2.  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

3.  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

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

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

6.  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 .3413 +.3413 =.6826 * 100% = 68.26%  So, the probability that a score will be between 45 and 55 is 68.26%!

7.  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

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

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

10.  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) .4772 -.3413 =.1359 * 100% = 13.59%  So, the probability that an IQ score will be between 55 and 60 is 13.59%!

11.  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

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

13.  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%.

14.  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

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

16.  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%

17.  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

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

19.  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%.