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.

Paragraph 1: Text, lorem, borem and so much more text, Text, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textvText, lorem, borem and so much more textvText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more text

Paragraph 2: Text, lorem, borem and so much more text, Text, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textvText, lorem, borem and so much more textvText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more textText, lorem, borem and so much more text

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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