We know that the length of time it takes a 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. When trying to find a parking spot in the library parking lot, 75.8% of the college students exceed what time?Question 2Select one:a.2.8 minutesb.3.2 minutesc.0.7 minutesd.4.2 minutes
Question
We know that the length of time it takes a 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. When trying to find a parking spot in the library parking lot, 75.8% of the college students exceed what time?Question 2Select one:a.2.8 minutesb.3.2 minutesc.0.7 minutesd.4.2 minutes
Solution
To answer this question, we need to understand the properties of a normal distribution. In a normal distribution, the mean (average) is the value around which the data is centered, and the standard deviation is a measure of the spread of the data.
The question is asking for the time that 75.8% of students exceed when trying to find a parking spot. This is equivalent to finding the 24.2 percentile point (since 100% - 75.8% = 24.2%) on the normal distribution.
We can use a standard normal distribution table (also known as a Z-table) to find this. The Z-table tells us what percentage of the data falls below a certain point in a standard normal distribution.
Looking up 24.2% in the Z-table, we find a Z-score of approximately -0.7.
The Z-score is a measure of how many standard deviations an element is from the mean. We can calculate the value associated with a Z-score using the formula:
Value = Mean + Z-score * Standard Deviation
Substituting the given values:
Value = 3.5 minutes + (-0.7) * 1 minute = 2.8 minutes
So, 75.8% of the college students take more than 2.8 minutes to find a parking spot. Therefore, the answer is a. 2.8 minutes.
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