Answer:
75% of college students exceed 6.63 minutes when trying to find a parking spot.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 10 minutes
Standard Deviation, σ = 5 minutes
We are given that the distribution of time for parking is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(X < x) = 0.25
We have to find the value of x such that the probability is 0.25.
P(X < x)
[tex]P( X < x) = P( z < \displaystyle\frac{x - 10}{5})=0.25[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z<-0.674) = 0.25[/tex]
[tex]\displaystyle\frac{x - 10}{5} = -0.674\\x = 6.63[/tex]
Hence, 75% of college students exceed 6.63 minutes when trying to find a parking spot.
