Lesson

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`=`n``p` and the mean number of failures $n-\mu=n(1-p)$`n`−`μ`=`n`(1−`p`) 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.

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×(1−12)=2.29.

According to the binomial formula, we need to calculate

$P(k=7)=\binom{21}{7}\left(\frac{1}{2}\right)^7\left(1-\frac{1}{2}\right)^{21-7}$`P`(`k`=7)=(217)(12)7(1−12)21−7

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 $np=21\times0.5=10.5$`n``p`=21×0.5=10.5, and the standard deviation is $\sqrt{np(1-p)}=\sqrt{10.5\times0.5}=2.29$√`n``p`(1−`p`)=√10.5×0.5=2.29.

The transformed interval is $\left(\frac{6.5-10.5}{2.29},\frac{7.5-10.5}{2.29}\right)\approx\left(-1.747,-1.31\right)$(6.5−10.52.29,7.5−10.52.29)≈(−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$0.4599−0.4049=0.055 which is close to the result from the binomial formula.

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 $np=500\times0.9=450$`n``p`=500×0.9=450 and the standard deviation is $\sqrt{500\times0.9\times0.1}\approx4.5$√500×0.9×0.1≈4.5. Thus, we need the area above $\frac{449.5-450}{4.5}\approx-0.11$449.5−4504.5≈−0.11 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+0.0438≈0.54.

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.

Summary

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

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