the length of times it takes college students to find a parking spot im the library parking lot follows a normal distribution with a mean of 5.5 minutes and a standard deviation of 1 minute. find the cut off time which 75.8% of the college students exceed when trying to find a parking spot in the library parking 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/probability (.758) and its Z score. Insert data into equation above to calculate score.
To find the cut off time at which 75.8% of college students exceed when trying to find a parking spot in the library parking lot, we need to find the z-score corresponding to this probability.
Step 1: Convert the given probability to a z-score using the standard normal distribution table (also known as the z-table or standard normal table).
The z-score can be calculated using the formula:
z = (X - μ) / σ
Where:
X is the given value (cut off time),
μ is the mean,
σ is the standard deviation.
Step 2: Use the z-table to find the z-score that corresponds to a cumulative probability of 75.8% or 0.758.
Step 3: Substitute the values back into the z-score formula and solve for X.
Now, let's calculate the cut off time:
Step 1: Calculate the z-score.
z = (X - μ) / σ
0.758 = (X - 5.5) / 1
Step 2: Use the z-table to find the z-score corresponding to a cumulative probability of 0.758. From the table, we find that the z-score is approximately 0.75.
Step 3: Substitute the values back into the z-score formula and solve for X.
0.75 = (X - 5.5) / 1
0.75 * 1 = X - 5.5
0.75 = X - 5.5
0.75 + 5.5 = X
X = 6.25
Therefore, the cut off time at which 75.8% of college students exceed when trying to find a parking spot in the library parking lot is approximately 6.25 minutes.
To find the cutoff time which 75.8% of college students exceed when trying to find a parking spot in the library parking lot, we can use the properties of the normal distribution.
Step 1: Convert the given percentage to its corresponding z-score. The z-score represents the number of standard deviations a given value is from the mean. We can use a standard normal distribution table or a statistical calculator to find the z-score.
In this case, we need to find the z-score corresponding to the upper 75.8% of the distribution. Since the normal distribution is symmetric, we can find the z-score corresponding to the lower 24.2% (100% - 75.8%) and then take the negative of that z-score.
Step 2: Once we have the z-score, we can use the formula:
x = μ + z * σ
where:
x = cutoff time
μ = mean
z = z-score
σ = standard deviation
Plugging in the given values, we have:
μ = 5.5 minutes
σ = 1 minute
z = -z-score (from Step 1)
Step 3: Calculate the cutoff time (x) using the formula above.
Let's go through the calculations:
1. Find the z-score corresponding to the lower 24.2%:
z-score = -0.674 (approximately)
Note: You can use a standard normal distribution table or a statistical calculator to obtain this z-score.
2. Calculate the cutoff time (x):
x = 5.5 + (-0.674) * 1
x = 5.5 - 0.674
x ≈ 4.826
Therefore, the cutoff time which 75.8% of college students exceed when trying to find a parking spot in the library parking lot is approximately 4.826 minutes.