Download Normal Approximations to Binomial Distributions

Normal Approximations to Binomial Distributions
Premature babies are those born more than 3 weeks early. Newsweek (May 16, 1988) reported that
10% of the live births in the U.S. are premature. Suppose that 250 live births are randomly selected
and that the number X of the “preemies” is determined. What is the probability that there are between
15 and 30 preemies, inclusive?
1) Find this probability using the binomial distribution.
2) What is the mean and standard deviation of the above distribution?
3) If we were to graph a histogram for the above binomial distribution, what shape do you think it will
have?
4) Input the following in your calculator. Graph:
histogram
L1: seq(x,x,0,45)
xlist: L1
L2: binomialpdf(250,.1,L1)
freq: L2
Set window:
xmin:-0.5
xmax: 45
xscl: 1
ymin:0
ymax:0.2
yscl:1
5) What do you notice about the shape?
Input Y1: normalpdf(x, µ, σ) using the mean and standard deviation for the binomial distribution
above.
Normal distributions can be used to estimate probabilities for binomial distributions when:
1) the probability of success is close to .5
or
2) n is sufficiently large
Rule: if n is large enough, then np > 10 & n(1 –p) > 10
Since a continuous distribution is used to estimate the probabilities of a discrete distribution, a continuity
correction is used to make the discrete values similar to continuous values. (  0.5 to the discrete values)
6) Use a normal distribution with the binomial mean and standard deviation above to estimate the
probability that between 15 & 30 preemies, inclusive, are born in the 250 randomly selected babies.
Binomial
written as
P (15 < X < 30)

Normal (w/continuity correction)
P (14.5 < X < 30.5) = ________________
7) How does the answer in question 6 compare to the answer in question 1?
8) What is the probability that less than 20 preemies are born out of the 250 babies?
Binomial
Normal
P(X < 20)

9) What is the probability that at least 30 preemies are born out of the 250 babies?
Binomial
Normal
P(X > 30)

Since P(preemie) = .1 which
is not close to .5, is n large
enough?
10) What is the probability that less than 35 preemies but more than 20 preemies are born out of the
250 babies?
Binomial
Normal
P(20 < X < 35)

Why 10?
Normal distributions extend infinitely in both directions; however, binomial distributions are between 0
and n. If we use a normal distribution to estimate a binomial distribution, we must cut off the tails of
the normal distribution. This is OK if the mean of the normal distribution (which we use the mean of
the binomial) is at least three standard deviations (3σ) from 0 and from n.
We require:
Or
As binomial:
Square:
Simplify:
Since (1 - p) < 1:
  3  0
  3
np  3 np 1  p 
n 2 p 2  9np 1  p 
np  91  p 
np  9
Therefore, we say the np should be at least 10 and n (1 – p) should be at least 10.
Why is the continuity correction

.5?
Think about how discrete histograms are made. Each bar is centered over the discrete values. The bar
for “1” actually goes from 0.5 to 1.5 & the bar for “2” goes from 1.5 to 2.5. Therefore, by adding or
subtracting .5 from the discrete values, you find the actually width of the bars that you need to estimate
with the normal curve.