Suppose that 80%, percent of trees in a forest of 300 trees are infected with a virus, and we take an SRS of 5 trees to test for the virus. What is the probability that exactly 2 of the 5 trees have the virus? You may round your answer to the nearest hundredth. P(X=2)=P(X=2)=P, left parenthesis, X, equals, 2, right parenthesis, equals

0.05

Step-by-step explanation:

Given that:

Number of samples, n = 5

Proportion, p = 80% = 0.8

P(x = 2)

Using binomial distribution formula ;

P(X = x) = nCx * p^x * (1 - p)^(n-x)

P(x = 2) = 5C2 * 0.8^2 * (1 - 0.8)^(5-2)

P(x = 2) = 5C2 * 0.8^2 * 0.2^3

P(x = 2) = 10 * 0.64 * 0.008

P(x = 2) = 0.0512

= 0.05 ( nearest hundredth)

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)