Chapter02 solutions

# Chapter02 solutions - Student Solutions Manual Chapter 2...

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Student Solutions Manual 7 Chapter 2 Solutions When applicable, the first few problems in each section will be done following the appropriate step by step procedures outlined in the corresponding sections of the chapter. Following problems will provide key points and the answers to the questions, but all answers can be arrived at using the appropriate steps. Section 2.1 Exercises 2.1 Given n = 2,000, the minimum number of groups for a grouped data frequency distribution determined using the 2 k n guideline is: 2 k n or 11 2 2,048 2,000 =≥ Thus, use k = 11 groups. 2.3 Step 1: List the possible values. The possible values for the discrete variable are 0 through 12. Step 2: Count the number of occurrences at each value. The resulting frequency distribution is shown as follows: 2.5 Note that two classes have a width of 8.05 – 7.85. Thus, the class width equals (8.05 – 7.85)/ 2 = 0.10. The upper class boundary equals the lower class boundary plus the class width. So, as an example, the first class boundary equals 7.85 + 0.10 = 7.95, and so on. The first class has the same relative frequency and cumulative relative frequency. The frequency is obtained by multiplying the sample size times the relative frequency. So the first class frequency equals 50(0.12) = 6. The other class frequencies follow similarly. Each cumulative relative frequency is produced by adding the respective class relative frequency to the preceding cumulative relative frequency. So the second class relative frequency is obtained as 0.48 – 0.12 = 0.36. The other calculations follow similarly yielding

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Chapter 2 8 Class Frequency Relative Frequency Cumulative Relative Frequency 7.85 – < 7.95 6 0.12 0.12 7.95 – < 8.05 18 0.36 0.48 8.05 – < 8.15 12 0.24 0.72 8.15 – < 8.25 5 0.10 0.82 8.25 – < 8.35 9 0.18 1.00 2.7 a. Proportion of days in which no shortages occurred = 1 – proportion of days in which shortages occurred = 1 – 0.24 = 0.76 b. Less than \$20 off implies that overage was less than\$20 and the shortage was less than \$20 = (proportion of overages less \$20) – (proportion of shortages at most \$20) = 0.56 – 0.08 = 0.48 c. Proportion of days with less than \$40 over or at most \$20 short = Proportion of days with less than \$40 over – proportion of days with more than \$20 short = 0.96 – 0.08 = 0.86. 2.9 a. Step 1 and Step 2. Group the data into classes and determine the class width: The problem asks you to group the data. Using the 2 k n guideline we get: 2 k 60 so 2 6 60 Class width is: 33 . 1 6 2 10 # = = = Classes Minumum Maximum W which we round up to 2.0 Step 3. Define the class boundaries: Since the data are discrete, the classes are: C l a s s 2 - 3 4 - 5 6-7 8-9 10-11 Step 4. Count the number of values in each class: Class Frequency Relative Frequency 2-3 2 0.0333 4-5 25 0.4167 6-7 26 0.4333 8-9 6 0.1000 10-11 1 0.0167
Student Solutions Manual 9 b. The cumulative frequency distribution is: Class Frequency Cumulative Frequency 2-3 2 2 4-5 25 27 6-7 26 53 8-9 6 59 10-11 1 60 c.

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## This note was uploaded on 10/06/2008 for the course QM 2241 taught by Professor Smith during the Spring '08 term at Troy.

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Chapter02 solutions - Student Solutions Manual Chapter 2...

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