hw1 solution - Chapter 1 17 a. Number Nonconforming 0 1 2 3...

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Chapter 1 17 a. Number Nonconforming Frequency RelativeFrequency(Freq/60) 0 7 0.117 1 12 0.200 2 13 0.217 3 14 0.233 4 6 0.100 5 3 0.050 6 3 0.050 7 1 0.017 8 1 0.017 doesn't add exactly to 1 because relative frequencies have been rounded 1.001 b. The number of batches with at most 5 nonconforming items is 7+12+13+14+6+3 = 55, which is a proportion of 55/60 = .917. The proportion of batches with (strictly) fewer than 5 nonconforming items is 52/60 = .867. Notice that these proportions could also have been computed by using the relative frequencies: e.g., proportion of batches with 5 or fewer nonconforming items = 1- (.05+.017+.017) = .916; proportion of batches with fewer than 5 nonconforming items = 1 - (.05+.05+.017+.017) = .866. c. The center of the histogram is somewhere around 2 or 3 and it shows that there is some positive skewness in the data. Using the rule of thumb in Exercise 1, the histogram also shows that there is a lot of spread/variation in this data. 8 7 6 5 4 3 2 1 0 .20 .10 .00 Number Relative Frequency
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33 a. 57 . 192 x , 189 ~ x . The mean is larger than the median, but they are still fairly close together.
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This note was uploaded on 04/23/2011 for the course STAT 503 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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hw1 solution - Chapter 1 17 a. Number Nonconforming 0 1 2 3...

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