Calculating Probability of X~N Distribution

[xmac1x]

i)Let X be a random variable with X~N (Mean, sigma^2),
calculate Probability that (mean - 1.40Sigma < X < Mean + 0.6Sigma)


3. Hmm I've done questions like these before but the are a lot more simple and I've spent about an hour qriting pointless rubbish that's getting me nowhere. Anyone have any clue? It'd be much appreciated thanks

 

[cronxeh]

Well if mean=0 and sigma=1, then you are evaluating between x=-1.40 and x=0.60. The probability of that is 64.5% if you evaluate the integral.

 

[Tedjn]

Allow me to elaborate a little. The normal distribution is given by a certain function, which you can find in the right box under pdf at http://en.wikipedia.org/wiki/Normal_distribution.

To find the probability between $\mu - 1.40\sigma$ and $\mu + 0.60\sigma$, you would evaluate the integral of this function between those bounds. However, what this table probably wants you to do is standardize the bounds to get the appropriate z-values, and use a normal distribution table. This is essentially an equivalent, simplified process.

If any part of this process is confusing for you, please let us know so that we can help clear it up.

 

[xmac1x]

Allow me to elaborate a little. The normal distribution is given by a certain function, which you can find in the right box under pdf at http://en.wikipedia.org/wiki/Normal_distribution.

To find the probability between $\mu - 1.40\sigma$ and $\mu + 0.60\sigma$, you would evaluate the integral of this function between those bounds. However, what this table probably wants you to do is standardize the bounds to get the appropriate z-values, and use a normal distribution table. This is essentially an equivalent, simplified process.

If any part of this process is confusing for you, please let us know so that we can help clear it up.

Im not sure what you mean by stnadardise the bounds but I am familiar in z-values and how to read the tables. It's figuring out the bits in between that I am having trouble with. Your help is greatly appreciated!

 

[Tedjn]

By standardize the bounds, all I mean is to change the values $\mu - 1.40\sigma$ and $\mu + 0.60\sigma$, which denote integration boundaries for the curve $\mathcal{N}(\mu, \sigma)$, to z-values, which denote integration boundaries for the curve $\mathcal{N}(0, 1)$, for which you can look up in a table. How do you change whatever values you have to z-values?

 

[cronxeh]

N(0,1) is standardized distribution with mean=0 and variance=1.

Z = (X - mean)/Sigma

Z = (X - 0)/1 = X/1

Z = X so you can rewrite that probability inequality as follows:

Probability(mean-1.40*sigma < Z < mean+0.60*sigma)=

Probability(-1.40 < Z < 0.60) =

Now refer to http://business.statistics.sweb.cz/normal01.jpg z-table

These probability are into the left of the z-value, meaning that all the points from -infinity to z are counted in that probability (-inf,z). When you have a range for z value you are counting between z=-1.40 and z=0.60 - a narrow band of probability density.

When z=0.60, probability=0.7257 that also includes all points from -infinity to z=0.60. So now you want to exclude all the points that lay between -infinity and z=-1.40 because you are interested in the range of -1.40 < Z < 0.60
When z=-1.40 its the same as when (1-P(z=1.40)) = 1-0.9192 = 0.0808

So now subtract P(z<0.60)-P(z<-1.40)
Probability = 0.7257 - 0.0808 = 0.6449

Or you could have simply integrated Using WolframAlpha

 

[xmac1x]

Thanks a million for the help, that's the way i was trying to do it all along, just wasnt sure whether i had the correct approach. Much appreciated