Question

Suppose 60% of a herd of cattle is infected with a particular disease. let p = the number of non diseased cattle in a sample of size 5. the distribution of y is

Answer 1

The distribution of y is $P(y) = ^5C_y * (60\%)^y * (1 - 60\%)^{5-y}$

How to determine the distribution?

The given parameters are:

Probability of success, p = 60%

Sample size, n = 5

The distribution of y is represented as:

$P(y) = ^nC_y * p^y * (1 - p)^{n-y}$

So, we have:

$P(y) = ^5C_y * (60\%)^y * (1 - 60\%)^{5-y}$

Hence, the distribution of y is $P(y) = ^5C_y * (60\%)^y * (1 - 60\%)^{5-y}$

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