6-3 Binomial Distributions Day 2

6-3 Binomial Mean and SD, Independence (the 10% Condition), Normal Approximations of Binomial (Large Counts)

If a count X has the binomial distribution with number of trials n and probability of success p, the mean and standard deviation of X are

 $\mu_{X\ }=np\ \ \ \ \ \ \sigma_{X\ }=\ \sqrt{np\left(1-p\right)}$  

Mean and Standard Deviation of Binomial Random Variables

1.    The makers of a diet cola claim that its taste is indistinguishable from the full-calorie version of the same cola. To investigate, Courtney prepared small samples of each type of soda in identical cups. Then she had volunteers taste each cola in a random order and try to identify which was the diet cola and which was the regular cola. Overall, 23 of the 30 subjects made the correct identification.

If we assume that the volunteers really couldn’t tell the difference, then each one was guessing with a 1/2 chance of being correct. Let X = the number of volunteers who correctly identify the colas. 

Open Ended

(a) Explain why X is a binomial random variable.

Multiple Choice

(c) Does this give convincing evidence that the volunteers can taste the difference between the diet and regular colas?

• Yes, there is such a small chance of guessing 23 out of 30 correctly if they really couldn't tell the difference, so this result is convincing evidence.

• No, there is such a small chance of guessing 23 out of 30 correctly if they really couldn't tell the difference, so this result does not give convincing evidence.

Open Ended

Check Your Understanding

To introduce her class to binomial distributions, Mrs. T-C gives a 10-item, multiple-choice quiz. The catch is, students must simply guess an answer (A through E) for each question. Mrs. T-C uses her computer’s random number generator to produce the answer key, so that each possible answer has an equal chance to be chosen. David is one of the students in this class. Let X = the number of David’s correct guesses. 

(a) Find the mean and interpret this value in context.

(b) Find the standard deviation and interpret this value in context.

(c) Find probability that the number of David's correct guesses is more than 4.53 (2 SD about the mean).

Open Ended

Independence in Sampling

2. You have a drawer with 8 batteries - 6 of which are good. What is the probability that you choose 4 batteries that all work?

a) use the binomial probability formula

b) use conditional probabiity without replacement.

Are the results the same? Why or why not?

Open Ended

Now, suppose the drawer contains 100 AAA batteries where 75 are good. What is the probability that you randomly select 4 batteries and that all 4 of them will work? The binomial probability won't change, but what about the conditional probability? What do you notice?

When taking an SRS of size n from a population of size N, you can use a binomial distribution to model the count of successes in the sample as long as                            

 $n\le10\%N$  

This is known as the 10% condition for independence in sampling

Normal Approximations for Binomial Distributions

As n gets larger, something interesting happens to the shape of a binomial distribution. The figures below show histograms of binomial distributions for different values of n and p.

What do you notice as n gets larger?

Open Ended

In a survey of 506 teenagers aged 14 to 18, subjects were asked a variety of questions about personal finance One question asked teens if they had a debit card. Suppose that exactly 10% of teens aged 14 to 18 have debit cards.

(c) Use a Normal distribution to estimate the probability that 40 or fewer teens in the sample have debit cards AND use the binomial probability to estimate the probability that 40 or fewer teens have debit cards.

Answer

• mean = .1(506) = 50.6

• SD =

   $\sqrt{506\cdot0.1\cdot0.9}$  = 6.748

• P(X<=40) = binomcdf(n=506, p=.1, x = 40) = 0.0175 = 0.064

• P(X<=40) = normalcdf(lower -9999, upper 40, mean 50.6, SD 6.748) =0.058