Statistical Methods for Psychology

standard deviation from the mean, if the total area under the curve is taken to be 1.00. (Re-

member that we care about areas because they translate directly to probabilities.) We al-

ready have seen that zscores represent standard deviations from the mean, and thus we

know that we want to find the area above z 5 1.

Only the positive half of the normal distribution is tabled. Because the distribution is

symmetric, any information given about a positive value of zapplies equally to the correspon-

ding negative value of z. (The table in Appendixzalso contains a column labeled “y.” This is

just the height [density] of the curve corresponding to that value of z. I have not included it

here to save space and because it is rarely used.) From Table 3.1 (or Appendixz) we find the

row corresponding to z 5 1.00. Reading across that row, we can see that the area from the

mean to z 5 1 is 0.3413, the area in the larger portionis 0.8413, and the area in the smaller

portionis 0.1587. If you visualize the distribution being divided into the segment below

z 5 1 (the unshaded part of Figure 3.7) and the segment above z 5 1 (the shaded part), the

meanings of the terms larger portionand smaller portionbecome obvious. Thus, the answer

to our original question is 0.1587. Because we already have equated the terms areaand prob-

ability, we now can say that if we sample a child at random from the population of children,

and if Behavior Problem scores are normally distributed, then the probability that the child

will score more than one standard deviation above the mean of the population (i.e., above 60)

is .1587. Because the distribution is symmetric, we also know that the probability that a child

will score more than one standard deviation belowthe mean of the population is also .1587.

Now suppose that we want the probability that the child will be more than one standard

deviation (10 points) from the mean in either direction. This is a simple matter of the sum-

mation of areas. Because we know that the normal distribution is symmetric, then the area

below z 52 1 will be the same as the area above z 51 1. This is why the table does not

contain negative values of z—they are not needed. We already know that the areas in which

we are interested are each 0.1587. Then the total area outside z 56 1 must be 0.1587 1

0.1587 5 0.3174. The converse is also true. If the area outside z 56 1 is 0.3174, then the

area between z 51 1 and z 52 1 is equal to 1 2 0.3174 5 0.6826. Thus, the probability

that a child will score between 40 and 60 is .6826.

To extend this procedure, consider the situation in which we want to know the proba-

bility that a score will be between 30 and 40. A little arithmetic will show that this is sim-

ply the probability of falling between 1.0 standard deviation below the mean and 2.0

standard deviations below the mean. This situation is diagrammed in Figure 3.8. (Hint: It is

always wise to draw simple diagrams such as Figure 3.8. They eliminate many errors and

make clear the area(s) for which you are looking.)

74 Chapter 3 The Normal Distribution

–3

z

f(

X

)

0.40

0.30

0.20

0.10

0

–2 –1 0 1 2 3

0.1587

0.3413

0.5000

0.8413

Figure 3.7 Illustrative areas under the normal distribution