You are interested in finding how many hours a person is willing to wait for a plane. It is found that the time people are willing to wait has a $μ = 5.2$ and a $σ = 1.1$. What is the probability a person is willing to wait more than seven hours?

Using the standard score formula, $z = (x - μ)/σ$, the z-score equals to two decimal places $1.64$, which corresponds to a probability of $0.9495$ on my z-score tables. However, on the answer key provided, it states the probability is $0.0505$ or what would correspond to a z-score of $-1.64$.

Is this a trick of the wording involved in the question or are my z-tables (or standard score formula) wrong?


1 Answer 1


You've computed correctly, but interpreted the number incorrectly. What you've found is the probability that someone is willing to wait at most $7$ hours; the probability you want is

$$1 - 0.9495 = 0.0505$$

as the answer key states.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .