Answers

2015-02-20T13:14:47+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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:

P(X = k) = \frac{e^{-\lambda\ \lambda^k}}{k!}\\

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

P(X \le k) \le \frac{e^{-\lambda\ (e\ \lambda)^k}}{k^k}\\

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)
=e^{-40}*[\frac{40^0}{0!} + \frac{40^1}{1!} + \frac{40^2}{2!} + \frac{40^3}{3!} + \frac{40^4}{4!} + \frac{40^5}{5!} ]\\\\=4.125 * 10^{-12}

==========================
If you assume the normal distribution function then:

   μ = mean = 40

P(

3 5 3
could you please answer this http://brainly.in/question/81863
Comment has been deleted