Algebra 2. The Normal Distribution.
PARKING: Over several years, Bertram conducted a study of how far into parking spaces people tend to park by measuring the distance from the end of a parking space to the front fender of a car parked in the space. He discovered that the distribution of the data closely approximated a normal distribution with mean 8.5 inches. He found that about 5% of cars parked more than 11.5 inches away from the end of the parking space. What percentage of cars would you expect parked less than 5.5 inches away from the end of the parking space?

- Angel_Kitty12

- katieb

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- amistre64

what do we have to work with? calculators or tables?

- Angel_Kitty12

I don't understand how to do this. I missed a few classes. I think caculatoes

- Angel_Kitty12

Wait no tables I'm sorry I don't have a graphic caculators

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## More answers

- amistre64

well i think this is assessing your knowledge about the properties of a the normal distribution ....

- amistre64

what do you know about a normal distribution?

- Angel_Kitty12

I know the formula for it and such but this question confused me because there is no standard deviation

- amistre64

do you know what a zscore is?

- Angel_Kitty12

yes

- Angel_Kitty12

I believe so

- amistre64

what is your zscore formula ... we can work with that; and also we need to know that a normal distribution is symettric about the mean ...
the percentage of data higher than a zscore of say: n
is going to be equal to the data less than a zscore of: -n

- amistre64

so if the data points referenced are the same distance from the mean, then we can compare the information to determine a solution

- Angel_Kitty12

My formula is z= X-n/ o
The n and o are Greek letters though. The n I believe is pronounced as mew but I forgot the other words name

- amistre64

sigma is the other one

- Angel_Kitty12

Yes that's the name thank you it slipped my mind

- Angel_Kitty12

|dw:1443902151868:dw|thats the formula I use

- amistre64

if
\[\frac{x_1-\mu}{\sigma}=-\frac{x_2-\mu}{\sigma}\]
then
\[x_1-\mu=\mu-x_2\]
\[x_1+x_2=2\mu\]
does x1+x2 = 2u?

- Angel_Kitty12

They shouldn't considering the variables are different.

- amistre64

what is 11.5 + 5.5?

- Angel_Kitty12

the mean is 8.5 and there is a few other numbers thrown in there like 11.5 and 5 percent but then they ask for the last number used what would you expect less than 5.5 so the numbers that would be used is 8.5 and 5.5 obviously.

- Angel_Kitty12

Is the five percent irrelevant and used to trick you?

- amistre64

8.5 is the mean of the distribution
11.5 is one of the data points
5.5 is the other
5% is relevant only if 11.5 + 5.5 = 2(8.5)

- Angel_Kitty12

oh okay i understand so you plug these numbers into the formula and solve it correct?

- amistre64

well, the formula is used to calculate a zscore for each data point.
a normal distribution has a symmetric property; the amount of data below and above +- z is the same.
so if x1 and x2 are the same distance from the mean ... then data below and above them will be equal with respect to the conditions needed

- Angel_Kitty12

okay

- amistre64

|dw:1443902705692:dw|
A=B

- amistre64

so, by using the formula
\[\frac{11.5-8.5}{\sigma}=-\frac{5.5-8.5}{\sigma}\]
we can see if they share a common 'z' value

- amistre64

some algebra just simplifies this to 11.5+5.5 = 2(8.5)
of if 8.5 is the average of the 2

- amistre64

we know B=5%, we are asked to find A
if the condition holds that 11.5+5.5 = 2(8.5), then A=B

- Angel_Kitty12

Wait I'm a bit lost

- Angel_Kitty12

Don't we need to find the z value to figure out the sigma though and how did you get a 2 from this equation?

- amistre64

sigma is irrelevant ... it cancels out in the process

- amistre64

do you agree that we are looking to find:
\[\frac{x_1-\mu}{\sigma}=-\frac{x_2-\mu}{\sigma}\]???

- Angel_Kitty12

|dw:1443903219200:dw| This is what I got from simplifying the problem you had shown but I'm confused how you got 11.5+5.5=2(8.5)

- Angel_Kitty12

And to solve this problem, a Z value would be needed

- amistre64

work some algebra
\[\sigma \frac{x_1-\mu}{\sigma}=-\sigma\frac{x_2-\mu}{\sigma}\]
\[x_1-\mu=-(x_2-\mu)\]
\[x_1-\mu=-x_2+\mu\]
\[x_1-\mu+\mu=-x_2+\mu+\mu\]
etc ...

- amistre64

we dont need a proper z value, we just need to know of 11.5 is the same distance from 8.5, as 5.5 is from 8.5
z values just tell us how many standard deviations (how far from the mean) a data point is
if 2 data points are the same distance, relative to the mean, they have the same z value but differ by their signs

- Angel_Kitty12

Okay i understand now

- amistre64

since 5.5 is 3 away from the 8.5
and 11.5 is 3 away from 8.5
they both have the same z values, for whatever the standard deviation is ...

- Angel_Kitty12

Ohhhhhhh

- amistre64

since they are the same distance relative to the mean, they share the same tail value

- amistre64

directionwise that is
left tail is equal to right tail

- Angel_Kitty12

Okay I understand that

- Angel_Kitty12

so then 11.5+5.5=2(8.5) because both 11.5 and 3.5 have the same z value from 8.5 thus multiplying it twice? Okay I'm a bit lost here....sorry. I understood everything else you said though.

- amistre64

its just working the math
think of it this way.
if the mean is the midpoint (the average) of 2 extremes, then the extremes are the same distance from the mean

- Angel_Kitty12

Oh alright i understand

- amistre64

we can normalize this (hence the name normal distribution) by making the mean equal to 0
5.5 < 8.5 < 11.5
well, 8.5-8.5 = 0, so lets subtract 8.5 from all parts
5.5-8.5 < 8.5-8.5 < 11.5-8.5
all we have done is reposition the data to be centered at 0 instead of 8.5
-3 < 0 < 3
now it should be easier to see that -3 and 3 are the same distance from 0

- Angel_Kitty12

Okay yes i got that

- amistre64

the rest of the solution is just knowing that the normal distribution is symmetric about the mean; it looks the same on the left as it does the right ...
|dw:1443903925176:dw|
one side is a mirror of the other

- Angel_Kitty12

the bell curve yes

- amistre64

|dw:1443903974936:dw|

- Angel_Kitty12

Oh

- amistre64

normalising the data gives us the questions
He discovered that the distribution of the data closely approximated a normal distribution with mean 0 inches.
He found that about 5% of cars parked more than 3 inches away from the end of the parking space.
What percentage of cars would you expect parked less than -3 inches away from the end of the parking space?

- Angel_Kitty12

5% ?

- amistre64

yep

- Angel_Kitty12

Oh thank god I thought I did that wrong. Wow its much simpler that it looks.

- amistre64

:) yeah

- Angel_Kitty12

Thank you so much, I appreciate you helping me even if i was so confused.

- amistre64

stats can be confusing, i think its the speed at which it is presented ...

- amistre64

good luck

- Angel_Kitty12

Thanks a lot :)

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