A new machine is known to reduce pollutants. However, it is expensive. The EPA estimates that one-third of plants has this new machine. What is the probability that fewer than 10% of the plants in the random sample own this machine?

0.05%

Step-by-step explanation:

The sample size is missing, which I investigated and seems to be 45, the same if it is not 45, you can change this value nothing else, the procedure is totally the same.

sample size = n = 45

population proportion = 0.3333

compute Pr (p <= 0.10) from the information, we calculate the standard deviation like this:

sd = (p * (1-p) / n) ^ (1/2)

replacing we have:

sd = (0.333 * (1-0.3333) / 45) ^ (1/2)

sd = 0.0702

Here notice that

np = 45 * (0.3333) = 15 => 10

n * (p-1) = 45 * (0.6676) = 30 => 10

which indicate that the assumption for normal approximation for the sampling distribution is met:

Pr (p <= 0.10) = Pr [(p * - 0.3333) /0.0702 <= (0.10 - 0.3333) /0.0702]

Pr (p <= 0.10) = Pr (z <= -3.32)

this value represents in the table of z (attached) 0.0005

Therefore the required probability is 0.05%

rotation point l

step-by-step explanation: