# Find the probability that at most 5 defective fuses will be found in a box of 200, if experience

shows that 20% of such fuses are defective.

1
by issh 20.02.2015

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shows that 20% of such fuses are defective.

1
by issh 20.02.2015

Log in to add a comment

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For such problems we have to assume that the samples of the fuses follow the standard normal probability distribution function or a poisson's probability function.

The number of defective fuses found in a sample of size N follows the Poisson's distribution function or standard normal distribution function.

p = 20% = 1/5

N = 200

Let X be the number of defective fuses in a box of N fuses.

Expected number of defective fuses in a box of 200 fuses is Np = λ = mean

E(X) = λ = 200 * 1/5 = 40

Probability density as per Poisson's distribution function. The probability that the number of defective fuses is k is equal to:

Also, the probability that the number of defective fuses is less than of equal to k is equal to:

Substitute λ = 40 and k = 5 in this expression :

P (X <= 5 ) <= 2.06 * 10⁻¹¹ very small probability.

It is expected as, in the box we expect around 40 defective fuses. So there being only 5 or less defective fuses is very very small.

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You could also calculcate

P (X <= 5) = P(X=0) + P(X = 1)+P(X=2) + P(X=3) + P(X=4) +P(X=5)

==========================

If you assume the normal distribution function then:

μ = mean = 40

P(

The number of defective fuses found in a sample of size N follows the Poisson's distribution function or standard normal distribution function.

p = 20% = 1/5

N = 200

Let X be the number of defective fuses in a box of N fuses.

Expected number of defective fuses in a box of 200 fuses is Np = λ = mean

E(X) = λ = 200 * 1/5 = 40

Probability density as per Poisson's distribution function. The probability that the number of defective fuses is k is equal to:

Also, the probability that the number of defective fuses is less than of equal to k is equal to:

Substitute λ = 40 and k = 5 in this expression :

P (X <= 5 ) <= 2.06 * 10⁻¹¹ very small probability.

It is expected as, in the box we expect around 40 defective fuses. So there being only 5 or less defective fuses is very very small.

===========================================

You could also calculcate

P (X <= 5) = P(X=0) + P(X = 1)+P(X=2) + P(X=3) + P(X=4) +P(X=5)

==========================

If you assume the normal distribution function then:

μ = mean = 40

P(