Question

Please solve the following question.

Answer 1

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:

$p = \frac{x}{N}$

So, we have:

p = 4/30

p = 0.13

The probability distribution is calculated as:

$P(x) = ^nC_x * p^x *(1 - p)^{n - x}$

So, we have:

$P(0) = ^2C_0 * 0.13^0 *(1 - 0.13)^{2 - 0} = \frac{19}{25}$

$P(1) = ^2C_1 * 0.13^1 *(1 - 0.13)^{2 - 1} =\frac{23}{100}$

$P(2) = ^2C_2 * 0.13^2 *(1 - 0.13)^{2 - 2} = \frac{1}{100}$

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:

$E(x) = \sum x * P(x)$

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:

brainly.com/question/15246027

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