Question

We suppose that the number of errors Z within the first year for a fabrication is Poisson distributed with waiting value 2.5. We also know that the number of errors Z wont be overrided or a replacement is needed for the fabrication.

a. Find the probabilility that within that year the errors will occur within the probability ?? = ?? (1 ≤ ?? <5).

b. The manufacturer wants to replace a maximum. 10% of all of the products. What is the minimum errors Z when this occurs?

Answer 1

given data:

and this follows poission distribution

=0.2052

   = 0.2565

=0.2137

=0.1336

= 0.0668

a) PP = pp(1<=xx<5)

= P(1)+P(2)+P(3)+P(4)

not 5 because quastion asked less than 5

PP = pp(1<=xx<5)

= P(1)+P(2)+P(3)+P(4)

=0.2052 + 0.2565 +  0.2137 + 0.1336

=0.809

Probabilility that within that year the errors will occur within the probability ?? = ?? (1 ≤ ?? <5) is 0.809, ie,. 80.9%

B) if manufacturer replace max 10% of the total product. means the probability for failure is 0.10.

P(x)=0.1

put hit and trial method.

P(0) = 0.0821= 8.21%

here we can see

if we have to get max 1 error. the probability is (0.0821+0.2052) 0.2873.

and it increase with more errors.

so if there s only 10% error allowed. so min error should be Zero.