For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 ⇐ p ⇐ 0.95), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation σ=np(1 − p). Explain this approximation making use the Central Limit Theorem.

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For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 ⇐ p ⇐ 0.95), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation   σ=np(1 − p). Explain this approximation making use the Central Limit Theorem.

Asma Vang · Accepted Answer

Step 1 Alternatively prove that: Theorem: If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the variablez =x − μσ − N(0, 1) where μ=np σ2=np(1 − p) Proof: It can be prove using Moment generating function for binomial distribution. It&#039;s given as, Mx(θ)=(q + peθ)n where q=1 − p. Step 2 By the linear transformation properties of the moment generating function. Mz(θ)=e − μ θσMx(θσ)=e − μ θσ(q + pe θσ)n Taking the natural log of both sides, and then expanding the power series of eθ/σ Then, ln⁡ Mz(θ)= − μ θσ + n ln⁡(q + pe θσ)= − μ θσ + n ln⁡(q + p ∑k=1∞1k!(θσ)k) Since p + q=1. = − μ θσ + n ln⁡(1 + p ∑k=1∞1k!(θσ)k) If n is made sufficiently large σ=npq can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1. Let t=p∑m=1∞1m!(θσ)m Thus for sufficiently large n, |t| &amp;lt; 1. The ln term in the previous expression is ln⁡(1 + t) where |t| &amp;lt; 1, and so we may expand this term as follows: ln⁡(1 + t)=∑k=1∞(−1)k − 1tkk Step 3 This means that ln⁡ Mz(θ)= − μ θσ + n ln⁡(1 + t)= − μ θσ + n ∑k=1∞(−1)k1

For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 Leftarrow p Leftarrow 0.95), the binomial distribution can be

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