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Unformatted text preview: 351Prob. & Statisitcs for EngineersSolutions for EvenNumbered ProblemsChapter 1Section 1.2 : 2, 6, 9, 10, 12, 15, 18.2.(a) Using the same stem and leaf units as in Exercise 1, a comparative stemandleaf display of this data is:data from Exercise 1 (n=27)data from Exercise 2 (n=20)9 5 833588 6 1600234677889 7 012488127 8 133590779 27871036811212613141From this display, the cylinder data appears to be even more positively skewed than the data from Exercise 1. The data value 14.1 appears to be an outlier. From the stemandleaf display, there are 3 values in the cylinder data that have stems of 11 or larger, so there the proportion of cylinder strengths that exceed 10 is 3/20 = .15, or, 15%.(b) Both data sets have approximately the same representative value of about 8 MPa and both stemandleaf displays exhibit positive skewness. The spread of the cylinder data is larger than that of the beam data and the cylinder data also appears to contain an outlier.6.A MINITAB stemandleaf display in which each stem appears only once is:Stemandleaf of C1 N = 40Leaf Unit = 1.09 6 03466789917 7 00122244(19) 8 00111112234455578991543212010yFrequency4 9 0358A MINITAB stemandleaf display in which each stem appears twice is:Stemandleaf of C1 N = 40Leaf Unit = 1.03 6 0349 6 66789917 7 0012224417 7 (12) 8 00111112234411 8 55578994 9 032 9 58In the display with repeated stems it is apparent that there is a gap in the data at the second '7' stem. This means that there are noexam scores between 75 and 79, which seems strange compared to the rest of the scores. 10. The following Pareto chart was constructed using MINITAB:From this chart, the three most frequently occurring injury categories (A, B, and C) account for 90.8% of all injuries.12. (a)A histogram of the y data appears below. From this histogram, the number of subdivisions having no culdesacs (i.e., y = 0) is 17/47 = .362, or 36.2%. The proportion having at least one culdesac (y 1) is (4717)/47 = 30/47 = .638, or 63.8%. Note that subtracting the number of culdesacs with y = 0 from the total, 47, is an easy way to find the number of subdivisions with y 1.2EDCBA252 12 32 52 .66 .627 .63 0 .33 2 .91 0 0 .09 7 .490 .86 3 .23 2 .97 06 05 04 03 02 01 01 0 08 06 04 02 0D e f e c tC o u n tP e r c e n tC u m %PercentCount876543211 05zF re que ncy(b) A histogram of the z data appears below. From this histogram, the number of subdivisions with at most 5 intersections (i.e., z 5) is 42/47 = .894, or 89.4%. The proportion having fewer than 5 intersections (z < 5) is 39/47 = .830, or 83.0%.18. (a)The classes overlap. For example, the classes 2030 and 3040 both contain the number 30, which happens to coincide with one of the data values, so it would not be clear which class to put this observationin....
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 Summer '08
 Palaniappan

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