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Chapter 3 Unit Test Solutions Question #1 a. Mean = 5(3) 6( 2) 8(8) 9(7) 10(5) 25 204 8.16 25 b. Median – 25 numbers / 2= 12.5, so the 13th number is the median Median = 8 c. Q1 = Median of bottom half = 5 5 5 6 6 8 8 8 8 8 8 8 = 8 d. Standard Deviation 3(5 8.16) 2 2(6 8.16) 2 8(8 8.16) 2 7(9 8.16) 2 5(10 8.16) 2 25 61.36 1.567 25 Question #2 • Box Plot A – range of 38, IQR of 22, median of 70 • Box Plot B – range of 44, IQR of 21, median of 56 • Box Plot C – range of 60, IQR of 31, median of 65 a. A median of 56 = Box Plot B b. A range of 60 = Box Plot C c. Most Consistent = Box Plot B (has smallest IQR) Question #3 • Mean = 1(3) 2(4) 3(9) 4(10) 5(6) 6(1) 33 3.45 Question #4 a. The class median would be raised by 3 marks. The person’s mark that would have been the middle remains the same, however, it will now be 3 marks higher. b. The standard deviation remains the same. The mean is 3 marks higher, however, every distance from the mean is the same as before the mark change … the value under the square root sign will remain the same, and the number of marks will be the same. Question #5 • Z-Score = 1 (1 std. dev) – using table = 84.13% • Z-Score = 2 (2 std. dev) – using table = 97.72% 97.72% – 84.13% = 13.59% y 0.25 0.2 0.15 0.1 0.05 x x x 2 x Question #6 Mean = 5, std dev = 10 X = mean – std dev – std dev X = 5 – 10 – 10 X = -15 Therefore X would have a value of -15 at -2 standard deviations from the mean. Question # 7a) Z-Score of 65 popsicles = 65 80 2.14 7 Z-Score of 90 popsicles = 90 80 1.43 7 1.62% 93.36% P(selling 65 to 90) = 93.36% - 1.62% = 90.74% 90.74% * 365 days = 331 days Therefore, you could expect to sell between 65 and 90 popsicles 331 days of the year. Question # 7b • I would think that the sales would follow a cyclical / yearly pattern (Peaking in the summer). So the date would be a better prediction (time series, line graph) • I would think that sales would be correlated with temperature (i.e. more sales on hot days). So the forecast / temperature could also be a good predictor. (Scatterplot with line of best fit). • Normal Distribution is not a good predictor in this case. Question #8a) 80 72 0.67 using table = 74.86% Z-Score = 12 This is the amount below 80%, need to subtract it from 100 to get the amount above 80%. 100 – 74.86 = 25.14% Therefore 25.14% of the students are achieving above the expectations. Question #8b) • Q1 = 25% -- using z-score table = -0.67 • Q3 = 75% -- using z-score table = 0.67 Q1 Value Q1 72 0.67 12 Q 72 0.67(12) 1 12 12 8.04 72 Q1 72 72 63.96 Q1 Q3 Value Q1 72 12 Q 72 0.67(12) 1 12 12 8.04 72 Q1 72 72 0.67 80.04 Q1 Therefore, the IQR contains marks from 63.96% to 80.04% which gives a range size of 16.08% Question #9 Jasmine Joel Measures of Central Tendency: Mean = 69.7 Mean = 70.4 Std Dev. = 62.69 Std. Dev. = 20.18 Q1 = 20 Q1 = 60 Median = 40 Median = 74.5 -All very close but Joel’s median is $74.50 compared to Jasmine’s $40. This means that 50% of the time Jasmine earns less than $40 a night. Joel only earned less than $40 one night. Q3 = 120 Q3 = 85 IQR = 100 IQR = 25 If tips are a measure of satisfaction, Joel’s customers seem a lot more satisfied than Jasmine’s. Question #9 - continued • Joel is more consistent with a standard deviation of $20.12 compared to Jasmine’s $62.70. Similarly, his tips are within a $25 range (IQR) 50% of the time, where Jasmine’s IQR is $100. • Joel’s Q1 is higher than Jasmine’s median, which means that 75% of the time he earns more than her median of $40. See Box Plots on Board Question #9 - continued Conclusions: I would have Joel. He is far more consistent which is important in a restaurant to ensure customer satisfaction. Although Jasmine has some great nights ($200, $150) she has many other nights of adequate satisfaction compared to Joel. Limitations: working same nights? Food quality affects tips? Number of tables? Sections? (smaller groups = smaller bill = smaller tip) Short period of time, how accurate a picture?