NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Normal approximation to binomials

When graphing binomial distributions, we saw that the shape of a binomial distribution depends on the number of Bernoulli trials in the experiment ($n$n) and the probability of obtaining a success on any particular trial ($p$p).

In studying binomial distributions, we saw that if $n$n was too small or $p$p was too close to $0$0 or $1$1, the distributions were skewed. However, binomial distributions are approximately normal when $n$n is large enough and $p$p is not too close to $0$0 or $1$1.

Have a play around with this applet now and investigate at what values of $n$n and $p$p the normal approximation works well. Discuss as a class.

As a general rule, both the mean number of successes $\mu=np$μ=np and the mean number of failures $n-\mu=n(1-p)$nμ=n(1p) must be greater than $5$5 for a good approximation by the normal curve.


The normal approximation to a binomial distribution is used because calculation with the help of a standard normal distribution table is often easier than calculation with the binomial formula, particularly when $n$n is large.

Example 1

Suppose a balanced coin is tossed $21$21 times. We wish to determine the probability of getting exactly $7$7 heads. In this case $p=\frac{1}{2}$p=12 and $\sigma=\sqrt{21\times\frac{1}{2}\times\left(1-\frac{1}{2}\right)}=2.29$σ=21×12×(112)=2.29.

According to the binomial formula, we need to calculate


This is easy enough to do with a good calculator but would be more difficult by hand as the numbers in the binomial coefficient are large. We find $P(k=7)=0.0554$P(k=7)=0.0554.

To work this out using the normal approximation, we have to represent the number of successes $(k=7)$(k=7) by the interval $(6.5,7.5)$(6.5,7.5). (This is called a continuity correction.)

Next, we obtain the $z$z-scores corresponding to $6.5$6.5 and $7.5$7.5 when the mean is $np=21\times0.5=10.5$np=21×0.5=10.5, and the standard deviation is $\sqrt{np(1-p)}=\sqrt{10.5\times0.5}=2.29$np(1p)=10.5×0.5=2.29.

The transformed interval is $\left(\frac{6.5-10.5}{2.29},\frac{7.5-10.5}{2.29}\right)\approx\left(-1.747,-1.31\right)$(6.510.52.29,7.510.52.29)(1.747,1.31).

We use the table for the standard normal probability density function to obtain the probability under the curve between $-1.747$1.747 and $-1.31$1.31. By symmetry, this is the same as the probability between $1.31$1.31 and $1.747$1.747. (You can find a table here.)

You should check that the probability obtained this way is $0.4599-0.4049=0.055$0.45990.4049=0.055 which is close to the result from the binomial formula.


Example 2

An aerosol spray has been designed to kill house flies. An individual fly exposed to the aerosol under test conditions has a probability of dying within one minute of $0.9$0.9. If $500$500 flies are exposed to the aerosol, what is the probability that at least $450$450 of them will be dead within one minute? 

We need to add the probabilities of $450,451,452,...,500$450,451,452,...,500 flies dying. This would be a very cumbersome calculation if done with the binomial formula. We use, instead, the normal approximation and look for the area beneath the normal density curve above $449.5$449.5 (using the continuity correction as before).

The mean of the binomial distribution is $np=500\times0.9=450$np=500×0.9=450 and the standard deviation is $\sqrt{500\times0.9\times0.1}\approx4.5$500×0.9×0.14.5. Thus, we need the area above $\frac{449.5-450}{4.5}\approx-0.11$449.54504.50.11 in the standard normal distribution.

This area is $0.5$0.5 plus the area between $0$0 and $0.11$0.11. So, according to the table, the probability must be $0.5+0.0438\approx0.54$0.5+0.04380.54.

If needed, we can work out from this that the probability that fewer than $450$450 flies will be dead after $1$1 minute is $0.46$0.46.



If $n$n is sufficiently large and $p$p is not too close to $0$0 or $1$1, (both $np>5$np>5 and $n(1-p)>5$n(1p)>5) then the binomial distribution will be approximately normal, with a mean of $\mu=np$μ=np and a standard deviation of $\sigma=\sqrt{np\left(1-p\right)}$σ=np(1p).



Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal


Apply probability distributions in solving problems

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