MA 1125 Lecture 18 - Normal Approximations to Binomial Distributions. Objectives: Compute probabilities for a binomial as a normal distribution.

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1 MA 25 Lecture 8 - Normal Approximations to Binomial Distributions Friday, October 3, 207 Objectives: Compute probabilities for a binomial as a normal distribution.. Normal Approximations to the Binomial Distribution As the number of trials in a binomial distribution goes to infinity, the shape of the corresponding histograms approach the shape of the normal curve. In other words, binomial distributions are approximately normally distributed. We can compute probabilities for a biniomial distribution exactly, but as you may have noticed, the denominators for the probabilities can be quite large for even five or six trials. Very quickly, the computations become intractable. Luckily, as the direct computations get worse, the approximation to the normal gets better, and we can get very good approximations using the normal table. That s what we ll do today. 2. Comparing a binomial distribution to a normal distribution Let s consider the binomial experiment of tossing a coin ten times and counting up heads. The probabilities for successes and failures are p = q = 2, so each outcome in this experiment is () The appropriate row of Pascal s triangle is 2 = 2 = 0 (2)

2 2 Knowing the numbers of ways and the probabilities for each outcome, we can construct the probability distribution. Number of Heads Probability (3) We know how to find the mean and standard deviation for a binomial distribution. Since n = 0, p = 2, and q = 2, we have (4) µ = np = 20 2 = 0, and (5) σ = npq = 20 2 = 5. I claim that this distribution is approximately normally distributed. In general terms, this means that you re much more likely to get around 5 heads than you are around 0 heads or 0 heads. Specifically, how do the probabilities work out? The table states that P( head) = 0. I can compute the z-score for head as follows. (6) x µ σ = 5.58 = 2.53 Since the standard normal is a continuous distribution, if we try to find a probability of something like P( 2.53), we get zero. That s definitely not equal to 0. The problem is that the binomial is an example of a discrete probability distribution that has non-zero probabilities for single values of x. How do we deal with that? If you think back to when we were working with frequency distributions and histograms, we spread the classes out to the class boundaries. After doing this, for example, the histograms didn t have gaps between them. We ll do the same thing here.

3 MA 25 Lecture 8 - Normal Approximations to Binomial Distributions 3 We ll turn this discrete probability distribution into a continuous one by making the following change. (7) P(x = ) becomes P(0.5 x.5). Now we re back to computations like the ones when we first saw normally distributed populations. OK. We have to convert x = 0.5 and x =.5 to z-scores. (8) = = I ll let you draw the picture. Both of these z-scores are to the left, so (9) P(0.5 x.5) = P( 2.85 z 2.22) = = 0.00 How does this compare to the exact probability? Let s see. (0) P( head) = 0 = We re off by = About a tenth of a percent. I ll do one more, and then you can do a couple. How about P(x = 7) = P(6.5 x 7.5)? First we convert to z-scores. () = = That looks a little weird, but whatever. (2) P(6.5 x 7.5) = P(0.95 z.58) = = 0.40 The exact probability is (3) P(x = 7) = 20 = 0.72 Our approximations aren t super close, but they re not too bad. Keep in mind that with only ten tosses, the histogram is going to be pretty jagged, and so the normal curve isn t going to fit perfectly. The approximations get much better with larger n s.

4 4 3. Quiz 8 My brother Andy is an 80% freethrow shooter, and he s going to try 20 freethrows. This makes µ = np = = 6, and σ = npq = =.789. Use the normal approximation to find the following probabilities.. P(Andy makes 6). 2. P(Andy makes less than or equal to 4). 4. Homework 8 For problems -7, suppose we have a twenty-five question multiple choice test with questions that have five answer choices. We ll assume that someone takes the test by guessing.. Find µ. 2. Find σ. 3. Find P(x = 5) using the normal approximation. Round to four decimal places. (Use P(4.5 x 5.5).) 4. Find P(x = 0) using the normal approximation. Round to four decimal places. (Use P(9.5 x 0.5).) 5. To find the probability of getting less than five right using the normal approximation (this is 0,, 2, 3, or 4 right), you would compute... (a) P(x 4.5) (b) P(x 4) (c) P(x < 4) (d) P(x 5.5) (e) none of these 6. Find the probability from Problem 5. Round to four decimal places. 7. To find the probability of getting at least ten right using the normal approximation (this means 0,, 2, or more), you would compute... (a) P(x 9.5) (b) P(x 0) (c) P(x > 0) (d) P(x 0.5) (e) none of these 8. Find the probability from Problem 7. Round to four decimal places. (continued on next page)

5 MA 25 Lecture 8 - Normal Approximations to Binomial Distributions 5 9. To find the probability of getting more than nine right using the normal approximation, you would compute... (a) P(x 8.5) (b) P(x 9) (c) P(x > 9) (d) P(x 9.5) (e) none of these 0. Find the probability from Problem 9. Round to four decimal places. Answers: ) µ = 5. 2) σ = 2. 3) P( 0.25 z 0.25) = ) P(2.25 z 2.75) = ) a) P(x 4.5). 6) P(z 0.25) = ) a) P(9.5 x). 8) P(2.25 z) = ) d) P(9.5 x). 0) Same as Problem 8.

MA 1125 Lecture 12 - Mean and Standard Deviation for the Binomial Distribution. Objectives: Mean and standard deviation for the binomial distribution.

MA 1125 Lecture 12 - Mean and Standard Deviation for the Binomial Distribution. Objectives: Mean and standard deviation for the binomial distribution. MA 5 Lecture - Mean and Standard Deviation for the Binomial Distribution Friday, September 9, 07 Objectives: Mean and standard deviation for the binomial distribution.. Mean and Standard Deviation of the