The expected number of defective sample is 0.25
The probability distribution
The given parameters are:
- Population, N = 30
- Sample, n = 2
- Selected, x = 4
Start by calculating the defective proportion using:
[tex]p = \frac{x}{N}[/tex]
So, we have:
p = 4/30
p = 0.13
The probability distribution is calculated as:
[tex]P(x) = ^nC_x * p^x *(1 - p)^{n - x}[/tex]
So, we have:
[tex]P(0) = ^2C_0 * 0.13^0 *(1 - 0.13)^{2 - 0} = \frac{19}{25}[/tex]
[tex]P(1) = ^2C_1 * 0.13^1 *(1 - 0.13)^{2 - 1} =\frac{23}{100}[/tex]
[tex]P(2) = ^2C_2 * 0.13^2 *(1 - 0.13)^{2 - 2} = \frac{1}{100}[/tex]
So, the probability distribution is:
x 0 1 2
P(x) 19/25 23/100 1/100
The expected number of defective sample
This is calculated using:
[tex]E(x) = \sum x * P(x)[/tex]
So, we have:
E(x) = 0 * 19/25 + 1 * 23/100 + 2 * 1/100
Evaluate
E(x) = 0.25
Hence, the expected number of defective sample is 0.25
Read more about binomial distribution at:
https://brainly.com/question/15246027
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