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Adapted from prepublication materials for Statistics with the TI-83, Meridian Creative Group, 1997 NORMAL APPROXIMATION FOR BINOMIAL DISTRIBUTION Example: Suppose someone tosses 50 coins and counts the resulting number of heads. Use the list features of the TI-83 to create a probability distribution for the binomial random variable X, representing the number of heads. Then display a probability histogram for this distribution and compute the mean and standard deviation. Solution: Clear L1 and L2 for storing the values of X and the associated probabilities. The number of heads, and thus the values of X, can range from 0 to 50 inclusive, so you need to enter the integers 0, 1, 2, . . . 50 in L1. To define L1 using seq, execute the following home screen instruction: seq( X, X, 0, 50, 1) STO> L1. Note: seq( is option 5 under the LIST OPS menu. We will assume that the coins are fair, so the probability of a head p is .5. The values of the random variable X are the entries of L1. To fill L2 with the associated binomial probabilities, execute the home screen instruction: binompdf( 50, .5, L1) STO> L2 When the computations are complete, examine L2 to observe very small probabilities associated with small and large values of X. Note that the probability of 25 heads, the most likely value of X, is approximately .1123. Display sum(L2) to confirm that the calculated probabilities add to exactly 1. To display the probability histogram for this distribution, define a stat plot to be a histogram with L1 as the Xlist and L2 as the frequency list. A good viewing window for this histogram is obtained by setting Xmin=12.5, Xmax=37.5, Xscl=1, Ymin= , Ymax=.12. This histogram has rectangles centered over the most likely values of X. Now use 1-Var Stats L1, L2 to display the mean and standard deviation associated with this probability distribution. Note that the mean value of 25 is the product = 50∗.5. The standard deviation value of 3.5355... is . Example: Show graphically how a normal curve can be used to approximate the probability histogram for the number of heads when 50 coins are tossed. Then compare normal approximations for the probability of exactly 25 heads and the probability of between 20 and 30 heads inclusive with the exact probabilities. Solution: The approximating normal distribution has the same mean and standard deviation as the binomial distribution. Press Y= to display the Y= editor. Move to the function of your choice and press 2nd [DISTR] 1 to paste normalpdf( next to Yn=. Complete the function definition to display normalpdf( X, 25, ). Then press GRAPH to overlay the normal curve. Notice that edges of rectangles extend beyond the smooth curve. Experiment with the WINDOW settings to see more clearly how close the normal curve is to the probability histogram. You can now determine the exact probability of 25 heads by tracing along the histogram until the cursor is on the rectangle extending from 24.5 to 25.5. According to the information displayed on the graph, the probability of 25 heads is .112275. To obtain the normal approximation for this probability, you need to evaluate the area under the normal curve from 24.5 to 25.5. This can be accomplished with normalcdf( 24.5, 25.5, 25, ). The normal approximation is . 11246.... Gloria Barrett, NC School of Science and Mathematics June, 1999 Adapted from prepublication materials for Statistics with the TI-83, Meridian Creative Group, 1997 To obtain the exact probability of obtaining 20 to 30 heads inclusive when 50 coins are tossed, execute the following instruction: binomcdf( 50, .5, 30) - binomcdf( 50, .5, 19). Then execute normalcdf( 19.5, 30.5, 25, ) to observe the normal approximation. Exercises 1. Suppose someone tosses 200 coins and counts the resulting number of heads. Compare the normal approximations for the probability of 100 heads and for the probability of between 90 and 110 heads inclusive with the exact probabilities. How do the normal approximations for 200 coins compare to the approximations for 50 coins? 2. Suppose a teacher gives a 100 question multiple choice test with four answer choices for each question. (a) Use the binomcdf option of the DISTR menu to determine the probability that a student could get 60 or more of the questions correct just by random guessing. (b) Use the normal approximation to the binomial distribution to estimate this probability. Gloria Barrett, NC School of Science and Mathematics June, 1999