Basic Statistics

THE NORMAL APPROXlMATiON TOTHE BINOMIAL 129

Figure 10.4 Proportion of p from .08 to .32.

10.3 THE NORMAL APPROXIMATION TOTHE BINOMIAL

When the sample size is large, one more statement can be made about the distribution

of sample proportions: It is well approximated by a normal distribution with the

proper mean and standard deviation. In this book no use is made of binomial tables;

instead, the normal tables are always used in problems involving proportions.

10.3.1

The normal tables are used as an approximation to the binomial distribution when
n, the sample size, is sufficiently large, and the question naturally arises as to what
“sufficiently large” means. One rule of thumb often given is that we may use the
normal curve in problems involving sample proportions whenever nr and n( 1 - T)
are both as large as or larger than 5. This rule reflects the fact that when r is $, the
distribution looks more like a normal curve than when r is close to 0 or close to 1; see
Figure 10.3, where (b) looks more nearly normal than (a), and (d) looks more nearly
normal than (c). Note that this rule is not conservative and should be considered a
minimum size.
Use of the normal distribution approximation for the distribution of sample propor-
tions may be illustrated by the following problem: A sample of size 25 is to be drawn
from a population of adults who have been given a certain drug. In the population,
the proportion of those who show a particular side effect is .2. What percentage of
all sample proportions for samples of size 25 lies between .08 and .32, inclusive?
The normal approximation may be used in solving this problem since nT =
25(.2) = 5 and n(1 - n) = 25(.8) = 20 are both as large as 5. The mean of
all possible p’s is .2, and an approximate answer to the problem is the shaded area in
Figure10.4. For the normal curve approximation, we use a normal curve that has a
mean of .2 and a standard deviation of p of

Use of the Normal Approximation to the Binomial

J- = Jm = J6?E = .08

To find the area from .08 to .32, we find the z corresponding to .32. Here z =

(.32 - .20)/.08 = .12/.08 = 1.5. Using Table A.2, we find that .9332 or 93.32%

of the sample proportions lie below .32, so that 6.68% of the sample proportions lie

> .32. By symmetry, 6.68% of the sample proportions lie < .08. This means that