Assignment02_OROUR - Slide50 a UpperQuartile(Q3(3/4(n...

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Slide 50 a) Upper Quartile (Q3): (3/4)(n+1) = (3/4)(16) = 12 The twelfth measurement is 9 b) 80 th  Percentile:  (p(n+1) )/100 = (80 x 16) / 100 = 12.8 The thirteenth measurement is 9 Slide 52 a) z  score for 10 = z = (x – (bar x)) / s = (10-7.6)/2.32 = 1.034 z score is positive, so I conclude the score of 10 lies a distance of 1.034 standard  deviations above. b) z  score for 10 = z = (x – (bar x)) / s = (5-7.6)/2.32 = -1.121 c) z score is negative, so I conclude the score of 5 lies a distance of -1.121 standard  deviations below. 2.34 a)
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Stem and Leaf of Radioactive LichenN = 9 Leaf Unit = 0.10 -4 1 2 5 6 9 -5 0 0 6     -6 0         c) Histogram d) I personally like the histogram graph as being the most informative. You can really see the  difference in the number of radioactive lichen with a bar chart. The dot plot is the second most  informative chart, as it also shows the trend of radioactive lichen. The least useful was the stem  and leaf display. e) 44.444 percent of the measurements have a radioactivity level of -5.00 or lower.  2.90 a) bar x, the mean, is 206/25 which equals 8.24. s^2, the sample variance is 61.3696/24 which equals 2.557.
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This note was uploaded on 07/28/2011 for the course STATISTICS BA 240 taught by Professor Winnieli during the Spring '11 term at Bellevue College.

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Assignment02_OROUR - Slide50 a UpperQuartile(Q3(3/4(n...

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