NOTE: For the binomial distribution, the values to the right of each = sign are primarily
included for illustrative purposes. The equalities which hold in the binomial distribution do not
hold in the normal distribution, because there is a gap between consecutive values of a. The
normal approximation deals with this by “splitting” the difference.
For example, in the binomial, P(X ≤ 6 ) = P(X < 7), since 6 is the next possible value of
X that is less than 7. In the normal, we approximate this by finding P(X ≤ 6.5). And, in the
binomial, P(X ≥ 6) = P(X > 5), because 6 is the next value of X that is greater than 5. In the
normal, we approximate this by finding P(X ≥ 5.5)
EXAMPLES.
1. Suppose 50% of the population approves of the job the governor is doing, and that 20
individuals are drawn at random from the population. Solve the following, using both the
binomial distribution and the normal approximation to the binomial.
a. What is the probability that exactly 7 people will support the governor?
b. What is the probability that 7 or fewer people will support the governor?
c. What is the probability that exactly 11 will support the governor?
d. What is the probability that 11 or fewer will support the governor?
SOLUTION: Note that N = 20, p = .5, so µ = Np = 10 and σ5 = Npq = 5, σ = 2.236. Since Npq
≥ 3, it is probably safe to assume that X has approximately a N(10,5) distribution.
a. For the binomial, find P(X = 7). Appx. E, Table II shows P(7) = .0739.
For the normal, find P(6.5 ≤ X ≤ 7.5). We convert 6.5 and 7.5 to their corresponding z-scores
(-1.57 and -1.12), and the problem becomes finding P(-1.57 ≤ Z ≤ -1.12) = F(1.57) - F(1.12) =
.9418 - .8686 = .0732.
b. To use the binomial distribution, find P (X ≤ 7). Using Appx. E, we get
P(7) + P(6) + P(5) + P(4) + P(3) + P(2) + P(1) + P(0) =
.0739 + .0370 + .0148 + .0046 + .0011 + .0002 + 0 + 0 = .1316.
To use the normal approximation to the binomial, find P(X ≤ 7.5). As noted above, the z-
score that corresponds to 7.5 is -1.12. F(-1.12) = 1 - F(1.12) = 1 - .8686 = .1314.
c. For the binomial, find P(X = 11). Appx. E shows P(11) = .1602.
For the normal, find P(10.5 ≤ X ≤ 11.5). If we convert 10.5 and 11.5 to their corresponding z-
scores, the problem becomes a matter of finding P(.22 ≤ Z ≤.67) = F(.67) - F(.22) = .7486 - .5871
= .1615.
d. For the binomial, find P(X ≤ 11). From Appx E Table 2, you can determine that
this is .7483. For the normal, find P(X ≤ 11.5). The z-score that corresponds to 11.5 is .67, and
F(.67) = .7486.
In all of the above, note that the results obtained using the binomial distribution and the
normal approximation to the binomial are almost identical.
Normal distribution - Page 9