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 #1
avatar+130561 
+5

The data set I have is 14.2, 15.6, 16.3, 17.1, 17.9, 18.6, 20.0

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Let's see how much of statistics I remember!!

First, let's find the mean (average) of this data set. We just sum the numbers and divide by the number of them that we have, (7)........

This gives:   ( 14.2+ 15.6 +16.3 + 17.1 +17.9 + 18.6 + 20.0 ) / 7   = 17.1

The, we subtract each of the values from the mean and square the result....note that we can disregard 17.1 itself !! (Subtracting it from the mean = 0)

So we have

(17.1 - 14.2)2 = (2.9)2  = 8.41

(17.1 - 15.6)2 = (1.5)2  = 2.25

(17.1 - 16.3)2 = (.8)2    =   .64

(17.1 - 17.9)2 = (-.8)2   =   .64

(17.1 - 18.6)2 = (-1.1)2  = 1.21

(17.1 - 20.0)2 = (2.9)2  =  8.41

Now, sum the values on the right hand side

We have (8.41 + 2.25 + .64 + .64 + 1.21 + 8.41) = 21.56

Now, since this is a "sample," we need to divide this by (the number of values - 1) = (7-1) = (6)

This gives us (21.56/6) ≈ 3.59

Now, take the square root of this   SQRT(21.56/6) = 1.895

And, assuming I didn't make any math errors, that's the standard deviation ... (S.D.)

Now....you might have the same question that I had when I first saw S.D......So what???

Well....what this tells us is how far (on average) each score is from the mean.  A "high" S.D. indicates a "dispersed" data set and a "low" S.D. indicates a data set that is bunched closely around the mean. This data set seems to be fairly closely bunched around the mean.......we don't have any "large" outlier values that would tend to "skew" the S.D. very much. Of course, the terms "high," low" and "closely bunched" are all relative to the situation.

Note....because this process is lengthy, it's usually better to use some sort of application to find S.D.

Here's an app to help you with that:

http://www.wolframalpha.com/input/?i=standard+deviation+98.17%2C+112.3%2C+102.6%2C+94.3%2C+108.1

This will find S.D. for samples or populations

Apr 28, 2014
 #6
avatar+118725 
+5

I put it through Wolfram|Alpha - here is the address

https://www.wolframalpha.com/input/?i=sin%282x%29%3D3%2F2

I don't think there are any complex solutions.  

Someone else will probably comment on this.

---------------------------------------------------

$$$$Now anonymousperson,

If you go to the insert dropdown list - choose special character (second from the bottom) ≤ this symbol is in the second line. 

or if you open the LaTex box and type in 5\le 6  the display will be as shown below.

$$5\le 6$$

the \ indicates that le is a functions and le stands for less than or equal to

I really want people to learn LaTex because you can write almost any math thing in there and once you get used to it it is quick quick.  

Apr 28, 2014
 #1
avatar
+5
Apr 28, 2014
Apr 27, 2014
 #1
avatar+33666 
+5

Picture the triangle made by a line from the centre of the earth through to the top of the lighthouse (which I'll call c); a line from the top of the lighthouse to the ship (which I'll call b); and a line from the ship back to the centre of the earth (which I'll call a).  I would prefer to show a diagram here, but I can't as we can't yet upload files!

Since the angle of depression is 30°, the angle, A, between lines b and c is 90° - 30° or 60°.  We can write the cosine rule for this triangle as a2 = b2 + c2 -2bccos(A).

Now the average radius of the Earth is 6371km (according to Wikipedia), so line a is 6371km long.  Line c is the radius of the Earth plus 325ft (=99.06m) so c is 6,371,000 + 99.06 = 6,371,099.06m.  The only unknown in the cosine rule above is b, and we can solve for b to find two mathematical solutions.  These are 6371km and 198.12m.

Clearly it is the smaller value we want here!  So b = 198.12m.

Now we need angle B, the angle between lines a and c. We get this from the sine rule:  sin(B)/b = sin(A)/a. So:

sin(B)=sin(60°)*198.12/6371000 = 2.693*10-5 

Angle B = sin-1(2.693*10-5) or B = 2.693*10-5 radians (when an angle is very small its value in radians is approximately equal to its sine).  The distance from the ship to the lighthouse along the curve of the Earth is just B*a where B is in radians.  So $${\mathtt{distance}} = {\mathtt{2.693}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{5}}\right)}{\mathtt{\,\times\,}}{\mathtt{6\,371\,000}} = {\mathtt{distance}} = {\mathtt{171.571\: \!03}}$$ metres

This is 562.9ft to the nearest tenth of a foot.

 Alternatively

 Since the distances are relatively small we could assume the surface of the Earth to be flat near the lighthouse, in which case the distance of the ship from the lighthouse is given by 325*tan(60°)ft or

$${\mathtt{distance}} = {\mathtt{325}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{60}}^\circ\right)} = {\mathtt{distance}} = {\mathtt{562.916\: \!512\: \!459\: \!925}}$$  feet

or 562.9ft.  

Apr 27, 2014

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