6.5: Normal Approximation to the Binomial Distribution
- Page ID
- 13602
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As you saw in Chapter 5, calculating binomial probabilities could become quite tedious, even when using the nCr function, if we needed to add more than a few probabilities together to get the answer. With better technology, functions such as binompdf() and binomialcdf() in our TI-83/84 calculators or binomdist() in most spreadsheet applications took away the tedium of the process. However, before that, an alternate method of using the standard normal distribution to approximate answers to a binomial distribution was used for the discrete random variable questions when \(n\) (the number of trials) was large and the experiment met the requirements for a Binomial Probability Distribution.
- The procedure has a fixed number of trials.
- Each trial must have all outcomes classified into two categories
- The trials must be independent
- The probability of a success remains the same in all trials.
Since Binomial distributions are found by using n and p these are a few guidelines to be aware of:
- When p is approximately 0.5 and n increases, the shape of a binomial distribution becomes similar to a normal distribution.
- For smaller values of n and for values of p closer to 0 or 1, the normal approximation to the binomial is not very good.
- A rule of thumb is that a normal approximation should only be used when \(np \leq 5\) and \(nq \leq 5\), though the approximation is better if they are both closer to 10.
A correction for continuity is also needed since binomial distributions use discrete data and normal distributions are used for continuous data. Remember that for a continuous random variable that uses area under the curve to calculate the probability, \(P(x = a)\) is 0 since there is not area for a line. But if this was a discrete random variable, then a probability value calculated using \(P(x = r) = \binom {n}{r} p^r (q)^{n - r}\) is possible.
Continuity Corrections:
| When finding: | Use: |
|---|---|
| \(P(x = a)\) |
\(P(a - 0.5 < x < a +0.5)\) By stretching from the desired point on either side, we create a small area for our normalcdf() function. |
|
\(P(x \geq a)\) Probability of \(a\) or more. |
\(P(x > a - 0.5)\) Stretching the interval to left of the desired point allows us to include \(a\) in the calculation. |
|
\(P(x > a)\) Probability of more than \(a\). |
\(P(x > a + 0.5)\) Stretching the interval to right of the desired point allows us to exclude \(a\) from the calculation. |
|
\(P(x \leq a)\) Probability of \(a\) or fewer. |
\(P(x < a + 0.5)\) Stretching the interval to right of the desired point allows us to include \(a\) from the calculation. |
|
\(P(x < a)\) Probability of less than \(a\). |
\(P(x < a - 0.5)\) Stretching the interval to left of the desired point allows us to exclude \(a\) in the calculation. |
Notice that in the corrections column, we never use an equality in any of the symbols since it doesn't add anything to the area.
Normal Approximation to the Binomial Distribution
In practice, technology should be used to find an exact answer for a binomial distribution question. However, in this introductory course, using the approximation to the binomial distribution is beneficial in solidifying understanding between a discrete random variable and a continuous random variable.
Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.
Find the probability that at least 150 favor a charter school.
Solution
Let \(x =\) the number that favor a charter school for grades K through 5.
Make sure that it meets the binomial conditions.
Conditions 1: There is a fixed number of trials, where \(n = 300\)
Condition 2: There are two outcomes, favor the charter school or don't favor the charter school.
Condition 3: We can reasonably assume that the 300 surveyed is less than 5% of the population in the area, so we can treat the events as independent.
Condition 4: Due to independence, we know that the probability of a success, \(p = 0.53\), will remain the same for each selection.
Since we are working with very large numbers, we want to see if we can use the normal approximation to the binomial instead of adding all of the discrete probabilities from 150 to 300, for example.
\(np = 300(0.53) = 159\) and \(nq = 300(0.47) = 141\)
Since \(np > 5\) and \(nq > 5\), we can use the normal approximation to the binomial.
Remember, the formulas for the mean and standard deviation of a binomial distribution are \(\mu = np\) and \(\sigma = \sqrt{npq}\).
The mean is 159 and the standard deviation is 8.6447.
