Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 7 - Section 7.2 - Relative Frequency - Exercises - Page 465: 30a

Answer

See the picture.

Work Step by Step

Total: $N=20+21+46+13=100$ Relative frequency of "0-14": $P(0-14)=\frac{fr(0-14)}{N}=\frac{20}{100}=0.20$ Relative frequency of "15-29": $P(15-29)=\frac{fr(15-29)}{N}=\frac{21}{100}=0.21$ Relative frequency of "30-64": $P(30-64)=\frac{fr(30-64)}{N}=\frac{46}{100}=0.46$ Relative frequency of ">64": $P(\gt64)=\frac{fr(\gt64)}{N}=\frac{13}{100}=0.13$
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