Question 1206708: Let x be a hypergeometric random variable with N = 16, n = 3, and M = 4. (Suppose you are given
p(0) = 0.39, p(1) = 0.47, p(2) = 0.13, p(3) = 0.01.
Furthermore, you are given that
𝜇 = E(x) = 0.75 and 𝜎2 = 0.48750.)
What proportion of the population of measurements fall into the interval (𝜇 ± 2𝜎)? (Round your answer to two decimal places.)
What proportion of the population of measurements fall into the interval (𝜇 ± 3𝜎)? (Round your answer to two decimal places.)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! ## Understanding the Problem and Given Information
We are given a hypergeometric random variable X with parameters N = 16, n = 3, and M = 4. We also have the probability mass function (PMF) of X:
* P(X = 0) = 0.39
* P(X = 1) = 0.47
* P(X = 2) = 0.13
* P(X = 3) = 0.01
Additionally, we are given the mean (μ) and variance (σ²) of X:
* μ = E(X) = 0.75
* σ² = Var(X) = 0.4875
## Calculating the Intervals
1. **Interval (μ ± 2σ):**
* Lower limit: μ - 2σ = 0.75 - 2 * sqrt(0.4875) ≈ -0.54
* Upper limit: μ + 2σ = 0.75 + 2 * sqrt(0.4875) ≈ 2.04
2. **Interval (μ ± 3σ):**
* Lower limit: μ - 3σ = 0.75 - 3 * sqrt(0.4875) ≈ -1.3
* Upper limit: μ + 3σ = 0.75 + 3 * sqrt(0.4875) ≈ 2.8
## Finding the Proportions
Since X is a discrete random variable, we need to find the probabilities of X falling within these intervals and sum them up.
**For the interval (μ ± 2σ):**
* The only possible values of X within this interval are 0, 1, and 2.
* So, the proportion of measurements within this interval is:
P(X = 0) + P(X = 1) + P(X = 2) = 0.39 + 0.47 + 0.13 = 0.99
**For the interval (μ ± 3σ):**
* All possible values of X (0, 1, 2, and 3) fall within this interval.
* So, the proportion of measurements within this interval is:
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1
**Therefore:**
* The proportion of measurements within (μ ± 2σ) is **0.99**.
* The proportion of measurements within (μ ± 3σ) is **1.00**.
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