Angel_Kitty12
  • Angel_Kitty12
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
what do we have to work with? calculators or tables?
Angel_Kitty12
  • Angel_Kitty12
I don't understand how to do this. I missed a few classes. I think caculatoes
Angel_Kitty12
  • Angel_Kitty12
Wait no tables I'm sorry I don't have a graphic caculators

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amistre64
  • amistre64
well i think this is assessing your knowledge about the properties of a the normal distribution ....
amistre64
  • amistre64
what do you know about a normal distribution?
Angel_Kitty12
  • Angel_Kitty12
I know the formula for it and such but this question confused me because there is no standard deviation
amistre64
  • amistre64
do you know what a zscore is?
Angel_Kitty12
  • Angel_Kitty12
yes
Angel_Kitty12
  • Angel_Kitty12
I believe so
amistre64
  • 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
  • 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
  • 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
  • amistre64
sigma is the other one
Angel_Kitty12
  • Angel_Kitty12
Yes that's the name thank you it slipped my mind
Angel_Kitty12
  • Angel_Kitty12
|dw:1443902151868:dw|thats the formula I use
amistre64
  • 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
  • Angel_Kitty12
They shouldn't considering the variables are different.
amistre64
  • amistre64
what is 11.5 + 5.5?
Angel_Kitty12
  • 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
  • Angel_Kitty12
Is the five percent irrelevant and used to trick you?
amistre64
  • 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
  • Angel_Kitty12
oh okay i understand so you plug these numbers into the formula and solve it correct?
amistre64
  • 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
  • Angel_Kitty12
okay
amistre64
  • amistre64
|dw:1443902705692:dw| A=B
amistre64
  • 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
  • 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
  • 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
  • Angel_Kitty12
Wait I'm a bit lost
Angel_Kitty12
  • 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
  • amistre64
sigma is irrelevant ... it cancels out in the process
amistre64
  • amistre64
do you agree that we are looking to find: \[\frac{x_1-\mu}{\sigma}=-\frac{x_2-\mu}{\sigma}\]???
Angel_Kitty12
  • 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
  • Angel_Kitty12
And to solve this problem, a Z value would be needed
amistre64
  • 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
  • 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
  • Angel_Kitty12
Okay i understand now
amistre64
  • 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
  • Angel_Kitty12
Ohhhhhhh
amistre64
  • amistre64
since they are the same distance relative to the mean, they share the same tail value
amistre64
  • amistre64
directionwise that is left tail is equal to right tail
Angel_Kitty12
  • Angel_Kitty12
Okay I understand that
Angel_Kitty12
  • 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
  • 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
  • Angel_Kitty12
Oh alright i understand
amistre64
  • 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
  • Angel_Kitty12
Okay yes i got that
amistre64
  • 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
  • Angel_Kitty12
the bell curve yes
amistre64
  • amistre64
|dw:1443903974936:dw|
Angel_Kitty12
  • Angel_Kitty12
Oh
amistre64
  • 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
  • Angel_Kitty12
5% ?
amistre64
  • amistre64
yep
Angel_Kitty12
  • Angel_Kitty12
Oh thank god I thought I did that wrong. Wow its much simpler that it looks.
amistre64
  • amistre64
:) yeah
Angel_Kitty12
  • Angel_Kitty12
Thank you so much, I appreciate you helping me even if i was so confused.
amistre64
  • amistre64
stats can be confusing, i think its the speed at which it is presented ...
amistre64
  • amistre64
good luck
Angel_Kitty12
  • Angel_Kitty12
Thanks a lot :)

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