Final answer:
The waiting time at the sandwich shop that exceeds the 80% probability threshold is approximately 5.19 minutes, calculated using the given mean, standard deviation, and the associated z-score for 80% probability in a normal distribution.
Step-by-step explanation:
The student is asking about the probability that the waiting time to fill an order at a sandwich shop is over a certain length of time given that the wait times are normally distributed. We are given the mean (4.1 minutes) and standard deviation (1.3 minutes) of the distribution. To find the time such that there is an 80% chance that the actual waiting time exceeds it, we need to look up the z-score that corresponds to the top 20% of the normal distribution since 1 - 0.80 = 0.20. This z-score is approximately 0.84. Next, we use the formula for the z-score in a normal distribution:
Z = (X - μ) / σ
Where Z is the z-score, X is the value from the normal distribution we are looking for, μ is the mean and σ is the standard deviation. We rearrange this to solve for X:
X = Z*σ + μ
Plugging in the values we get:
X = 0.84 * 1.3 + 4.1
X = 5.192
Thus, the probability is 80% that the waiting time is greater than approximately 5.19 minutes.