The random variable for the binomial distribution is \(x\). We will use \(Y\) for the normal distribution: \(Y \sim N(159, 8.6447)\). See The Normal Distribution for help with calculator instructions.
You include 150 and shade to the right, so our interval needs to stretch to the left:
\(P(x \geq 150) \approx P(Y > 149.5) = \text{normalcdf}(149.5,10^{99},159,8.6447) = 0.8641\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first. Use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{149.5 - 159}{8.6447} = -1.09893...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x \geq 150) \approx P(Y > 149.5) = \text{normalcdf}(-1.0989,10^{99}) = 0.8649\)
There is about an 86% chance that 150 or more people favor a charter school for grades K through 5.
Using the information from Example \(\PageIndex{1}\), find the probability that at most 160 favor a charter school.
- Answer
-
Our \(x =\) the number that favor a charter school for grades K through 5. We already know that this experiment meets the binomial distribution requirements and meets the normal approximation to the binomial conditions as well.
We want to include 160 and shade to the left so our interval needs to stretch to the right:
\(P(x \leq 160) \approx P(Y < 160.5) = \text{normalcdf}(-10^{99},160.5,159,8.6447) = 0.5689\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first. Use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{160.5 - 159}{8.6447} = 0.173516...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x \leq 160) \approx P(Y < 160.5) = \text{normalcdf}(-10^{99},0.173516) = 0.56887...\)
There is about a 57% chance that 160 or fewer people favor a charter school for grades K through 5.
Using the information from Example \(\PageIndex{1}\), find the probability that more than 155 favor a charter school.
Solution
Our \(x =\) the number that favor a charter school for grades K through 5. We already know that this experiment meets the binomial distribution requirements and meets the normal approximation to the binomial conditions as well.
You exclude 155 and shade to the right, so our interval needs to stretch to the right:
\(P(x > 155) \approx P(Y > 155.5) = \text{normalcdf}(155.5,10^{99},159,8.6447) = 0.6572\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first. Use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{155.5 - 159}{8.6447} = -0.40487...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x > 155) \approx P(Y > 155.5) = \text{normalcdf}(-0.40487,10^{99}) = 0.6572\)
There is about an 66% chance that more than 155 people favor a charter school for grades K through 5.
Using the information from Example \(\PageIndex{1}\), find the probability that fewer than 147 favor a charter school.
- Answer
-
Our \(x =\) the number that favor a charter school for grades K through 5. We already know that this experiment meets the binomial distribution requirements and meets the normal approximation to the binomial conditions as well.
We want to exclude 147 and shade to the left so our interval needs to stretch to the left:
\(P(x < 147) \approx P(Y < 146.5) = \text{normalcdf}(-10^{99},146.5,159,8.6447) = 0.0741\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first, then use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{146.5 - 159}{8.6447} = -1.44597...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x < 146.5) \approx P(Y < 160.5) = \text{normalcdf}(-10^{99},-1.44597) = 0.0741...\)
There is about a 7% chance that fewer than 147 people favor a charter school for grades K through 5.
Using the information from Example \(\PageIndex{1}\), find the probability that exactly 175 favor a charter school.
Solution
Our \(x =\) the number that favor a charter school for grades K through 5. We already know that this experiment meets the binomial distribution requirements and meets the normal approximation to the binomial conditions as well.
We want to include only 175 so our interval needs to stretch to the left and the right to create two dimensional shape we can find the area for:
\(P(x = 175) \approx P(174.5 < Y < 175.5) = \text{normalcdf}(174.5,175.5,159,8.6447) = 0.0083\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first for both endpoints. Use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{174.5 - 159}{8.6447} = 1.793006...\)
\(z = \dfrac{175.5 - 159}{8.6447} = 1.908683...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x = 175) \approx P(174.5 < Y < 175.5) = \text{normalcdf}(1.793006,1.908683) = 0.0083...\)
There is about a 0.8% chance that exactly 175 people favor a charter school for grades K through 5.
In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that
- Exactly 304 people favor Dawn Morgan for Mayor.
- At least 250 favor Dawn Morgan for mayor.
