Question

Suppose that 65% 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

The probability that the proportion surviving for at least five years

will exceed 80% is 0.00084

Step-by-step explanation:

The given is:

1. Suppose that 65% of all dialysis patients will survive for at least

   5 years

2. A random sample has 100 new dialysis patients

We need to find the probability that the proportion surviving for at

least five years will exceed 80%

At first find σ from the rule below

∵ σ = $\sqrt{\frac{P(1-P)}{n}}$

∵ P = 65% = 0.65

∵ n = 100

∴ σ = $\sqrt{\frac{0.65(1-0.65)}{100}}$

∴ σ = 0.04770

Now find z-score from the rule:

∵ z = (x - μ)/σ

∵ x = 80% = 0.80

∵ μ = P = 0.65

∵ σ = 0.04770

- Substitute these values in the rule

∴ z = $\frac{0.80-0.65}{0.04770}$ = 3.14

Use the normal distribution table for z-score to find the

corresponding area of z = 3.14

∵ The corresponding area is 0.99916

∵ For P(x > 80%) the area to the right is needed

∵ P(x > 80%) = 1 - 0.99916

∴ P(x > 80%) = 0.00084

The probability that the proportion surviving for at least five years

will exceed 80% is 0.00084

Learn more:

You can learn more about the random sample in brainly.com/question/5510873

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