[...]Normally distributed Standard deviation Probability? | Reading One[...]…

For any normal random variable X with mean ? and standard deviation ? , X ~ Normal( ? , ? ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( ? , ?² ). Most software denotes the normal with just the standard deviation.)

You can translate into standard normal units by:
Z = ( X – ? ) / ?

Moving from the standard normal back to the original distribuiton using:
X = ? + Z * ?

Where Z ~ Normal( ? = 0, ? = 1). You can then use the standard normal cdf tables to get probabilities.

If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.

If a sample of size is is drawn from a population with mean ? and standard deviation ? then the sample average xBar is normally distributed

with mean ? and standard deviation ? /?(n)

An applet for finding the values

calculator

how to read the tables

In this question we have
X ~ Normal( ?x = 950 , ?x² = 48400 )
X ~ Normal( ?x = 950 , ?x = 220 )

Find P( X > 1400 )
P( ( X – ? ) / ? > ( 1400 – 950 ) / 220 )
= P( Z > 2.045455 )
= P( Z < -2.045455 )
= 0.02040503

Find P( X < 500 )
P( ( X – ? ) / ? < ( 500 – 950 ) / 220 )
= P( Z < -2.045455 )
= 0.02040503