If we know that the length of time it takes a university student to find a parking space in the university car park follows a normal distribution with a mean of 3.5 minutes and a standard deviation of 1 minute,find the probability that a randomly selected university student will find a parking space in the car park in less than 3 minutes.
A) 0.3085
B) 0.3551
C) 0.2674
D) 0.1915

Question

Quizplus · Accepted Answer

To find the probability of a student finding a parking space in less than 3 minutes, we use the Z-score formula: Z = (X - μ) / σ, where X is the value (3 minutes), μ is the mean (3.5 minutes), and σ is the standard deviation (1 minute). Plugging in the values, we get Z = (3 - 3.5) / 1 = -0.5. Looking up the Z-score of -0.5 in a standard normal distribution table gives us a probability of approximately 0.3085.

If We Know That the Length of Time It Takes