Normally distributed Standard deviation Probability?
sleeeplessinsf asked:
*Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. Based on this information, what is the probability of a student reading at more than 1400 words per minute after finishing the course?
and
*Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words per minute and a standard deviation equal to 220 words per minute. If the two students were selected at random, what is the probability that they would both read at less than 500 words per minute?
speed reading
Tags: Speed Reading Course, Standard Deviation, Words Per Minute
June 18th, 2009 at 3:13 pm
speed reading
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