# 1.5 standard deviations within the mean

Asked Feb 11, 2010, 07:47 PM — 3 Answers
All my professor gave me was what proportion of measurements from a normal distribution falls within 1.5 standard deviations of the mean. No mean value was given and I do not understand how to solve the question. He goes on to ask the same thing only for is more than 1.5 above the mean, is more than 1.5 below the mean, and is between 1.5 and 1.8 above the mean. I am very frustrated

 galactus Posts: 2,272, Reputation: 1436 Ultra Member #2 Feb 12, 2010, 12:36 PM
This is a question about Chebychev's Theorem.

The portion of the data set lying within k standard deviations (k>1) of the mean is at least

$1-\frac{1}{k^{2}}$

Just plug in k=1.5.
 elscarta Posts: 116, Reputation: 103 Junior Member #3 Feb 14, 2010, 06:24 AM
Actually this question is about the Normal Distribution function, not about Chebychev's Theorem! You need to have a calculator capable of calculating Normal Distribution values or a Table which lists Normal Distribution values.

The reason that you were not given the mean (and by the way nor were you given the standard deviation) is because any Normal Distribution can be standardized to distribution with a mean of zero and a standard deviation of one.

Given that Z is the Standardized Normal Distribution function then the questions become:

P(-1.5 <Z<1.5), P(Z>1.5), P(Z<-1.5) and P(1.5<Z<1.8)

http://en.wikipedia.org/wiki/Standard_normal_table gives a table of values for P(Z<z)

Since the Normal Distribution is symmetrical the following property is true.

Property 1: P(Z<-z) = P(Z>z)

Also since the Normal Distribution is a probability density function the following also hold true.
Property 2: P(Z<∞) = 1
Property 3: P(Z>z) = 1 – P(Z<z)
Property 4: P(a<Z<b) = P(Z<b) – P(Z<a)

So P(-1.5 <Z<1.5)
= P(Z<1.5) – P(Z<-1.5) (Property 4)
= P(Z<1.5) – P(Z>1.5) (Property 1)
= P(Z<1.5) – (1 – P(Z<1.5)) (Property 3)
= P(Z<1.5) – 1 + P(Z<1.5)
= 2 x P(Z<1.5) – 1
= 2 x 0.9332 -1 (value from table)

= 0.8664

You should be able to do the remainder of the questions.
 galactus Posts: 2,272, Reputation: 1436 Ultra Member #4 Feb 14, 2010, 06:59 AM
Yeah, I reckon you're correct about that. I thought Chebyshev's theorem would have been close enough for this though. Doing the z-score is certainly better. The way you showed, all one has to do is look them up in the z-table. Much easier than fooling with Chebyshev's theorem and more accurate. after all, it says 'at least'.

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