The manager at a local movie theater has collected data for a long period of time and has concluded that the revenue from concession sales during the first show each evening is normally distributed with a mean equal to $336.25 and a standard deviation equal to $80. Based on this information, what are the chances that the revenue on the first show will be between $300 and $500?
A) About 0.3062
B) Approximately 0.6534
C) 0.1736
D) Approximately 0.4798

Question

Quizplus · Accepted Answer

To find the probability that the revenue will be between $300 and $500, we use the Z-score formula: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. For $300: Z = (300 - 336.25) / 80 = -0.4531. For $500: Z = (500 - 336.25) / 80 = 2.046875. Using the Z-table, we find the probabilities corresponding to these Z-scores and subtract the smaller from the larger to find the probability of being between these two values. This calculation gives us approximately 0.6534, indicating that there's about a 65.34% chance the revenue will be between $300 and $500.

The Manager at a Local Movie Theater Has Collected Data