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 5.5 minutes and a standard deviation of 1 minute. Find the
probability that a randomly selected college student will take between 4.0 and 6.5 minutes to find a
parking spot in the library lot.
Ah, the perpetual struggle of finding a parking spot! Well, according to my calculations, the mean time it takes for college students to find a parking spot is 5.5 minutes, with a standard deviation of 1 minute. Now, if we want to find the probability that a randomly selected student will take between 4.0 and 6.5 minutes to find a spot, we can use the normal distribution.
First, let's calculate the z-scores for both 4.0 and 6.5. The z-score formula is (x - mean) / standard deviation. So, for 4.0 minutes: (4.0 - 5.5) / 1 = -1.5, and for 6.5 minutes: (6.5 - 5.5) / 1 = 1.0.
Now, using a z-table or a calculator, we can find the area under the normal curve between these two z-scores. The probability that a randomly selected student will take between 4.0 and 6.5 minutes to find a spot is approximately the same as the area under the curve between -1.5 and 1.0.
So, the final answer is... drumroll, please... approximately 0.7247, or about 72.47%. That's quite a high chance! Hopefully, this means more students will be able to find a spot and spend less time circling the lot.
To find the probability that a randomly selected college student will take between 4.0 and 6.5 minutes to find a parking spot in the library lot, we need to calculate the z-scores for both values and then find the area under the normal curve between those z-scores.
Step 1: Calculate the z-score for 4.0 minutes.
The z-score formula is: z = (x - μ) / σ
where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
z1 = (4.0 - 5.5) / 1 = -1.5
Step 2: Calculate the z-score for 6.5 minutes.
z2 = (6.5 - 5.5) / 1 = 1.0
Step 3: Look up the z-scores in the standard normal distribution table or use a calculator to find the area under the curve between these z-scores.
Using a standard normal distribution table, the area to the left of z = -1.5 is 0.0668.
The area to the left of z = 1.0 is 0.8413.
Step 4: Find the area between the z-scores.
To find the probability between these two z-scores, we subtract the smaller area from the larger one:
P(4.0 < x < 6.5) = 0.8413 - 0.0668 = 0.7745
Therefore, the probability that a randomly selected college student will take between 4.0 and 6.5 minutes to find a parking spot in the library lot is approximately 0.7745 or 77.45%.
To find the probability that a randomly selected college student will take between 4.0 and 6.5 minutes to find a parking spot in the library lot, we need to calculate the area under the normal distribution curve between those two values.
The first step is to standardize the values using the z-score formula:
z = (x - μ) / σ
Where:
x = the value we want to standardize
μ = the mean of the distribution
σ = the standard deviation of the distribution
For the lower value, 4.0 minutes:
z1 = (4.0 - 5.5) / 1 = -1.5
For the upper value, 6.5 minutes:
z2 = (6.5 - 5.5) / 1 = 1.0
Next, we need to find the area under the standard normal distribution curve between these two z-scores. We can use a standard normal distribution table or a calculator to find the corresponding probabilities.
Using a standard normal distribution table:
The probability for z1 is P(z < -1.5)
From the table, P(z < -1.5) is approximately 0.0668.
The probability for z2 is P(z < 1.0)
From the table, P(z < 1.0) is approximately 0.8413.
Finally, to find the probability that a randomly selected college student will take between 4.0 and 6.5 minutes to find a parking spot in the library lot, we subtract the probability for the lower value from the probability for the upper value:
P(4.0 < x < 6.5) = P(z < 1.0) - P(z < -1.5)
P(4.0 < x < 6.5) ≈ 0.8413 - 0.0668
P(4.0 < x < 6.5) ≈ 0.7745
So, there is approximately a 77.45% probability that a randomly selected college student will take between 4.0 and 6.5 minutes to find a parking spot in the library lot.
4-5.5 = -1.5
z =-1.5/1 = -1.5 or 1.5sigma below mean
6.5 - 5.5 = 1
z = 1/1 = 1 sigma above mean
so what fraction of a normal distribution lies between z = -1.5 and z = + 1?
from table
F(-1.5) = .067 are below -1.5
F(1) = .841 are below +1
so .842-.067 are between