Answer to Question #278772 in Statistics and Probability for BBJ

Question

2% of the trees in a plantation are known to have a certain disease.what is the probability that,in a sample of 250 trees.

(I) less than 1% are diseased

Answer 1

$n=250 \\ p=0.02 \\ q = 1-p=1-0.02=0.98 \\ \hat{p} → N(\mu_{\hat{p}}, \sigma_{\hat{p}}) \\ \mu_{\hat{p}} = p = 0.02 \\ \sigma_{\hat{p}} = \sqrt{\frac{p \times q}{n}} \\ = \sigma{\frac{0.02 \times 0.98}{250}} = 0.0089$

P[less than 1% are diseased] = P(p<0.01)

$= P(Z< \frac{0.01-0.02}{0.0089}) \\ = P(Z< -1.12) \\ = 0.1314$

The required probability is 0.1314