Normal Approximation to the Binomial
Example:
20% of customers at a bakery will buy a brownie. 500 customers arrive at the bakery in a
day. Assume that individual customers make their purchases independently.
(a) What is the probability that more than 110 customers buy a brownie?
(b) How many brownies does the bakery need to have in stock so that the probability of
selling out in a day is 1%?
Solution:
Let X = number of customers who buy a brownie in a day. Assuming that the purchases
of individual customers are iid Bernoulli trials with “probability of success” being 20%,
X ∼ Bin(500, .2). The computations needed to solve the problems given using the Binomial
distribution directly are very cumbersome because we have to calculate quantities like
500 X
500
(.2)i (.8)500−i
i
i=111
Thus we will use the CLT to approximate the Binomial with a Normal distribution. If
X ∼ Bin(n, p), the CLT tells us that X ∼
˙ N (np, np(1 − p)) for np > 5 and n(1 − p) > 5 and
that the approximation is very good for np > 20 and n(1 − p) > 20. Here the approximation
is very good and gives
X ∼
˙ N (500 × 0.2, 500 × .2 × .8)
∼
˙ N (100, 80)
That we have E[X] = µ = 100 and V ar[X] = σ 2 = 80.
(a) The required probabilty is
P (X > 110) = 1 − P (X ≤ 110)
110 − 100
√
≈ 1−Φ
80
= 1 − Φ(1.12)
= 1 − 0.86864 = .1314
1
from the Normal cdf tables
(b) Let t = the number of brownies in stock. The bakery runs out of brownies if X ≥ t.
To answer the question we need to find t such that P (X ≥ t) = .01
P (X > t) = 1 − P (X ≤ t − 1)
t − 1 − 100
√
≈ 1−Φ
80
We set
to .01 and use algebra
and the
Normal cdf table to find t:
this equal
t−1−100
t−1−100
√
√
= .01 is equivalent to Φ
= .99; thus we find the 99th percentile
1−Φ
80
80
√
√
of the normal distribution and equate it to t−1−100
.
We
get
2.33
×
80 = t − 1 − 100
80
giving t − 1 = 100 + 2.33 × 8.944272 and thus t = 120.8 + 1 ≈ 122. That is, they need
to stock 122 brownies to meet their policy that there is only a 1% chance that they
will run out.
2