normal distribution help!!!!!!!!!?
March 10th, 2010 by admin | Posted in Single Dating Services
A woman posts an advertisement on an internet dating service. She is seeking to date men who are between 68 and 74 inches ….Assume the mean height of men is 69 inches and the standard deviation is equal to 1.5 inches
what proportion of men is she willing to date?
for prob of x >75 i found it to be 4 but dont know how to convert that using the chart
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Tags: advertisement, internet dating service, Normal Distribution, prob, proportion, standard deviation, Woman
For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)
You can translate into standard normal units by:
Z = ( X – μ ) / σ
Moving from the standard normal back to the original distribuiton using:
X = μ + Z * σ
Where Z ~ Normal( μ = 0, σ = 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 μ and standard deviation σ then the sample average xBar is normally distributed
with mean μ and standard deviation σ /√(n)
An applet for finding the values
http://www-stat.stanford.edu/~naras/jsm/FindProbability.html
calculator
http://stattrek.com/Tables/normal.aspx
how to read the tables
http://rlbroderson.tripod.com/statistics/norm_prob_dist_ed9.html
In this question we have
X ~ Normal( μx = 69 , σx² = 2.25 )
X ~ Normal( μx = 69 , σx = 1.5 )
Find P( 68 < X < 74 )
= P( ( 68 – 69 ) / 1.5 < ( X – μ ) / σ < ( 74 – 69 ) / 1.5 )
= P( -0.6666667 < Z < 3.333333 )
= P( Z < 3.333333 ) – P( Z < -0.6666667 )
= 0.999571 – 0.2524925
= 0.7470784
i think what u have to do is set the equation as 68<x<74 and the figure it out like u did for x<74
Convert 68 and 74 to Z-scores.
Z=(x-mean)/sd
P(68 < X < 74) = P[ (68-69)/1.5 < Z < (74-69)/1.5]
=P(-0.667 < Z < 3.333)
There are several types of normal probability tables and your text probably has one. You may use the source if you wish.
Using the table, you find the area from 0 to 3.3333 and also the area from 0 to 0.667 (same as -0.667) and add them.
0.2486+0.4999 = 0.7485
She is willing to date about 75 % of men.