 A standardized value  A number of standard deviations a given value, x, is above or below the mean  z = (score (x) – mean)/s (standard deviation)

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Presentation transcript:

 A standardized value  A number of standard deviations a given value, x, is above or below the mean  z = (score (x) – mean)/s (standard deviation)  A positive z-score means the value lies above the mean  A negative z-score means the value lies below the mean  Round to 2 decimals

 The mean for IQ scores is 50 with a standard deviation of 5 with a normal distribution  What is the probability that scores will be between 45 and 55?  Calculate z-score first  Use the Normal Distribution (z) statistical table

 Score (x) = 45  Mean = 50  s = 5  z = (score (x) – mean)/s (standard deviation)  z = ( )/5 (standard deviation) o = -5/5, = -1 o Wait! There’s a second value to consider.

 Score (x) = 55  Mean = 50  s = 5  z = (score (x) – mean)/s (standard deviation)  z = ( )/5 (standard deviation) o = 5/5, = 1

 Using the normal distribution (z) statistical table:  Determine the area from the mean:  1 s up (mean to z) =.3413  1 s down (mean to z) =.3413  Add the 2 values together  =.6826 * 100% = 68.26%  So, the probability that a score will be between 45 and 55 is 68.26%!

 The mean for IQ scores is 50 with a standard deviation of 5 with a normal distribution  What is the probability that an IQ score will be between 55 and 60?  Calculate z-score first  Use the Normal Distribution (z) statistical table

 Score (x) = 55  Mean = 50  s = 5  z = (score (x) – mean)/s (standard deviation)  z = ( )/5 (standard deviation) o = 5/5, = 1 o Wait! There’s a second value to consider.

 Score (x) = 60  Mean = 50  s = 5  z = (score (x) – mean)/s (standard deviation)  z = ( )/5 (standard deviation) o = 10/5, = 2

 Using the normal distribution (z) statistical table:  Determine the area from the mean:  1 s up (mean to z) =.3413  2 s up (mean to z) =.4772  Subtract the 2 values (because we only want the distance from 1z to 2z, not the mean to 2z)  =.1359 * 100% = 13.59%  So, the probability that an IQ score will be between 55 and 60 is 13.59%!

 The mean for IQ scores is 100 with a standard deviation of 15 with a normal distribution  What percentage of scores will lie below 100?  Calculate z-score first  Use the Normal Distribution (z) statistical table

 Score (x) = 100  Mean = 100  s = 15  z = (score (x) – mean)/s (standard deviation)  z = ( )/15 (standard deviation) o = 0/15 = 0

 Using the normal distribution (z) statistical table:  Determine the area from the mean:  0 s down (larger portion) =.5000  So, the probability that a student’s IQ score will be below 100 is 50%.

 The mean for IQ scores is 100 with a standard deviation of 15 with a normal distribution  What percentage of scores will lie below 115?  Calculate z-score first  Use the Normal Distribution (z) statistical table

 Score (x) = 115  Mean = 100  s = 15  z = (score (x) – mean)/s (standard deviation)  z = ( )/15 (standard deviation) o = 15/15 = 1

 Using the normal distribution (z) statistical table:  Determine the area from the mean:  1 s up (larger portion) =.8413  So, the percentage of scores that will be below 115 is 84.13%

 The mean for IQ scores is 100 with a standard deviation of 15 with a normal distribution  What percentage of scores will lie above 115?  Calculate z-score first  Use the Normal Distribution (z) statistical table

 Score (x) = 115  Mean = 100  s = 15  z = (score (x) – mean)/s (standard deviation)  z = ( )/15 (standard deviation) o = 15/15 = 1

 Using the normal distribution (z) statistical table:  Determine the area from the mean:  1 s up (smaller portion) =.1587  So, the probability that a student’s IQ score will be above 115 is 15.87%  Note: this area of 15.87% plus the area of scores below 115, 84.13%, equal 100%.