Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial: 20

Answer

0.2709, the observed rate could be possible.

Work Step by Step

p=0.00034 $q=1-p=1-0.00034=0.99966$ $n⋅p=420095⋅0.00034=143≥5.$ $n⋅q=420095⋅0.99966=419952≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdot p=420095\cdot0.00034=142.8323.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{420095\cdot0.00034\cdot0.99966}=11.9492.$ 135.5 is the first value more than 135, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{135.5-142.8323}{11.9492}=-0.61.$ By using the table, the probability belonging to z=-0.72: 0.2358, hence the probability: 0.2709. This probability is not too close to 0, therefore the observed rate could be possible.
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.