SOLUTION: For a normal distribution that has a mean of 100 and a standard deviation of 8. Determine the Z-score for each of the following X values: X=70 Z=-3.75 X=124

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Question 626509: For a normal distribution that has a mean of 100 and a standard deviation of 8. Determine the Z-score for each of the following X values:
X=70 Z=-3.75
X=124 Z=3
Use the information in 1 to determine the area or probability of
p(92 < x < 108)
p(84 < x < 116)
p(x < 84)
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
For a normal distribution that has a mean of 100 and a standard deviation of 8. Determine the Z-score for each of the following X values:
X=70 Z=-3.75
X=124 Z=3
Use the information in 1 to determine the area or probability of
p(92 < x < 108) = ?
z(92) = (92-100)/8 = -1
z(108) = (108-100)/8 = 1
P(92 < x < 108) = P(-1 < z < 1) = 0.6827
---------------------------------------------
p(84 < x < 116) = ?
z(84) = (84-100)/8 = -2
z(116) = (116-100)/8 = 2
P(84 < x < 116) = P(-2 < z < 2) = 0.9545
-----
p(x < 84) = ?
z(84) = -2
P(x < 84) = P(z < -2) = P(-100< z < -2) = 0.0228
===================
Cheers,
Stan H.
===================
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

Important to Understand z -values as they relate to the Standard Normal curve:

Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  

Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and %50 to the right



For the normal distribution: 

one  standard deviation from the mean accounts for about 68.2% of the set 

two standard deviations from the mean account for about 95.4%

and three standard deviations from the mean account for about 99.7%.

a normal distribution that has a mean of 100 and a standard deviation of 8.

p(92 < x < 108)= .682

 p(84 < x < 116) = .954

 p(x < 84) = (1-.954)/2 

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