Question

It is reported that 25% of the flights arriving at the Washington Dulles airport in January 2008 were late. Assume the population proportion is .25. What's the probability that the sample proportion will be within +-.03 of the population proportion if a sample size of 625 is selected?

Answer 1

`P(0.22<p<0.28)=P(-1.732<Z < 1.732) =P(Z<1.732)-P(Z<-1.732) = 0.958-0.042=0.916`

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

For this case the standard error is given by:

`SE =\sqrt{(p(1-p))/(n)} =\sqrt{(0.25(1-0.25))/(625)}=0.0173`

Solution to the problem

For this case we want to find the probability that the sample proportion will be within +-.03 of the population proportion like this:

`P(0.25-0.03 < p< 0.25+0.03) =P (0.22<p<0.28)`

And for this case we can use the following z score:

`z = (\hat p - p)/(SE_(p))`

And using this formula we got:

`P(0.22<p<0.28) = P((0.22 -0.25)/(0.0173) <Z< (0.28-0.25)/(0.0173)) = P(-1.732<Z <1.732)`

And we can find this probability like this:

`P(-1.732<Z < 1.732) =P(Z<1.732)-P(Z<-1.732) = 0.958-0.042=0.916`