Question

If 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, find the point in the distribution in which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot.
A. 2.8 minutes
B. 3.2 minutes
C. 3.4 minutes
D. 4.2 minutes

Answer 1

Answer:

A. 2.8 minutes

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

Let X the random variable that represent the length of time it takes a college student to find a parking spot in the library parking of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.5,1)$  

Where $\mu=3.5$ and $\sigma=1$

And we need to find a value a, such that we satisfy this condition:

$P(X>a)=0.758$   (a)

$P(X<a)=0.242$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.242 of the area on the left and 0.758 of the area on the right it's z=-0.700. On this case P(Z<-0.700)=0.242 and P(z>-0.700)=0.758

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.242$  

$P(z<\frac{a-\mu}{\sigma})=0.242$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.700<\frac{a-3.5}{1}$

And if we solve for a we got

$a=3.5 -0.700*1=2.8$

So the value of height that separates the bottom 24.2% of data from the top 75.8% is 2.8 minutes.