This would lead to a hypergeometric distribution: X follows Hypergeometric(K, N, n), where: 

  • K = number of defect units in one daily production
  • N = the population size = the number of units produced per day
  • n= the sample size = 25 draws, and in which
  • p = the proportion of success/defect units in a daily production:p=K/N

 

However, we assume that the daily production is more than 10 times the sample size: \(N >10n\Leftrightarrow N>250\) which means that we can approximate to the binomial distribution: \(X\sim Bin(n;p)\) with a 10% probability of defect units, expressed by p= 0.1, so we get: the expression, the expected value and the varianse:

   

\({X\sim \operatorname {Bin} (25;0.1)}\)
\({\operatorname {E} [X]=np}=2.5\) 

\({\operatorname {Var} (X)=np(1-p)} \Leftrightarrow {\operatorname {Var} (X)=250(1-0.1).} = 1.5\)  

 

The criteria for returning a one-day production to inspection is if there are 4 or more defect units in the daily sample, which can be expressed as: 

    

\(\displaystyle P(X\ge 4)=1-P(X\leq 3)\) 

\(\displaystyle\Rightarrow P(X\leq k)=\sum _{i=0}^{n}{n \choose i}p^{i}(1-p)^{n-i}\) 

\(\displaystyle\Leftrightarrow P(X\leq 3)=\sum _{i=0}^{25}{25 \choose i}0.1^{i}(1-0.1)^{25-i}\) 

 

Header 3, blue

Header 3, creyon

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yxTtherm

Paragraph 1: The Swedish company, yxTherm, produces thermostats and wish to revise their established quality assurance parameters>

  • K = number of defect units in one daily production
  • N = the population size = the number of units produced per day
  • n= the sample size = 25 draws, and in which
  • p = the proportion of success/defect units in a daily production:p=K/N a change of machinery.

 

here is a box

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However, we assume that the daily production is more than 10 times the sample size: \(N >10n\Leftrightarrow N>250\) which means that we can approximate to the binomial distribution: \(X\sim Bin(n;p)\) with a 10% probability of defect units, expressed by p= 0.1, so we get: the expression, the expected value and the varianse:

* \({X\sim \operatorname {Bin} (25;0.1)}\)

* \({\operatorname {E} [X]=np}=2.5\)

* \({\operatorname {Var} (X)=np(1-p)} \Leftrightarrow {\operatorname {Var} (X)=250(1-0.1).} = 1.5\)

The criteria for returning a one-day production to inspection is if there are 4 or more defect units in the daily sample, which can be expressed as:

\(\displaystyle P(X\ge 4)=1-P(X\leq 3)\)
\(\displaystyle\Rightarrow P(X\leq k)=\sum _{i=0}^{n}{n \choose i}p^{i}(1-p)^{n-i}\)
\(\displaystyle\Leftrightarrow P(X\leq 3)=\sum _{i=0}^{25}{25 \choose i}0.1^{i}(1-0.1)^{25-i}\)

The step-by-step sum of the probablilities of having 3 or less:

\(\displaystyle{\Pr(0{\text{ defect units}})=f(0)=\Pr(X=0)={25 \choose 0}0.1^{0}(1-0.1)^{25-0}=0.071789799}\)
\(\displaystyle{\Pr(1{\text{ defect unit}})=f(1)=\Pr(X=1)={25 \choose 1}0.1^{1}(1-0.1)^{25-1}=0.199416108}\)
\(\displaystyle{\Pr(2{\text{ defect units}})=f(2)=\Pr(X=2)={25 \choose 2}0.1^{2}(1-0.1)^{25-2}=0.265888144}\)
\(\displaystyle{\Pr(3{\text{ defect units}})=f(3)=\Pr(X=3)={25 \choose 3}0.1^{3}(1-0.1)^{25-3}=0.226497308}\)
\(\displaystyle\Leftrightarrow 1 – (0.071789799+0.199416108+0.265888144+0.226497308) = \underline{0.2364}\)

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