In the production of fabrics by the meter you will watch and control the quality

Question

In the production of fabrics by the meter you will watch and control the quality by looking at the number of errors per ten meters of fabric. After a trial production it has been discovered following probability distribution for the number of errors per ten meters of fabric. Results are based on 100 rolls of ten meters each.

a) Make a probability histogram and find the expected number of errors and the standard deviation in the number of errors, E (X) and SD (X).

b) During a control, there have been three errors in a roll, so we know that there are at least three errors. What is the probability that there are four errors in this roll?

c) During a period it has been taken out a random roll 12 times. In 4 of these it has been observed that there are 3 or 4 errors. What is the probability that this might happen?

produksjon av stoff som metervare vil man s ed se pa antall feil per ti meter av stoffet. en funnet folgende sannsynlighetsfordeling t offet. Resultatene er basert pa 100 ruller pa t T 0 1 2 3 4 p(T) 0.11 0,37 0,16 0,05 0,01 a) Lag et sannsynlighetshistogram og finn dar davviket i antall feil, E(X) og SD(X) b) Underveis i en kontroll har man sett tre det er minst 3 feil. Hva er sannsvnlig heter

Explanation / Answer

Solution

x

0

1

2

3

4

Total

p(x)

0.41

0.37

0.16

0.05

0.01

1.0

x.p(X)

0

0.37

0.32

0.15

0.04

0.88

x2.p(X)

0

0.37

0.64

0.45

0.16

1.62

Part (a)

Let X = number of errors.

Then, expected number of errors = E(X) = sum{x.p(x)} = 0.88 ANSWER 1[read form last column of above table]

Variance of number of errors = V(X) = E(X2) – {E(X)}2

=sum{x2.p(x)} – 0.882= 1.62[read form last column of above table] – 0.7744

= 0.8456 and

the standard deviation of the number of errors = sq.rt of V(X) = sq.rt(0.8456)

So, SD (X) = 0.9198 ANSWER 2

Part (b)

Number of errors follow Poisson distribution. Here, the parameter , which represents the average number of errors, is 0.88.

Given that 3 errors have been found, we want to find P(4 errors) = P(4 errors/X = 3) =

P(4 errors)/P(3 errors) [using the theorem, P(A/B) = P(A and B)/P(B)]

= {(e-0.88.0.884)/4!}/ {(e-0.88.0.883)/3!} = 0.88/4 = 0.22 ANSWER

[For Poisson variable X with parameter , P(X = x) = (e-. x)/x!]

Part (c)

P(3 or 4 errors) = P(3 errors) + P(4 errors) = 0.0471 + 0.0104 = 0.0575 [these probabilities can be read off from Standard Poisson Probability Tables or using Excel function with = 0.88 and x = 3, x = 4]

Now, P(3 or 4 errors happening 4 times) = {P(3 or 4 errors)}4 = 0.05754 = 0.000011 ANSWER

x

0

1

2

3

4

Total

p(x)

0.41

0.37

0.16

0.05

0.01

1.0

x.p(X)

0

0.37

0.32

0.15

0.04

0.88

x2.p(X)

0

0.37

0.64

0.45

0.16

1.62

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