- Answer
-
Part a:
Let \(x =\) the number that favor Dawn Morgan for Mayor.
Make sure that it meets the binomial conditions.
Conditions 1: There is a fixed number of trials, where \(n = 500\)
Condition 2: There are two outcomes, favor Dawn Morgan or not favor Dawn Morgan.
Condition 3: We can reasonably assume that the 500 surveyed is less than 5% of the population in the area, so we can treat the events as independent.
Condition 4: Due to independence, we know that the probability of a success, \(p = 0.46\), will remain the same for each selection.
Since we are working with very large numbers, we want to see if we can use the normal approximation to the binomial. Verify that is possible.
\(np = 500(0.46) = 230\) and \(nq = 500(0.54) = 270\)
Since \(np > 5\) and \(nq > 5\), we can use the normal approximation to the binomial.
Remember, the formulas for the mean and standard deviation of a binomial distribution are \(\mu = np\) and \(\sigma = \sqrt{npq}\).
The mean is 230 and the standard deviation is 11.14451.
The random variable for the binomial distribution is \(x\). We will use \(Y\) for the normal distribution: \(Y \sim N(230, 11.14451)\).
We want to include only 304 so our interval needs to stretch to the left and the right to create two dimensional shape we can find the area for:
\(P(x = 304) \approx P(303.5 < Y < 304.5) = \text{normalcdf}(303.5,304.5,230,11.14451) = 9.73091087E-12 = 0.00000000000973091...\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first for both endpoints. Use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{303.5 - 230}{11.14451} = 6.59518...\)
\(z = \dfrac{304.5 - 230}{11.14451} = 6.68491...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x = 304) \approx P(303.5 < Y < 304.5) = \text{normalcdf}(6.59518,6.68491) = 9.7309778E-12 = 0.00000000000973097...\)
There is about a 0.0000000010% chance that exactly 304 people favor Dawn Morgan.
Part b:
You include 250 and shade to the right, so our interval needs to stretch to the left:
\(P(x \geq 250) \approx P(Y > 249.5) = \text{normalcdf}(249.5,10^{99},230,11.14451) = 0.0401\).
or, if using the tables or are requested to use standard normal distribution, then need to find \(z\) first. Use \(z = \dfrac{x - \mu}{\sigma}\), where \(x\) is the corrected data value, \(Y\).
\(z = \dfrac{249.5 - 230}{11.14451} = 1.74974...\) If using tables, round to two decimal places, otherwise use plenty of decimal places.
\(P(x \geq 250) \approx P(Y > 249.5) = \text{normalcdf}(1.74974,10^{99}) = 0.0401\)
There is about an 4% chance that at least 250 people favor favor Dawn Morgan.
References
- Data from the Wall Street Journal.
- “National Health and Nutrition Examination Survey.” Center for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed May 17, 2013).
Glossary
- Binomial Distribution
- The probability for discrete random variables can be calculated using a binomial distribution when
- The number of trials, \(n\), is fixed.
- Each trial outcome can be classified as a success or failure.
- The trials independent.
- The probability of a success, \(p\), is the same for each trial.
- Mean
- a number that measures the central tendency; a common name for mean is "average." The term "mean" is a shortened form of "arithmetic mean." By definition, the mean for a sample (denoted by \(\bar{x}\)) is \(\bar{x} = \dfrac{\text{Sum of all values in the sample}}{\text{Number of values in the sample}}\), and the mean for a population (denoted by \(\mu\)) is \(\mu = \dfrac{\text{Sum of all values in the population}}{\text{Number of values in the population}}\).
- Normal Distribution
- a continuous random variable (RV) with pdf \(f(x) = \dfrac{1}{\sigma \sqrt{2\pi}}e^{\dfrac{(x - \mu)^{2}}{2\sigma^{2}}}\), where \(\mu\) is the mean of the distribution and \(\sigma\) is the standard deviation.; notation: \(X \sim N(\mu, \sigma)\). If \(\mu = 0\) and \(\sigma = 1\), the RV is called the standard normal distribution.