Question

Suppose 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.

What is the point in the distribution in which 95.44% of the college students exceed when trying to find a parking spot in the library parking lot? What about 99.7%? (Explain/show how you obtain your answer.) What statistical rule about normal curves does this problem illustrate?

Answer #1

We are given the distribution here as:

a) From standard normal tables, we have:

P(Z < -1.689) = 0.0456

Therefore, P(Z > -1.689) = 1 - 0.0456 = 0.9544

Therefore the point here is computed as:

= Mean - 1.689*Std Dev

= 3.5 - 1.689 = 1.811

**Therefore 1.811 is the point in the distribution in
which 95.44% of the college students exceed when trying to find a
parking spot in the library parking lot.**

b) From standard normal tables, we have:

P(Z < -2.748) = 0.003

Therefore, P(Z > -2.748) = 1 - 0.003 = 0.997

Therefore the point here is computed as:

= mean - 2.748*Std Dev

= 3.5 - 2.748 = 0.752

**Therefore 0.752 minute is the required point on the
distribution in which 99.7% of the college students exceed when
trying to find a parking spot in the library parking
lot.**

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