Question

A production line produces a variable number N of times each day.Suppose each item produced has the same probability p of not conforming to standards.If N has a Poisson distribution with mean λ,then the number of nonconforming items in one day’s production Y has a Poisson distribution with mean λp.The average number of resistors produced by a facility in one day has a Poisson distribution with mean 100. 5% of the resistors produced not meet specification.
a.Find the expected number of resistors not meeting specification
b.Find the probability that all resistor will meet specifications on a d

Answer 1

The mean number of non-conforming resistors produced in a day, Y, is given by
\[Y=\lambda p=100\times0.05=\lambda _{1}\]
The expected value of Y is given by
\[E(Y)=\lambda _{1}\]

Answer 2

The probability that all resistors produced on a day will meet specification is the same as the probability that Y = 0:
\[P(Y=0)=\frac{e ^{\lambda _{1}} \lambda _{1}^{0}}{0!}\]

Answer 3

@loai71 Do you understand? Can you calculate the value of\[\lambda _{1}\]

Answer 4

@loai71 Are you there?

Answer 5

soory i go to take the lunch. just now i come

Answer 6

Would You Please Explain To Me?

Answer 7

The expected value of Y is found from:
E(Y) = 100 * 0.05 = ?

Answer 8

And how we Find the probability that all resistors will meet the specifications on a given day?

Answer 9

@kropot72 And how we Find the probability that all resistors will meet the specifications on a given day?

Answer 10

The probability that all resistors produced on a given day will meet specification is the same as the probability that Y = 0
\[P(Y=0)=\frac{e ^{-5}5^{0}}{0!}\]

Answer 11

This reduces to
\[P(Y=0)=e ^{-5}\]

Answer 12

@loai71 Can you calculate P(Y = 0) ?

Answer 13

No

Answer 14

Well, if you cannot do such a simple calculation, why are you studying statistics at this level?

Answer 15

this subject is prerequestt in my instituet

Answer 16

You can calculate e raised to the power of -5 with any scientific calculator.
\[e ^{-5}=0.00674\]

Answer 17

@kropot72 And How we Find the probability that more than five resistors fail to meet specifications on a given day

Answer 18

@kropot72 And How we Find the probability that more than five resistors fail to meet specifications on a given day

Answer 19

\[P(Y>5)=1-[P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4)+P(Y=5)]\]
\[P(Y>5)=1-(.00674+0.03369+0.08422+0.14037+0.17547+0.17547)\]
\[P(Y>5)=1.00000-0.61595=you\ can\ calculate\]