Question

In 2005, 0.76% of all airline flights were on-time. If we choose a simple random sample of 2000 flights, find the probability that... (to four decimal places, using normal chart, no continuity correction)

(a) at least 79% of the sample's flights were on time

(b) at most 1580 of the sample's flights were on time

(c) the sample proportion of on-time flights (p-hat) differs from the truth by more than three percent

Answer 1

a) We have to find here:

Using the z-score formula, we have:

  

Now using the standard normal table, we have:

(b) at most 1580 of the sample's flights were on time

Answer:

We have to find here

Using the z-score formula, we have:

  

Now using the standard normal table, we have:

(c) the sample proportion of on-time flights (p-hat) differs from the truth by more than three percent

Answer: We have to find here:

Using the z-score formula, we have:

  

Now using the standard normal table, we have: