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NAME: _________________________________ Describe each distribution (shape, center, spread): 1) mean = 0.2 std dev =0.677 Median = 0 Q1 = -0.7 Q3 = 0.9 min = -1.8 max = 2.4 2) mean = 1982 std dev = 25 Median = 1995 Q1 = 1985 Q3 = 2001 min = 1950 max = 2009 Unimodal, roughly symmetric Center at mean of 0.2 units Std deviation of 0.677 units Range of (-1.8, 2.4) units No outliers Unimodal, left skewed Center at median of 1995 years IQR of 16 years Range of (1950, 2009) years No outliers Gap at approx. 1955 years 3) mean = 2.9 std dev = 0.3 Median = 3.35 Q1 = 2.7 Q3 = 3.6 min = 1.6 max = 4.5 Bimodal, left skewed Center at median of 3.35 points IQR of 0.9 points Range of (1.6, 4.5) points possible outlier at 4.5 points Unimodal, slight left skew Center at median of 5 units IQR of 4.5 units Range of (0,16) units outlier at 16 units 4) mean = 7.2 std dev = 2.7 Median = 5 Q1 = 3 Q3 = 7.5 min = 0 max = 16 5) Go back to the picture/summary stats in #4. Check to see if the highest observation is an outlier. LF = 3 – (1.5 * 4.5) = -3.75 UF = 7.5 + (1.5 * 4.5) = 14.25 Normal data = (-3.75, 14.25) Yes, the observation at 16 is an outlier 6) Using the following summary statistics, how many outliers are there? min 2 a. b. c. d. e. Q1 18 Med 22 Q3 25 Max 40 Work: LF = 18 – (1.5*7) = 7.5 UF = 25 + (1.5*7) = 35.5 Normal data = (7.5, 35.5) Min (2) is an outlier Max (40) is an outlier None 1 2 At least 1 At least 2 7) Create a frequency histogram of the list AGEMO. This is a list of the ages in months of the students in our class. 13 10 8 # 6 3 190 200 210 220 Ages of students in months 8) Create a relative frequency histogram of the list PRES. These are the ages of death of all US presidents. Then describe the distribution. 18.9 % 16.2 5.4 2.7 45 50 55 60 65 75 85 Presidents Ages of Death 95 Description: Right skewed, unimodal. Center at the Median of 67 years. Spread is an IQR of 18 years, and a range of (46, 93) years. 9) Use the list HT (heights of students in our class) a. Find the Mean, Median, Range, Q1, Q3, IQR, and Standard deviation. mean 67.865 std dev 4.178 min 58 Q1 66 Med 68 Q3 71 Max 76 Range (58, 76) IQR 5 b. Create a boxplot of the data -------------------------------------------------------------------58 59 66 68 71 76 Heights of students in our class c. Suppose we add the data point 78” to this set of data. Indicate how each of the statistics in part (a) would change: increase, decrease, or stay about the same. mean std dev Increase Increase min Same Q1 Same Med Same Q3 Same Max Range Increase Increase IQR Same d. Going back to the original set of data: suppose we add the data point 50” to this set of data. Indicate how each of the statistics in part (a) would change: increase, decrease, or stay about the same. mean std dev min decrease Increase decrease Q1 Same Med Same Q3 Same Max Same Range Increase IQR Same 10) Give a set of numbers that have a standard deviation of 0. 5, 5, 5, 5, 5, 5, 5 (any set of numbers where they are all the same) 11) Create parallel boxplots of the lists TST1F and TST1M. These are test scores for females and males. FEMALES MALES ----------------------------------------------------------------------------------20 45 55 62 68 76 82 90 92 TEST SCORES 12) Based solely on the mean and median given, decide on the shape of the distribution, and what measure of center and spread you would report. (a) Mean = 100 (b) Mean = 20 (c) Mean = 934 Median = 98 Median = 41 Median = 850 Approx. symm. Mean & std dev. Left skew Median & IQR right skew median & IQR 13) The following are quiz scores from two Algebra 1 Classes. Class 1: {68, 93, 53, 100, 77, 86, 91, 88, 72, 74, 66, 82} and Class 2: {77, 91, 82, 68, 75, 72, 85, 65, 70, 79, 94, 86} a. Find the means of both classes. Which would you rather be in? Mean of Class 1 = 79.167 Mean of Class 2 = 78.667 I would rather be in Class 1- they have the higher mean b. Find the standard deviations of both classes. Which class was more consistent in their scores? Why? Std. Dev. of Class 1 = 13.361 Std. Dev. of Class 2 = 9.188 Class 2 had the more consistent scores. I would now rather be in class 2 because their mean was not that much lower and they were more consistent. Class 1 was more spread out.