3.6 Central limit theorem: using the central limit theorem  (Page 3/8)

Historically, being able to compute binomial probabilities was one of the most important applications of the Central Limit Theorem. Binomial probabilities were displayed in a table in a book with a small value for $n$ (say, 20). To calculate the probabilities with large values of $n$ , you had to use the binomial formula which could be very complicated. Using the Normal Approximation to the Binomial simplified the process. To compute the Normal Approximation to the Binomial, take a simple random sample from a population. You must meet the conditionsfor a binomial distribution :

• there are a certain number $n$ of independent trials
• the outcomes of any trial are success or failure
• each trial has the same probability of a success $p$

Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K - 5. A simple random sample of 300 is surveyed.

1. Find the probability that at least 150 favor a charter school.
2. Find the probability that at most 160 favor a charter school.
3. Find the probability that more than 155 favor a charter school.
4. Find the probability that less than 147 favor a charter school.
5. Find the probability that exactly 175 favor a charter school.

Let $X=$ the number that favor a charter school for grades K - 5. $X$ ~ $B(n,p)$ where $n=300$ and $p=\mathrm{0.53}.$ Since $np>5$ and $nq>5,$ use the normal approximation to the binomial. The formulas for the mean and standard deviation are $\mu =np$ and $\sigma =\sqrt{npq}.$ The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is $Y$ . $Y~N(159,\mathrm{8.6447})$ . See The Normal Distribution for help with calculator instructions.

For Problem 1., you include 150 so $P(x\ge 150)$ has normal approximation $P(Y\ge 149.5)=\mathrm{0.8641}$ .

normalcdf $(149.5,\mathrm{10^99},159,8.6447)=0.8641$ .

For Problem 2., you include 160 so $P(x\le 160)$ has normal approximation $P(Y\le 160.5)=\mathrm{0.5689}$ .

normalcdf $(0,160.5,159,8.6447)=0.5689$

For Problem 3., you exclude 155 so $P(x> 155)$ has normal approximation $P(y> 155.5)=\mathrm{0.6572}$ .

normalcdf $(155.5,\mathrm{10^99},159,8.6447)=0.6572$

For Problem 4., you exclude 147 so $P(x< 147)$ has normal approximation $P(Y< 146.5)=\mathrm{0.0741}$ .

normalcdf $(0,146.5,159,8.6447)=0.0741$

For Problem 5., $P\left(x=175\right)$ has normal approximation $P(174.5<y<175.5)=\mathrm{0.0083}$ .

normalcdf $(174.5,175.5,159,8.6447)=0.0083$

Because of calculators and computer software that easily let you calculate binomial probabilities for large values of $n$ , it is not necessary to use the the Normal Approximation to the Binomial provided you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators and they easily calculate probabilities for the binomial. In an Internet browser, if you type in "binomial probability distribution calculation," you can find at least one online calculator for the binomial.

For Example 3 , the probabilities are calculated using the binomial ( $n=300$ and $p=0.53$ ) below. Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial.

$P(x\ge 150)$ : 1 - binomialcdf $(300,0.53,149)=0.8641$

$P(x\le 160)$ : binomialcdf $(300,0.53,160)=0.5684$

$P(x> 155)$ : 1 - binomialcdf $(300,0.53,155)=0.6576$

$P(x< 147)$ : binomialcdf $(300,0.53,146)=0.0742$

$P\left(x=175\right)$ : (You use the binomial pdf.) binomialpdf $(175,0.53,146)=0.0083$

**Contributions made to Example 2 by Roberta Bloom