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between 15 and 22 =  p(15<x<22) =p(15-23.5)/3.6 < (22-23.5)/3.6) = p(z<-1.53)+p(z>0.42) =0.04003 ( how as 0.04003 CALCULATED ?

Mar 10th, 2015

OK, breaking this down step by step to what each part is doing:

=(15-23.5)/3.6  and (22-23.5)/3.6 are giving you the z-scores. The part in the bracket is taking the value you observe (15 or 22) and finding out how far from the mean it is (here the mean seems to be 23.5). So 15 and 22 are:

-8.5 and -1.5 units from the mean.

But to convert these into a z-score, we divide them by the standard deviation (3.6):

- 1.53 and -0.42

Remember, the z-score is the value on the horizontal axis of the standard normal distribution - the simplest normal, with mean = 0 and standard deviation 1. So we can look up these value in a table to find the associated p value.

Every z score has a p-value, which is the area of the standard normal to the left of it. So as you move right (z gets bigger) the area to its left gets greater, until eventually it reaches 1. So z=-infinity has probability 0, z=0 has probability 0.5 (since 0 is the mean, so it's exactly in the middle) and z=+infinity has probability 1.

Nowadays, you don't actually need a table - those are used because calculating p(z) without a computer is hard! But with a computer we can easily do it, using the cumulative normal distribution function. For example in Excel:

Norm.dist(z = -1.53, mean = 0, sd = 1, cdf  = 1)

= Norm.dist(-1.53,0,1,1) = 0.063008364 (check this with a z-table, you should get a similar number!)

and Norm.dist(z = -0.42, mean = 0, sd = 1, cdf  = 1)

0.337242727

So to find the distance between these values, you'd find everything to the left of your upper value and subtract everything that's to the left of the smaller one:

= 0.337242727 - 0.063008364

0.272987239 (The question above gets muddled towards the end; it mixes up the sign on the -0.42 and then tries to find the area outside that range, plus I think it makes more errors after that).

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By the way, we tend to convert things into z (the scale for the standard normal) by dividing by the standard deviation. That was so we could use the standard tables! But we don't actually have to. We don't have to divide, so long as we use the standard deviation in the formula. So:

Norm.dist(z = -1.53, mean = 0, sd = 1, cdf  = 1) is the same as

Norm.dist(z = -1.53 x 3.6, mean = 0, sd = 3.6, cdf  = 1)

And the mean doesn't have to be zero either! Instead of finding the difference between our value and the mean, we can just enter them directly in the formula:

Norm.dist(z = 15-23.5, mean = 0, sd = 1, cdf  = 1) is the same as:

Norm.dist(z = 15, mean = 23.5, sd = 1, cdf  = 1)

Hopefully this helps (it might take a few readings; also you may find it useful to draw out the normal distribution and mark the areas you are trying to find as you go along).

Mar 10th, 2015

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Mar 10th, 2015
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Mar 10th, 2015
Dec 11th, 2016
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