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HMWK1solutions

# HMWK1solutions - Chapter 1 19 a From this frequency...

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Chapter 1 19. a. From this frequency distribution, the proportion of wafers that contained at least one particle is (100-1)/100 = .99, or 99%. Note that it is much easier to subtract 1 (which is the number of wafers that contain 0 particles) from 100 than it would be to add all the frequencies for 1, 2, 3, … particles. In a similar fashion, the proportion containing at least 5 particles is (100 - 1-2-3- 12-11)/100 = 71/100 = .71, or, 71%. b. The proportion containing between 5 and 10 particles is (15+18+10+12+4+5)/100 = 64/100 = .64, or 64%. The proportion that contain strictly between 5 and 10 (meaning strictly more than 5 and strictly less than 10) is (18+10+12+4)/100 = 44/100 = .44, or 44%. c. The following histogram was constructed using Minitab. The data was entered using the same technique mentioned in the answer to exercise 8(a). The histogram is almost symmetric and unimodal; however, it has a few relative maxima (i.e., modes) and has a very slight positive skew. 15 10 5 0 .20 .10 .00 Number of particles Relative frequency

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23. a. b. The histogram is skewed right, with a majority of observations between 0 and 300 cycles. The class holding the most observations is between 100 and 200 cycles. c. [proportion 100] = 1 – [proportion < 100] = 1 – .21 = .79 900 800 700 600 500 400 300 200 100 0 30 20 10 0 brkstgth Percent 900 600 500 400 300 200 150 100 50 0 0.004 0.003 0.002 0.001 0.000 brkstgth Density
33. a 57 . 192 = x , 189 ~ = x . The mean is larger than the median, but they are still fairly close together. B Changing the one value, 71 . 189 = x , 189 ~ = x . The mean is lowered, the median stays the same. C 0 . 191 = tr x . 07 . 14 1 = or 7% trimmed from each tail. D For n = 13, Σ x = (119.7692) x 13 = 1,557 For n = 14, Σ x = 1,557 + 159 = 1,716 5714 . 122 14 1716 = = x or 122.6 81. Assuming that the histogram is unimodal, then there is evidence of positive skewness in the data since the median lies to the left of the mean (for a symmetric distribution, the mean and median would coincide). For more evidence of skewness, compare the distances of the 5th and 95th percentiles from the median: median - 5th percentile = 500 - 400 = 100 while 95th percentile -median = 720 - 500 = 220. Thus, the largest 5% of the values (above the 95th percentile) are further from the median than are the lowest 5%. The same skewness is evident when comparing

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