If we know that the length of time it takes a university student to find a parking space in the university car park follows a normal distribution with a mean of 4.5 minutes and a standard deviation of 1.25 minutes, then the point in the distribution in which 67% of the university students exceed when trying to find a parking space in the car park is单项选择题
A
4.74 minutes
B
3.95 minutes
C
5.05 minutes
D
3.81 minutes

Question

If we know that the length of time it takes a university student to find a parking space in the university car park follows a normal distribution with a mean of 4.5 minutes and a standard deviation of 1.25 minutes, then the point in the distribution in which 67% of the university students exceed when trying to find a parking space in the car park is单项选择题

A

4.74 minutes

B

3.95 minutes

C

5.05 minutes

D

3.81 minutes

BlackTom AI · Accepted Answer

We are given a normal distribution for the time to find a parking space with mean mu = 4.5 minutes and standard deviation sigma = 1.25 minutes.

Step 1: Interpret the phrasing. If 67% of students exceed a certain time, that means the remaining 33% are at or below that time. In other words, the target time corresponds to the 33rd percentile (0.33) of the distribution, because P(X ≤ x) = 0.33 when 67% exceed x.

Step 2: Find the z-score corresponding to the 33rd percentile. For a standard normal distribution, the z-value at the 0.33 cumulative probability is negative, approximately z ≈ -0.44 (since P(Z ≤ -0.44) ≈ 0.33).

Step 3: Transform back to the original scale. Use x = mu + z * sigma. Substituting, x = 4.5 + (-0.44) * 1.25.

Step 4: Compute the value. (-0.44) * 1.25 ≈ -0.55, so x ≈ 4.5 - 0.55 = 3.95 minutes.

Now evaluate the answer options:
- 4.74 minutes: This would correspond to a z around +0.96 (since 4.74 ≈ 4.5 + 0.96*1.25). That would place the percentile well above 50%, not at 33rd percentile, so it would imply far fewer than 67% exceed it, which is inconsistent with the given condition.
- 3.95 minutes: This matches the 33rd percentile calculation above, yielding the correct interpretation that 67% exceed this time.
- 5.05 minutes: This is about 0.44 standard deviations above the mean (4.5 + 0.44*1.25 ≈ 5.05), which would correspond to a large lower tail proportion, not the 33rd percentile, so it does not fit the requirement.
- 3.81 minutes: This is slightly below the 3.95 value, corresponding to a z about -0.28, which would give a higher lower-tail probability than 0.33, meaning less than 67% exceed it, again not matching the stated condition.

In summary, the correct interpretation requires locating the 33rd percentile of the normal distribution and converting back to minutes, which yields approximately 3.95 minutes. The other options correspond to different z-scores that do not satisfy the condition that 67% exceed the time.

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