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Sagot :
Answer: 0.0208
Step-by-step explanation:
Given: The length of time to find a parking spot on the UW campus is normally distributed with a mean of 4.75 minutes and a standard deviation of 1.35 minutes.
Let x = time taken to find parking spot
The probability that a randomly selected driver takes less than 2 minutes to find a parking spot on the UW campus will be :
[tex]P(x<2)=P(\dfrac{x-\mu}{\sigma}<\dfrac{2-4.75}{1.35})\\\\=P(Z<-2.037)\ \ \ [z=\dfrac{x-\mu}{\sigma}]\\\\=1-P(Z<2.037)\\\\=1- 0.9792=0.0208\ \ \ \text{[By p-value table]}[/tex]
Hence, the probability than a randomly selected driver takes less than 2 minutes to find a parking spot on the UW campus.= 0.0208
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