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Assume that it takes a college student an average of 10 minutes to find a parking spot in the main parking lot. Assume also that this time is normally distributed with a standard deviation of 5 minutes. What time is exceeded by approximately 75% of the college students when trying to find a parking spot in the main parking lot?

Respuesta :

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.