business statistics

Question

On a busy day, the average roller coaster wait time at a large amusement park is 27 minutes. Suppose the standard deviation of the wait time is 11.9 minutes and the distribution of wait times is approximately normally distributed

If it is a busy day, approximately what percentage of people at the amusement park will have a wait time that is at least 30 minutes? Round to the nearest whole percent

Answer 1

Hello Senthamil,

Let X represent the randomn variable "roller coaster wait time". We need to find the probability that X is at least 30 minutes, that is P(X ≥ 30). Given that X ∼ normal with mean 27 and standard deviation 11.9, we should standardize X ≥ 30. In other words, we find the z-score for 30 as follows:

(X - μ) ÷ σ ≥ (30 - 27) ÷ 11.9

Which simplifies to Z ≥ 0.25, so that P(X ≥ 30) = P(Z ≥ 0.25).

However, since the standard normal distribution table only yields areas (percentages) to the left of a z-score, we use the complement rule:

P(Z ≥ 0.25) = 1 - P(Z ≤ 30) = 1 - 0.5987 ≈ 0.4013 ≈ 40%

Note: that if you are using statistical software, there is no need to standardize nor use the complement rule.