Solution_Assign1_Fall_2010

Solution_Assign1_Fall_2010 - • Sample median =...

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MATH144: Applied Statistics (L2, L3) - Fall 10/11 Solutions to Assignment 1: 1. (a) The frequency distribution with the given class intervals is as follows: Class interval Class midpoint Frequency Relative Cumulative Relative cumulative frequency frequency frequency 305-309 307 3 0.06 3 0.06 310-314 312 9 0.18 12 0.24 315-319 317 10 0.2 22 0.44 320-324 322 15 0.3 37 0.74 325-329 327 10 0.2 47 0.94 330-334 332 3 0.06 50 1 (b) Sample mean = 320.0800 Sample median = sample middle quartile = 320 Sample Lower quartile = 316 Sample Upper quartile = 325 Sample variance = 44.9731 Sample standard deviation = 6.7062 (c) There are 31 observations lying within one standard deviation from the mean [313.3738, 326.7862], and 48 observations lying within two standard deviation from the mean [306.6676, 333.4924]. 2. (a) Sample mean = 6.9719 Sample Lower quartile = (6+6.1)/2 =6.05
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Unformatted text preview: • Sample median = (6.9+7.0)/2 = 6.95 • Sample Upper quartile = (8.4+8.5)/2 = 8.45 • Sample interquartile range (IQR) = 8.45 - 6.05 = 2.4 • Sample variance = 8.7208 • Sample standard deviation = 2.9531 (b) See the next page. (c) Note that Lower quartile - 1.5IQR = 2 . 45 >-5 . 3 = minimum point, and Upper quartile + 1.5IQR = 12 . 05 < 15 . 1 = maximum point. Thus, we have four potential outliers. They are -5.3, -4.9, 2.1, and 15.1. Since the data is the observed value of the lead concentration (in μg/m 3 ), it cannot be negative. In other words, -5.3 and -4.9 are outliers. 1-5 5 10 15 Figure 1: Boxplot for Q2b 3. • Sample mean = 590.4444 • Sample median = 572 • Mode = 572 2...
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This note was uploaded on 10/20/2011 for the course MATH 144 taught by Professor Yu during the Spring '11 term at HKU.

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Solution_Assign1_Fall_2010 - • Sample median =...

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