(adsbygoogle = window.adsbygoogle || []).push({}); Here is what I know, 3) An engineer has designed a modified welding robot. The robot will be considered good enough to manufacture if it misses only 1% of its assigned welds. And it will be judged a poor performer if it misses 5% of its welds. (In-between possibilities are not considered.) The new design will be accepted if the number of missed welds, R, is 2 or less and rejected otherwise. A test is performed using 100 welds.

a. What is the probability that a good design will be rejected

b. What is the probability that a poor design will accepted

It is good if [tex] X=x=R \leq 2 [/tex]

n = 100

p=1% is rated Good

p=5% is rated bad

Im guessing I have to use a Binomial distribution here??

[tex] p(x) = P(X=x) = \frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x} [/tex]

So,

[tex] X\sim Bin(100,0.01) [/tex]

The fact that we want 2 or fewer welds means

[tex] P(X \leq 2) = P(X=0-or-X=1-or-X=2) [/tex]

PART A)

The probability that a good deisgn will be rejected will be equal to:

[tex] 1 -p(x) = 1- P(X=x) = 1 -\frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x} [/tex]

[tex] 1 -\frac{ 100!}{x!(100-x)!} (.01)^x(1-.01)^{100-x} [/tex]

for x=0,1,2:

[tex] P(X=x \leq 2) = P(x=0)+P(x=1)+P(x=2) [/tex]

Because of the EXCESSIVE n value, its not going to work on my Ti-83 calculator and I don't feel like busting out matlab. So I am going to use the cumulative table provided by the prof online that I just found by mistake (nice of him to mention needing to use it :grumpy:)

[tex] p(x \leq 2)=0.92063 [/tex]

So the probability that a good design gets rejected is

[tex] 1 - p(x \leq 2) = 0.07937 [/tex]

which is equivalent to, 7.937% good parts accidentially rejected.

PART B)

The same thing as Part a, however the table now refers to values for

[tex] X\sim Bin(100,0.05) [/tex]

[tex] p(x) = P(X=x) = \frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x} [/tex]

[tex] \frac{ 100!}{x!(100-x)!} (.05)^x(1-.05)^{100-x} [/tex]

for x=0,1,2:

[tex] P(X=x \leq 2) = P(x=0)+P(x=1)+P(x=2) [/tex]

[tex] p(x \leq 2)=0.011826 [/tex]

so

1.1826% are accepted from the bad batch.

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# Homework Help: Stat #3

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