Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 6 - Normal Probability Distributions - 6-4 Sampling Distributions and Estimators - Beyond the Basics: 19

Answer

Yes.

Work Step by Step

Evaluating the expressions at x=0, 0.5, 1: $P(0)=\frac{1}{2(2-2\cdot0)!(2\cdot0)!}=\frac{1}{2\cdot2!\cdot0!}=\frac{1}{4}$ $P(0.5)=\frac{1}{2(2-2\cdot0.5)!(2\cdot0.5)!}=\frac{1}{2\cdot1!\cdot1!}=\frac{1}{2}=\frac{2}{4}$ $P(1)=\frac{1}{2(2-2\cdot1)!(2\cdot1)!}=\frac{1}{2\cdot0!\cdot2!}=\frac{1}{4}$ The probabilities are equal to the ones in exercise 15, therefore they describe the sampling distribution found there.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.