Statistics: Informed Decisions Using Data (4th Edition)

Published by Pearson
ISBN 10: 0321757270
ISBN 13: 978-0-32175-727-2

Chapter 2 - Section 2.3 - Assess Your Understanding - Vocabulary and Skill Building: 5b

Answer

5. Lower limit: 0 Upper limit: 9

Work Step by Step

The -5 in the horizontal axis does not represent any class. 5 is the midpoint of the first class. The class width is 10 and the midpoint of the first class is 5: $Lower~limit~of~the~first~class=midpoint~of~the~first~class-\frac{class ~width}{2}=5-\frac{10}{2}=5-5=0$ $Upper~limit~of~the~first~class=midpoint~of~the~first~class+\frac{class ~width}{2}-1=5+\frac{10}{2}-1=5+5-1=9$ The second class lower limit is 10 ($15-\frac{10}{2}$). So, the upper limit of the first class must be 9. Otherwise the classes would overlap. Also, notice that the first class covers 10 different ages: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
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