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Binomial Distribution

Consider the following exercise: Suppose that a student is taking a multiple-choice exam in which each question has four choices.   Assuming that she has no knowledge of the correct to any of the questions, she has decided on a strategy in which she will place four balls (marked A, B, C, and D) into a box.   She randomly selects one ball for each question and replaces the ball in the box.   The marking on the ball will determine her answer to the question.

Using this method (not recommended for Stat 216!), the probability of getting a correct answer would be 1/4 = .25 for each question.   This is a binomial situation with n=5 and p=.25.

Note that all of the properties of the binomial distribution are satisfied:

  1. The student is sampling with replacement.
  2. There are only two outcomes: A correct answer (Success) or an incorrect answer (Failure).   We are not concerned with which wrong answer it might be among the three possibilities.
  3. p=.25 is constant from draw to draw.
  4. The draws are independent- each one does not effect the next.

This page will contain the following solutions:

First, let's solve this problem by hand, using the binomial formula (page 52-53).

(a)(1) If there are five multiple-choice questions on the exam, what is the probability that she will get five questions correct?

$P(X=5) = {5 \choose 5}(.25)^5(.75)^0 = .00098$

(a)(2) What is the probability that she will get at least four questions correct?

$P(X \geq 4) = P(X=4) + P(X=5) =
{5 \choose 4}(.25)^4(.75)^1 + .00098 =$
.01465 + .00098 = .01563

(a)(3) What is the probability that she will get no questions correct?

$P(X=0) = {5 \choose 0}(.25)^0(.75)^5 = .23730$

(a)(4) What is the probability that she will get no more than two questions correct?

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) =$
$.23730 + {5 \choose 1}(.25)^1(.75)^4 + {5 \choose 2}(.25)^2(.75)^3 =$
.23730 + .39551 + .26367 = .89648

(b) The assumptions of the binomial that were previously mentioned.

(c) What are the average and the standard deviation of the number of questions that she will get correct in (a)?

$\mu = np = 5(.25) = 1.25$
$\sigma = \sqrt{np(1-p)} = \sqrt{5(.25)(.75)} = .96825$

(d) This cannot be done very easily by hand...

Now, let's go back and solve part (a) using the TI-83.

Assuming the lists L1, L2, and L3 are empty, put the sample space {0, 1, 2, 3, 4, 5} in L1, using the seq function under LIST OPS.   Simply insert "A" for the expression and variable, since we just want the sequence from 0 to 5.   Choose binompdf from the DISTR menu, giving (5, .25) for the (n, p) arguments.   Store this in L2.   Now choose binomcdf and supply the same arguments.   Store this in L3. Now go to STAT Edit to look at the lists.

TI-83 screen TI-83 screen

The pdf column, L2, contains the probabilities P(X=x), while the cdf column, L3, contains $P(X \leq x)$.

(a)(1) P(X=5) can be found in L2, the last entry: 9.8E-4 = .00098.

(a)(2) $P(X \geq 4)$ can be found either by adding P(X=4) + P(X=5) from L2, which is .01465 + .00098 = .01563, or by taking $P(X \geq 4) = 1 - P(X \leq 3) = 1 - .98437 = .01563$ from L3.

(a)(3) P(X=0) = .2373, from L2.

(a)(4) $P(X \leq 2) = .86948$, from L3.

Part (d) can be done easily with the calculator.

(d) Now we have n=50, p=.25, and we want $P(X \geq 30)$. Here, let's just get the cdf list, since adding up the pdf column from 30 to 50 would be almost as tedious as doing the problem by hand!
Put the sample space {0, 1, ..., 50} in L4. Put the cdf list in L5. Again, go to STAT Edit to look at the lists. Scroll down so that the "30" entry is in view.

TI-83 screen TI-83 screen

$P(X \geq 30) = 1 - P(X \leq 29) = 1 - 1 = 0$. So the probability is zero! Not much chance of passing the exam with this strategy.   Actually, this probability isn't exactly zero, but it is very close to zero.   The TI-83 ran out of significant digits to display.   This will be illustrated when we use Excel to do this problem next.

Lastly, let's see how parts (a) & (d) can be done in Excel.

Visit www.stat.wmich.edu/s216/ExcelTools/ to learn how to use the Excel software.

Choose Probability Distributions from the menu, then pick Binomial.

For part (a), enter 5 for the Sample Size.   The Probability of Success should be 0.25, then enter 0 and 5 for Outcomes From: and To:, respectively.   Enter an Output Title if you wish, and put a checkmark in the box next to Cumulative Probabilities.

ExcelTools screen

On the Binomial sheet, notice that the mean and standard deviation are given.   These match the results we obtained by hand.   Column B contains the possible outcomes, C lists the pdf or P(X=x) probabilities (L2 when we were using the TI-83), D lists the cdf or $P(X \leq x)$ probabilities (L3 from before), and columns E, F, and G contain the probabilities from other ways to list them.   Slightly overkill, but we can make use of column G in this problem.

(a)(1) P(X=5) = .000977, from column C.

(a)(2) $P(X \geq 4) = .015625$, from column G.

(a)(3) P(X=0) = .237305, from column C.

(a)(4) $P(X \leq 2) = .896484$, from column D.

(d) Re-do the Binomial Probability Distribution dialog box with n=50, p=.25, Outcomes from 0 to 50, and be sure the Cumulative Probabilities box is checked.

Excel screen

Scrolling down in order to view the probabilities when x=30, we see:

Excel screen

Now, $P(X \geq 30) = 1.64E-07 = .000000164$ can be found in column G.



 
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Updated: 20 August 2003