sta220 - statistics mr. smith room 310 class #12

Sta220 - Statistics Mr. SmithRoom 310Class #12

Section 4.3

4.3- The Binomial Random Variable

Many experiments result in responses for which there exist two possible alternatives, such as Yes-No, Pass-Fail, Defective-Nondefective or Male-Female.

These experiments are equivalent to tossing a coin a fixed number of times and observing the number of times that one of the two possible outcomes occurs. Random variables that possess these characteristics are called Binomial random variables.

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Definition

Example 4.10

The Heart Association claims that only 10% of U.S. adults over 30 years of age meet the minimum requirements established by President’s Council of Fitness, Sports and Nutrition. Suppose four adults are randomly selected and each is given the fitness test.

a) Find the probability that none of the four adults passes the test.

b) Find the probability that three of the four adults pass the test.

c) Let x represent the number of four adults who pass the fitness test. Explain why x is a binomial random variable.

d) Use the answers to parts a and b to derive a formula for p(x), the probability distribution of the binomial random variable x.*

a)

1) First step is define the experiment.

We are observing the fitness test results of each of the four adults: Pass(S) or Fail(F)

2) Next we need to list the sample points associated with the experiment.

Example: FSSS represents the sample point denoting that adult 1 fails, while adult 2, 3, and 4 pass the test.

There are 16 sample points.


Table 4.2

3) Now we have to assign probabilities to the sample points.

Since sample points can be viewed as the intersection of four adults’ test results, and assuming that the results are independent, we can obtain the probability of each sample point by the multiplicative rule.

Let find the probability that all four adults passed the fitness test.

P(SSSS) =

=

=

So, P(SSSS) = .0001

Another sample point.

P(FSSS)

So, P(FSSS) = .0009

4) We wanted to obtain the sample point that none of the four adults’ passed.

P(FFFF)

So , P(FFFF) = .6561

b) Probability that three of the four adults pass the test. Looking at the sample points, we have four sample points: FSSS, SFSS, SSFS, and SSSF.

P(3 of 4 adults pass)= P(FSSS) + P(SFSS) + P(SSFS) + P(SSSF)

= .0036

c.I. We can characterize this experiment as a consisting

on four identical trials: the four test results II. There are two possible outcomes to each trail, S of

F. III. The probability of passing , p = .1, is the same for

each trial. IV. Assuming that each adult’s test result is

independent of all others, so that the four trials are independent.

d. *Since time is limited in this class, we will not derive a formula for p(x) in class. However, I have attached it to the end of the PowerPoint.


Procedure

Example 4.11

Refer to Example 4.10. use the formula for a binomial random variable to find the probability distribution of x, where x is the number of adults who pass the fitness test. Graph the distribution.

For this application, we have n = 4 trails. Since a success S is defined as an adult who passes the test, p = P(S) = .1 and q = 1- p = .9

So we have the following:n = 4p = .1q = .9

p(0) =

p(1) =

p(2) =

p(3) =

p(4) =


Table 4.3


Figure 4.8 Probability distribution for physical fitness example: graphical form

Example 4.12

Refer to Examples 4.10 and 4.11 . Calculate and respectively, of the number of the four adults who pass the test. Interpret the results.

We first need to find the mean of a discrete probability distribution

We must refer back to the table

Thus, in the long run, the average number of adults (out of four) who pass the test is only .4.

The variance: =

Since the distribution is skewed right, we should apply Chevyshev’s rule to describe where most of the x-values fall. According to the rule, at least 75% of the x values will fall into the interval which is Since x cannot be negative, we expect the number of adults out of four who pass the fitness test to be less than 1.6.

If we look at the and in relation to binomial probability distribution ( that using n, p, and q), you need not use the expectation summation rule to calculate and for a binomial random variable.


Definition

Using Binomial Tables

Calculating binomial probabilities becomes tedious when n is large. For some values of n and p, the binomial probabilities have been tabulated to Table II of Appendix A. The entries in the table represent cumulative binomial probabilities.

Example (put this on you table)Let p = .10 and x = 2.


Figure 4.9 Binomial probability distribution for n=10 and p=.10, with highlighted( 2)P x

Example 4.13

Suppose a poll of 20 voters taken in a large city. The purpose is to determine x, the number who favor a certain candidate for mayor. Suppose that 60% of all the city’s voters favor the candidate.

a) Find the mean and the standard deviation of x. b) Use Table II of Appendix A to find the probability

that x less than equal 10c) Use Table II to find the probability x > 12

[P(x >12)]

d) Use Table II to find the probability that x = 11[P(x =11)]

e) Graph the probability distribution of x, and locate the interval on the graph.

First, n = 20p = .60q = .40

a)

The= 12 and= 2.19

b)Find the probability that k = 10p = .60n = 20

c)Find the probability that

d)Find the probability that

E. The probability distribution for x,


Figure 4.10 The binomial probability distribution for x in Example 4.13; n=20 and p=.6

The probability that x falls into the interval P(x = 8, 9, 10,…, 16) = =.984-.021=.963

Note that his probability is very close to the .95 given by the empirical rule. Thus, we expect the number of voters in the same of 20 who favor mayoral candidate to between 8 and 16.

sta220 - statistics mr. smith room 310 class #12

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