Answer
$\mu=E(X)=0.17 mm$
$\sigma^{2}=0.0002 mm^{2}$
Work Step by Step
Let $X=\frac{1}{100}y,y=15,16,17,18,19$
$\mu=E(X)=(\frac{1}{100}).\frac{19+15}{2}=0.17$
$\sigma^{2}=(\frac{1}{100}^{2}).\frac{(19-15+1)^{2}-1}{12}=0.0002$
Applied Statistics and Probability for Engineers, 6th Edition
by Montgomery, Douglas C.; Runger, George C.
Published by
Wiley
ISBN 10:
1118539710
ISBN 13:
978-1-11853-971-2
Chapter 3 - Section 3-5 - Discrete Uniform Distribution - Exercises - Page 79: 3-78
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