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The speed limit on Interstate 75 around Findlay, Ohio, is 65 mph. On a clear day with no construction, the mean speed of automobiles was measured at 63 mph with a standard deviation of 8 mph. If the speeds are normally distributed, what percentage of the automobiles is exceeding the limit?
The first step i did was (63-65/8)=-0.25
From the table the value i get for -0.25 is 0.4013
If i subtract 0.4013 from 1 I get the percentage of automobiles exceeding 63, whereas the question asks for the percentage of vehicles exceeding 65, so how do i find that!
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Hi Aisha;
Your .4013 is the correct answer for the fraction of vehicles exceeding 65. The percentage is 40.13%
If i subtract 0.4013 from 1 I get the percentage of automobiles exceeding 63
That is not correct. You get the fraction of automobiles travelling from 63 to 65.
Use the positive z scores.
http://www.regentsprep.org/Regents/math … ZChart.htm
You will see that from - infinity to + .25 is .5987 That is the fraction of cars going 65 or less. It is the area of the whole left side.
The area or fraction of cars going faster than 65 is 1 - .5987 = .4013. That is the probability.
This is the area you are interested in:
A nice little applet to check your work is here:
http://www.measuringusability.com/pcalc … .35&side=1
Here is another one that is good and gives you a different perspective.
http://www.mathsisfun.com/data/standard … table.html
The point is this is nothing but an area under curve calculation.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Ok now im confused, doesnt the z value i got(-0.25)correspond to the 63 ?
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Nope, you computed how many z scores 65 is from 63. With 8 being the SD. Then 2 is 1 / 4 of that.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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