Finding probability with a sample size and no SD or mean

I have a problem dealing with finding a desired percentage and I'm only familiar with finding these kinds of probabilities with a standard deviation and sample mean present. I believe I need to utilize the Z-scores to find this probability but I'm not sure how to get started. Any guidance to get me in the right direction would be great.

Here is the problem:

In 2017,6% of the flights were delayed. Suppose that a random sample of 80 of the flights for the first month of 2018 is selected. What is the probability that, in this sample, the percentage of delayed flights is less than 4%? Assume that the airline’s on time performance rate for the first month of 2018 is the same as it was in 2017.
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Less is more. Stay pure. Stay poor.
I agree with @obh - think of this as more of a flipping a coin problem. If the coin is fair, you should get heads 50% of the time, given a number of flips, what is the probability of getting say 25% heads - which may or may not be rare given the number of flips. But as flips approaches infinity, the observed percentage of heads should be close to the truth.
So if I used binomial distribution,would these be the proper inputs? N= 80, π=.06

I'm not sure if this is the proper method because I wouldn't know how to find a probability less than a certain value.
A very helpful question to always ask is; 'Are the observations continuous or discrete?' In this case the observations are pass/fail, on-time/delayed flights.

With only discrete distributions to pick from, the one that makes most sense is the binomial.

Because they want to know what is the probability that less than 4% of 80 flights are delayed, you are really asking what is the probability that 3 or few flights are delayed. 4% of 8 is 3.2, so round down to the nearest integer.

So sum up to probabilities that 0, 1, 2, or 3 flights are delayed.

P(x) = Combination(80,x) * .08^(x) * (1-.08)^(80-x)

P(0) + P(1) + P(2) + P(3) = .285788