Assume that it takes a college student an average of 5 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 2 minutes. What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?
A) 3.65 minutes
B) 5.75 minutes
C) 6.36 minutes
D) 9.21 minutes

Question

Quizplus · Accepted Answer

To find the time exceeded by approximately 75% of the students, we look for the 25th percentile in a normal distribution (since 100% - 75% = 25%). The Z-score corresponding to the 25th percentile is approximately -0.674 (from Z-tables or normal distribution calculators). Using the formula for converting Z-scores to raw scores: X = μ + Zσ, where μ is the mean, σ is the standard deviation, and Z is the Z-score, we get X = 5 + (-0.674)(2) = 5 - 1.348 = 3.652. This value is closest to 3.65 minutes, indicating that approximately 75% of the students take longer than this time to find a parking spot.

Assume That It Takes a College Student an Average of 5