Question

Suppose that 70% of all dialysis patients will survive for at least 5 years.

In a simple random sample of 100 new dialysis patients, what is the probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places?

Answer 1

We know that the sample proportion is distributed as:

$p \sim N(P,\frac{P(1-P)}{n}) \sim N(0.7,\frac{0.7(1-0.7)}{100}) \sim N(0.7,0.0021)$

Required probability =

= P(Z > 2.18) = 1 - P(Z < 2.18) 0.8 – 0.7 P(p > 0.8) = P(Z > V0.0021 = 1- 0.98537 = 0.01463