If Y has a geometric distribution with success probability .3, what is the largest value, y0, such that P(Y > y0) ≥ .1?
Step 1 of 3
Step1 of 2:
We have random variable ‘Y’ it follows geometric distribution with parameter ‘p = 0.3’.
Then the probability mass function of geometric distribution is given by:
x = random variable
p = probability of success(Parameter)
n = sample size
We need to find the largest value of such that P(Y > ) = 0.1.
Step2 of 2:
We have p = 0.3 then q = 1 - p
= 1 - 0.3
We need to find the value of such that:
Apply natural log on both side we get
Hence, the largest value of is 6.4570.
Textbook: Mathematical Statistics with Applications
Author: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer
Mathematical Statistics with Applications was written by and is associated to the ISBN: 9780495110811. This textbook survival guide was created for the textbook: Mathematical Statistics with Applications , edition: 7. The answer to “If Y has a geometric distribution with success probability .3, what is the largest value, y0, such that P(Y > y0) ? .1?” is broken down into a number of easy to follow steps, and 23 words. This full solution covers the following key subjects: distribution, geometric, largest, Probability, success. This expansive textbook survival guide covers 32 chapters, and 3350 solutions. The full step-by-step solution to problem: 76E from chapter: 3 was answered by , our top Statistics solution expert on 07/18/17, 08:07AM. Since the solution to 76E from 3 chapter was answered, more than 500 students have viewed the full step-by-step answer.