Question

our boss is a biologist who needs wood samples from long-leaf pine trees with a fungal disease which is only visible under a microscope, and she sends you on an assignment to collect the samples. She wants at least 50 different diseased samples. She tells you that approximately 28% of long-leaf pine trees currently have the fungal disease. If you sample 160 long-leaf pine trees at random, what is the probability you’ll have at least 50 diseased samples to return to your boss? (Use the normal approximation to calculate this probability and chose the closest answer to the question.)

Answer 1

Answer:

Step-by-step explanation:

In this scenario, the probability of success, p is 28% = 28/100 = 0.28

Number of samples, n = 160

Probability of failure, q = 1 - p = 1 - 0.28 = 0.72

Mean,µ = np = 0.28 × 160 = 44.8

Standard deviation, σ = √npq = √160 × 0.28 × 0.72 = 5.68

Let x be the random variable representing the number of wood samples from long-leaf pine trees with a fungal disease. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

the probability that you’ll have at least 50 diseased samples to return to your boss is expressed as

P(x ≥ 50) = 1 - P(x < 50)

For P(x < 50)

z = (50 - 44.8)/5.68 = 0.91

Looking at the normal distribution table, the probability corresponding to the z score is 0.819

Therefore,

P(x ≥ 50) = 1 - 0.819 = 0.181