# hw1_soln - Problem 1.7 (a) Number Nonconforming Frequency

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Problem 1.7 (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 Problem 1.11 (b) A histogram of this data, using classes of width 1000 separated at 0, 1000, 2000, and 6000 is shown below. The proportion of subdivisions with total length less than 2000 is (12+11)/47 = .489, or 48.9%. Between 2000 and 4000, the proportion is (10 + 7)/47 = .362, or 36.21%. 6000 5000 4000 3000 2000 1000 0 10 5 0 length Frequency Problem 1.13 (a) The frequency distribution is: Relative Relative Class Frequency Class Frequency 0-< 150 .193 900-<1050 .019 150-< 300 .183 1050-<1200 .029 300-< 450 .251 1200-<1350 .005 450-< 600 .148 1350-<1500 .004 600-< 750 .097 1500-<1650 .001 750-< 900 .066 1650-<1800 .002 1800-<1950 .002

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The relative frequency distribution is almost unimodal and exhibits a large positive skew. The typical middle value is somewhere between 400 and 450, although the skewness makes it difficult to pinpoint more exactly than this. (b) The proportion of the fire loads less than 600 is .193+.183+.251+.148 = .775. The proportion of loads that are at least 1200 is .005+.004+.001+.002+.002 = .014. Problem 1.19 (a) The density curve forms a rectangle over the interval [4, 6]. For this reason, uniform densities are also called rectangular densities by some authors. Areas under
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## This note was uploaded on 10/17/2008 for the course STAT 390 taught by Professor Marzban during the Spring '08 term at Washington State University .

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hw1_soln - Problem 1.7 (a) Number Nonconforming Frequency

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