To find the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot, we need to calculate the z-scores for both 2 minutes and 4.5 minutes and then find the area under the normal distribution curve between those two z-scores.
Using the z-score formula: z = (x - μ) / σ
For 2 minutes:
z1 = (2 - 3.5) / 1 = -1.5
For 4.5 minutes:
z2 = (4.5 - 3.5) / 1 = 1
Now, we need to find the area between these two z-scores using a standard normal distribution table or calculator. The probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot is the difference between the two areas:
P(2 < x < 4.5) = P(z < 1) - P(z < -1.5)
From a standard normal distribution table, P(z < 1) = 0.8413 and P(z < -1.5) = 0.0668
Thus,
P(2 < x < 4.5) = 0.8413 - 0.0668
P(2 < x < 4.5) = 0.7745
Therefore, the probability that a randomly selected college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot is 0.7745.
If we know that the length of time it takes a college student to find a parking spot in the library parking lot 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 college student will take between 2 and 4.5 minutes to find a parking spot in the library parking lot.
0.0919
0.2255
0.4938
0.7745
1 answer