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 = (45 - 50)/5 (standard deviation)

o = -5/5, = -1o Wait! There’s a second value to consider.

Score (x) = 55 Mean = 50 s = 5 z = (score (x) – mean)/s (standard

deviation) z = (55 - 50)/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

.3413 + .3413 = .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 = (55 - 50)/5 (standard deviation)

o = 5/5, = 1o Wait! There’s a second value to consider.

Score (x) = 60 Mean = 50 s = 5 z = (score (x) – mean)/s (standard

deviation) z = (60 - 50)/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) .4772 - .3413 = .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 = (100 - 100)/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 = (115 - 100)/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 = (115 - 100)/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%.