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Q.The mortality rate for a certain disease is 0.007. Using Poisson distribution, calculate the probability for 2 deaths in a group of 400 people

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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With mortality rate 0.0070.007 over 400400 people, λ=np=2.8\lambda=np=2.8; then P(X=2)=e−2.8(2.8)22!≈0.2384.P(X=2)=\dfrac{e^{-2.8}(2.8)^2}{2!}\approx0.2384.

λ=np\lambda=np, P(X=k)=e−λλkk!\quad P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}

where n=400n=400, p=0.007p=0.007 (mortality rate).

Steps

  1. Poisson parameter: λ=np=400×0.007=2.8.\lambda=np=400\times0.007=2.8.

  2. Required: P(X=2)=e−2.8(2.8)22!.P(X=2)=\dfrac{e^{-2.8}(2.8)^{2}}{2!}.

  3. Compute powers: (2.8)2=7.84(2.8)^{2}=7.84 and 2!=2.2!=2.

  4. Use e−2.8=0.060810.e^{-2.8}=0.060810.

  5. Substitute: P(X=2)=0.060810×7.842=0.060810×3.92.P(X=2)=\dfrac{0.060810\times7.84}{2}=0.060810\times3.92.

  6. Compute: P(X=2)=0.238376≈0.2384.P(X=2)=0.238376\approx0.2384. …

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