Problem 76E

If Y has a geometric distribution with success probability .3, what is the largest value, y0, such that P(Y > y0) ≥ .1?

Solution 76E

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:

Where,

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

= 0.7

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.