suppose it takes a college student an average of 5 minutes to find a parking space on the schools 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?Answer 3.7 minutes
from your Z table, 25% of the area means 0.674 std below the mean.
That means 5 - .674(2) = 3.65 which rounds to 3.7
Make a visit to
http://davidmlane.com/hyperstat/z_table.html
to play around with Z table stuff
To find the time exceeded by approximately 75% of college students when trying to find a parking spot in the main parking lot, you can use the concept of z-scores.
First, you need to find the corresponding z-score for the desired percentile. In this case, we want to find the z-score associated with the 75th percentile. You can use a standard normal distribution table or a calculator to find the z-score.
Using the standard normal distribution table, you would find the z-score that corresponds to a cumulative probability of 0.75. In this case, the z-score is approximately 0.674.
Next, you can use the formula for z-score to find the corresponding raw score (time in minutes):
X = μ + (z * σ)
where:
X = raw score (time in minutes)
μ = mean (average time to find a parking space)
z = z-score (calculated from the percentile)
σ = standard deviation
Plugging in the values:
X = 5 + (0.674 * 2)
Calculating this equation, you get:
X = 5 + 1.348 = 6.348 minutes
Therefore, approximately 75% of college students exceed a time of 6.348 minutes when trying to find a parking spot in the main parking lot. Rounded to one decimal place, the answer is approximately 6.3 minutes, not 3.7 minutes.