Probability distrubtion... anyone wanna help me out??

Question

Assume X is normally distributed with a mean of 10 and a standard deviation of 2. determine the value for x that solve each of the following:
a) P(X>x)=0.5
b) P(X>x)=0.95
c) P(x d) P(-x E) P(-x
Thanks!

Answer 1

The nice thing about a normal distribution is that everything you need to know is contained in just two numbers: the mean and the standard deviation. That's partly because the normal distribution is symmetric, with the mean = median, etc.

In the normal distribution, the fraction of the population inside N standard deviations is given by:

erf(n/sqrt(2))

where erf is the errof function. Here is a table of some values:

1 => 0.682689492137
2 => 0.954499736104
3 => 0.997300203937
4 => 0.999936657516
5 => 0.999999426697
6 => 0.999999998027

This means that 68.3% of the population is within 1 standard deviation of the mean, whatever it is, leaving the remaining 21.7% evenly split - half less than the median but more than one standard deviation away and half greater than the median.

This is all you need to know.

So, for example, question b.

1) let n be the number of standard deviations corresponding to x. We know that the population more than n standard deviations away and below the median is 5%. That means that the number more than n standard deviations away and above the median must also be 5%. (symmetry of the normal distribution)

2) So we need to find an n such that erf(n/sqrt(2)) = 0.90. Use a table or a statistical calculator. Here is one source:
http://en.wikipedia.org/wiki/Normal_distribution

3) Now that we have n in standard deviations:

x = mean - (n times standard deviation)