 16.5 Median = 10 Q 3 (third quartile) = 6 Inner fence (6 – 15.75 = -9.75, but never lower than the minimum value: +2 ) 34 30 26 22 18 14 10 6 2