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Check Your Progress · Q14

Q.A computer disk manufacturer tests disk quality on random basis before approving it. The approval is based on the number of errors in a test area on each disk and follows Poisson distribution with λ\lambda = 0.2. What is the percentage of test areas having two or a smaller number of errors?

Yanam CbseNCERTSubjective· 3mImportance★★★★★
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For λ=0.2\lambda=0.2, P(X≤2)=P(0)+P(1)+P(2)=0.99885P(X\le2)=P(0)+P(1)+P(2)=0.99885, so about 99.89%99.89\% of test areas have two or fewer errors.

P(X=k)=e−λλkk!P(X=k)=\dfrac{e^{-\lambda}\lambda^{k}}{k!}, P(X≤2)=P(0)+P(1)+P(2)\quad P(X\le2)=P(0)+P(1)+P(2)

where λ=0.2\lambda=0.2 errors per test area.

Steps

  1. λ=0.2\lambda=0.2; e−0.2=0.818731.e^{-0.2}=0.818731.

  2. P(X=0)=e−0.2=0.818731.P(X=0)=e^{-0.2}=0.818731.

  3. P(X=1)=e−0.2×0.2=0.163746.P(X=1)=e^{-0.2}\times0.2=0.163746.

  4. P(X=2)=e−0.2×0.222!=0.818731×0.042=0.016375.P(X=2)=e^{-0.2}\times\dfrac{0.2^{2}}{2!}=0.818731\times\dfrac{0.04}{2}=0.016375.

  5. Sum: P(X≤2)=0.818731+0.163746+0.016375=0.998852.P(X\le2)=0.818731+0.163746+0.016375=0.998852. …

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