The length of time it takes college students to find a parking spot in the library parking lot follows a normal distribution with a mean of 4.0 minutes and a standard deviation of 1 minute. Find the probability that a randomly selected college student will take between 2.5 and 5.0 minutes to find a parking spot in the library lot.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion between those two Z scores.
To find the probability that a randomly selected college student will take between 2.5 and 5.0 minutes to find a parking spot in the library lot, we need to calculate the area under the normal distribution curve between these two values.
First, we need to standardize the values using the z-score formula:
z = (x - μ) / σ
where x is the given value, μ is the mean of the distribution, and σ is the standard deviation.
For 2.5 minutes:
z1 = (2.5 - 4.0) / 1 = -1.5
For 5.0 minutes:
z2 = (5.0 - 4.0) / 1 = 1.0
Next, we use a standard normal distribution table or calculator to find the probabilities associated with each z-score.
The probability of z being less than -1.5 is the same as the probability of finding a value below -1.5 on the standard normal distribution table. From the table or calculator, we find this probability to be approximately 0.0668.
The probability of z being less than 1.0 is the same as the probability of finding a value below 1.0 on the standard normal distribution table. From the table or calculator, we find this probability to be approximately 0.8413.
To find the probability between 2.5 and 5.0 minutes, we subtract the first probability from the second probability:
P(2.5 ≤ X ≤ 5.0) = P(X ≤ 5.0) - P(X ≤ 2.5)
≈ 0.8413 - 0.0668
≈ 0.7745
Therefore, the probability that a randomly selected college student will take between 2.5 and 5.0 minutes to find a parking spot in the library lot is approximately 0.7745.