Question

Let μ=E(X), σ=standard deviation of X. Find the probability P(μ-σ ≤ X ≤ μ+σ) if X has a Geometric distribution with p = 0.61. Round your answer to 4 decimal places.

Answer 1

P(μ-σ ≤ X ≤ μ+σ) =
`0.61 + 0.1275 = 0.7375`

Explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

`f(x) = (1-p)^(x)p`

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

`\mu = (1-p)/(p)`

The standard deviation of the geometric distribution is given by the following formula:

`\sigma = \sqrt{(1-p)/(p^(2))`

In this problem, we have that:

`p = 0.61`.

So

`\mu = (1-p)/(p) = 0.6393`

`\sigma = \sqrt{(1-p)/(p^(2)) = 1.0234`

P(μ-σ ≤ X ≤ μ+σ) =
`P(-0.3841 \leq X \leq 1.6627) = P(0 \leq X \leq 1.6627) = f(0) + f(1.6627)`

`f(x) = (1-p)^(x)p`

`f(0) = (0.39)^(0)(0.61) = 0.61`

`f(1.6627) = (0.39)^(1.6627)(0.61) = 0.1275`

P(μ-σ ≤ X ≤ μ+σ) =
`0.61 + 0.1275 = 0.7375`