Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 1 of 21 Chapter 7 Section 5 The Normal Approximation to the Binomial Probability Distribution

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 2 of 21 Chapter 7 – Section 5 ●Learning objectives  Approximate binomial probabilities using the normal distribution 1

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 3 of 21 Chapter 7 – Section 5 ●Recall from section 6.2 that a binomial experiment is one where  An experiment is performed n independent times  Each time, or trial, has two possible outcomes – success or failure  The outcome of success – p – is the same for each trial

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 4 of 21 Chapter 7 – Section 5 ●The binomial probabilities are the probabilities of observing x successes in n trials ●If X has a binomial distribution, then P(X = x) = n C x p x (1 – p) n-x ●The binomial probabilities are the probabilities of observing x successes in n trials ●If X has a binomial distribution, then P(X = x) = n C x p x (1 – p) n-x ●We would like to calculate these probabilities, for example when both n and x are large numbers

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 5 of 21 Chapter 7 – Section 5 ●If both n and x are large numbers, then calculating the number of combinations n C x in P(X = x) = n C x p x (1 – p) n-x is time-consuming since we need to compute terms such as n! and x! ●If both n and x are large numbers, then calculating the number of combinations n C x in P(X = x) = n C x p x (1 – p) n-x is time-consuming since we need to compute terms such as n! and x! ●Calculations for probabilities such as P(X ≥ x) are even more time-consuming

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 6 of 21 Chapter 7 – Section 5 ●For example, if n = 1000, then P(X ≥ 700) = P(X = 700) + P(X = 701) + … + P(X = 1000) which is the sum of 301 terms ●For example, if n = 1000, then P(X ≥ 700) = P(X = 700) + P(X = 701) + … + P(X = 1000) which is the sum of 301 terms ●Each P(X = x) by itself is an involved calculation, and we need to do this 301 times, so this becomes an extremely long and tedious sum to compute!

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 7 of 21 Chapter 7 – Section 5 ●However, we can simplify this calculation ●As the number of trials, n, increases, then the distribution of binomial random variables becomes more and more bell shaped ●However, we can simplify this calculation ●As the number of trials, n, increases, then the distribution of binomial random variables becomes more and more bell shaped ●As a rule of thumb, if np(1-p) ≥ 10, then the distribution will be reasonably bell shaped ●However, we can simplify this calculation ●As the number of trials, n, increases, then the distribution of binomial random variables becomes more and more bell shaped ●As a rule of thumb, if np(1-p) ≥ 10, then the distribution will be reasonably bell shaped

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 8 of 21 Chapter 7 – Section 5 ●To show that the approximation is pretty good, the following chart shows the binomial with n = 40, p = 0.5, and its corresponding normal

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 9 of 21 Chapter 7 – Section 5 ●Even if the original binomial trial isn’t symmetric, such as for p = 0.3, the normal is still a good approximation (n = 60 in the chart below)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 10 of 21 Chapter 7 – Section 5 ●Because the binomial distribution is approximately normal when np(1-p) ≥ 10, the normal distribution can be used to approximate a binomial distribution ●Remember that a binomial distribution has  Mean = n p  Standard deviation = √ n p (1 – p)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 11 of 21 Chapter 7 – Section 5 ●It would make sense to use an approximation that has the same mean and standard deviation ●Thus to approximate a binomial random variable X (with parameters n and p), we use a normal random variable Y with parameters  μ = n p  σ = √ n p (1 – p)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 12 of 21 Chapter 7 – Section 5 ●This matches the mean and the standard deviation between X and Y ●We could then approximate the probabilities of X with probabilities of Y P(X ≤ a) = P(Y ≤ a) ●This matches the mean and the standard deviation between X and Y ●We could then approximate the probabilities of X with probabilities of Y P(X ≤ a) = P(Y ≤ a) ●However, there is a slightly better method ●This better method is called the correction for continuity

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 13 of 21 Chapter 7 – Section 5 ●First we will look at an example ●Consider a binomial variable X with  n = 60 trials and  p = 0.3 probability of success ●First we will look at an example ●Consider a binomial variable X with  n = 60 trials and  p = 0.3 probability of success ●This variable has  Mean = n p = 18  Standard deviation = √ n p (1 – p) = 3.55 ●First we will look at an example ●Consider a binomial variable X with  n = 60 trials and  p = 0.3 probability of success ●This variable has  Mean = n p = 18  Standard deviation = √ n p (1 – p) = 3.55 ●Thus we use a normal random variable Y with mean μ = 18 and standard deviation σ = 3.55

