Let 𝑋1,𝑋2,…,𝑋50∼i.i.d. Poisson(0.04)X 1 ,X 2 ,…,X 50 ∼i.i.d. Poisson(0.04) and let 𝑌=∑𝑖=150𝑋𝑖Y= i=1∑50 X i . Use Central Limit theorem to find 𝑃(𝑌>6)P(Y>6). Enter the answer correct to 2 decimal places.
Question
Let 𝑋1,𝑋2,…,𝑋50∼i.i.d. Poisson(0.04)X 1 ,X 2 ,…,X 50 ∼i.i.d. Poisson(0.04) and let 𝑌=∑𝑖=150𝑋𝑖Y= i=1∑50 X i . Use Central Limit theorem to find 𝑃(𝑌>6)P(Y>6). Enter the answer correct to 2 decimal places.
Solution
To solve this problem, we first need to understand the properties of the Poisson distribution and the Central Limit Theorem (CLT).
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The mean (μ) and variance (σ^2) of a Poisson distribution are both equal to the rate parameter, λ. In this case, λ = 0.04.
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According to the CLT, the sum of a large number of independent and identically distributed (i.i.d.) random variables will be approximately normally distributed. Here, we have 50 i.i.d. Poisson random variables.
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The mean of the sum of random variables is the sum of their means, and the variance of the sum of independent random variables is the sum of their variances.
So, for Y = ∑Xi from i=1 to 50, the mean E(Y) = 50 * E(Xi) = 50 * 0.04 = 2 and the variance Var(Y) = 50 * Var(Xi) = 50 * 0.04 = 2.
We want to find P(Y > 6). To use the CLT, we standardize Y to create a standard normal random variable Z = (Y - μ) / σ = (Y - 2) / sqrt(2).
So, P(Y > 6) = P(Z > (6 - 2) / sqrt(2)) = P(Z > 2.83) (approximately).
We can find this probability using the standard normal distribution table or a calculator with a normal distribution function. The result is 1 - P(Z < 2.83) = 1 - 0.9977 = 0.0023.
So, P(Y > 6) is approximately 0.0023, or 0.23% when expressed as a percentage. To two decimal places, this is 0.00 or 0.23%.
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