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?

Question

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?

amistre64 · Answer

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

angel_kitty12 · Answer

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

angel_kitty12 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

what do you know about a normal distribution?

angel_kitty12 · Answer

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

amistre64 · Answer

do you know what a zscore is?

angel_kitty12 · Answer

yes

angel_kitty12 · Answer

I believe so

amistre64 · Answer

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 · Answer

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 · Answer

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 · Answer

sigma is the other one

angel_kitty12 · Answer

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

angel_kitty12 · Answer

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

amistre64 · Answer

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 · Answer

They shouldn't considering the variables are different.

amistre64 · Answer

what is 11.5 + 5.5?

angel_kitty12 · Answer

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 · Answer

Is the five percent irrelevant and used to trick you?

amistre64 · Answer

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 · Answer

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

amistre64 · Answer

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 · Answer

okay

amistre64 · Answer

|dw:1443902705692:dw|

A=B