Download Suppose someone tosses 50 coins and counts the

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