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 14 of 21 Chapter 7 – Section 5 ●The graph shows the situation for P(X = 17)  The pink curve shows the binomial probabilities for X  The blue curve shows the normal approximation for Y

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 15 of 21 Chapter 7 – Section 5 ●A better approximation for P(X = 17)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 16 of 21 Chapter 7 – Section 5 ●A better approximation for P(X = 17) would be P(16 ½ ≤ Y ≤ 17 ½)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 17 of 21 Chapter 7 – Section 5 ●Thus a better approximation for P(X = a) is P(a – ½ ≤ Y ≤ a + ½) ●Thus a better approximation for P(X = a) is P(a – ½ ≤ Y ≤ a + ½) and a better approximation for P(X ≤ a) is P(Y ≤ a + ½) ●This is the correction for continuity

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 18 of 21 Chapter 7 – Section 5 ●Now we look back at the example ●We want to approximate P(X ≤ 17) for a binomial variable X with n = 60 trials and p = 0.3 probability of success ●Now we look back at the example ●We want to approximate P(X ≤ 17) for a binomial variable X with n = 60 trials and p = 0.3 probability of success ●X has mean 18 and standard deviation 3.55 ●We use a normal random variable Y with mean 18 and standard deviation 3.55

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 19 of 21 Chapter 7 – Section 5 P(X ≤ 17) ≈ P(Y ≤ 17.5) ●Since Y has mean μ = 18 and standard deviation σ = 3.55, the Z-score of 17.5 is and P(Z ≤ – 0.14) = ●Thus the normal approximation to the binomial is P(X ≤ 17) ≈

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 20 of 21 Chapter 7 – Section 5 ●From the normal, an approximation is P(X ≤ 17) ≈ ●The actual value (computed from Excel) is P(X ≤ 17) = ●These are quite close ●However, technology can compute the Binomial probabilities directly, so the normal approximation is not needed as much these days

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 21 of 21 Summary: Chapter 7 – Section 5 ●The binomial distribution is approximately bell shaped for large enough values of np(1 – p) ●The normal distribution, with the same mean and standard deviation, can be used to approximate this binomial distribution ●With technology, however, this approximation is not as needed as it used to be

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 22 of 21 Example: Chapter 7 – Section 5 ●in the United States never use the Internet. (Source: gallup.com) Suppose 200 people are randomly selected. Use the normal approximation to the binomial to  a. approximate the probability that exactly 54 of them will never use the Internet. (0.0635)  b. approximate the probability that more than 45 of them will never use the Internet. (0.9121)  c. approximate the probability that more than 60 of them will never use the Internet. (0.1503)  d. approximate the probability that the number of them who will never use the Internet is between 46 and 60, inclusive. (0.7618)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 23 of 21 Example: Chapter 7 – Section 5 ●Darren did not prepare for a multiple-choice test in his biochemistry class. He is forced to randomly guess on each question. Suppose the test has 60 questions with 4 possible answers for each question. Use the normal approximation to the binomial to  a. approximate the probability that he will get exactly 12 questions right. (0.0797)  b. approximate the probability that he will get more than 15 questions right. (0.4407)  c. approximate the probability that he will get more than 20 questions right. (0.0505)  d. approximate the probability that he will get more than 25 questions right. (0.0009)  e. approximate the probability that the number of questions he will get is between 12 and 17, inclusive. (0.6236)

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 7 Section 5 – Slide 24 of 21 Example: Chapter 7 – Section 5 ●Most nurseries guarantee their plants for one year. If the plant dies within the first year of the purchase date, it can be returned for credit. Experience has shown that about 5% of all plants purchased from Evergreen Nursery will die in their first year. Evergreen Nursery sold 250 walnut trees this season. Use the normal approximation to the binomial to  a. approximate the probability that exactly 12 walnut trees will be returned for credit. (0.1142)  b. approximate the probability that more than 15 walnut trees will be returned for credit. (0.1920)  c. approximate the probability that more than 20 walnut trees will be returned for credit. (0.0101)  d. approximate the probability that more than 25 walnut trees will be returned for credit. (0.0001)