Question

It is known that the probability of an item produced by a certain machine will be defective is 0.05. If the produced items are sent to the market in packets of 20, find De number of packets containing at least, exactly and at mow 2 defective items in a consignment of 1000 packets using i) binomial distribution and G) Poisson approximation to binomial distribution

Answer 1

Given :  probability of an item produced by a certain machine will be defective is 0.05.   the produced items are sent to the market in packets of 20,

To Find :   number of packets containing at least, exactly and at most 2 defective items in a consignment of 1000 packets

using (i) binomial distribution and (ii) Poisson approximation to binomial distribution

Solution:

p defective  = 0.05

q non defective = 1 - 0.05 = 0.95

packets of 20

=> n = 20

Exactly 2 defective

x = 2

P (x) = ⁿCₓpˣqⁿ⁻ˣ

=> p(2) = ²⁰C₂(0.05)²(0.95)¹⁸   =   0.1886768

188.67 = 189 defectives   in 1000

at most 2 defective items

= P(0) + P(1) + P(2)

=  ²⁰C₀(0.05)⁰(0.95)²⁰ +  ²⁰C₁(0.05)¹(0.95)¹⁹  +  ²⁰C₂(0.05)²(0.95)¹⁸

= 0.3584859 + 0.3773536 +  0.1886768

= 0.9245163

= 924.5  = 925  defectives  in 1000

containing at least 2   defective items

= 1  - P(0) - P(1)

=  1 - 0.3584859 - 0.3773536

= 0.2641605

264 defectives  in 1000

Poisson approximation to binomial distribution

P(x) = $e^{-\lambda} \dfrac{{\lambda}^x}{x!}$

Here  λ = 20(0.05) =  1

Exactly 2 defective

p(2)  =  e⁻¹(1)²/2!   = 0.1839397

184 defectives in 1000

at most 2 defective items

= P(0) + P(1) + P(2)

=   e⁻¹(1)⁰/0! +  e⁻¹(1)¹/1! + e⁻¹(1)²/2!      

=    0.36787944 + 0.36787944  +  0.1839397

= 0.91969858

= 920 defectives in 1000

containing at least  2   defective items

= 1  - P(0) - P(1)

= 1 - 0.36787944 - 0.36787944

= 0.26424112

264  defectives in 1000

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