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# Answer to Question #209146 in Chemical Engineering for Appu

Question #209146

In a component manufacturing industry, there is a small probability for any

component to be defective. The components are supplied in packets of . Use

Poisson approximation to calculate appropriately the number of packets containing

(i) at least one defective (ii) at most one defective in a consignment of 100 packets

1
2021-06-22T05:08:26-0400

The number of defective blades in a packet has binomial distribution 𝐵(𝑛, 𝑝) with

parameters 𝑛 = 100 and 𝑝 = 0.0002

The binomial distribution can be approximated using Poisson with parameter 𝑚 =

𝑛𝑝 = 0.02

Let 𝑋 equals the number of defective blades in a packet.

"\ud835\udc5d_0 = \ud835\udc43(\ud835\udc4b = 0) = \ud835\udc52^{-0.02} = 0.9802"

Using the formula "\ud835\udc5d_{\ud835\udc65+1 }= \ud835\udc5d\ud835\udc65 \u22c5\n\n\\frac{\ud835\udc5a\n}{\n\ud835\udc65 + 1}"

we have:

"\ud835\udc5d_1 = \ud835\udc5d_0 \u22c5 \\frac{\n0.02}{\n\n1} = 0.019604"

"\ud835\udc5d_2 = \ud835\udc5d_1 \u22c5\\frac{\n\n0.02}{\n\n2} = 0.009802"

"\ud835\udc5d_3 = \ud835\udc5d_2 \u22c5 \\frac{\n\n0.02}{\n\n3}\u2248 0"

Thus expected frequencies are:

(i) at least one defective

"\ud835\udc5b_1+\ud835\udc5b_2+\ud835\udc5b_3 =100 \u22c5 \ud835\udc5d_1+100 \u22c5 \ud835\udc5d_2+100 \u22c5 \ud835\udc5d_3 \u2248 1.96+0.9802+0 \u2248 2.9402"

ii) at most one defective

"\ud835\udc5b_0+\ud835\udc5b_1=100 \u22c5 \ud835\udc5d_0+100 \u22c5 \ud835\udc5d_1 \u2248 0.019604+0.009802 \u2248 0.029406"

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