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Lesson 8 - 1

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Lesson 8 - 1

Discrete Distribution

Binomial

- Describe the conditions that need to be present to have a binomial setting.
- Define a binomial distribution.
- Explain when it might be all right to assume a binomial setting even though the independence condition is not satisfied.
- Explain what is meant by the sampling distribution of a count.
- State the mathematical expression that gives the value of a binomial coefficient. Explain how to find the value of that expression.
- State the mathematical expression used to calculate the value of binomial probability.

- Evaluate a binomial probability by using the mathematical formula for P(X = k).
- Explain the difference between binompdf(n, p, X) and binomcdf(n, p, X).
- Use your calculator to help evaluate a binomial probability.
- If X is B(n, p), find µx and x (that is, calculate the mean and variance of a binomial distribution).
- Use a Normal approximation for a binomial distribution to solve questions involving binomial probability

- Binomial Setting – random variable meets binomial conditions
- Trial – each repetition of an experiment
- Success – one assigned result of a binomial experiment
- Failure – the other result of a binomial experiment
- PDF – probability distribution function; assigns a probability to each value of X
- CDF – cumulative (probability) distribution function; assigns the sum of probabilities less than or equal to X
- Binomial Coefficient – combination of k success in n trials
- Factorial – n! is n (n-1) (n-2) … 2 1

A binomial experiment with n independent trials and probability of success p has

Mean μx = np

Standard Deviation σx = √np(1-p)

Find the mean and standard deviation of a binomial distribution with n = 10 and p = 0.1

Mean:

μx = np = 10(0.1) = 1

Standard Deviation:

σx = √np(1-p)

= 10(0.1)(0.9)

= 0.9

= 0.9487

As binominal’s number of trials increases the formula for a binomial becomes unworkable (a situation alleviated with statistical software). So statisticians developed a procedure to use a continuous distribution, the normal, to estimate a discrete distribution.

This procedure is used later with proportions.

Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if shopping was often frustrating and time-consuming. Assume that 60% of all US adults would agree if asked the same question, what is the probability that 1520 or more of the sample would agree?

P(X ≥ 1520) = 1 – P(X ≤ 1519) = 1 – 0.7869 = 0.2131

using binomcdf(2500, 0.6, 1519)

E(X) = 2500 (0.6) = 1500

V(X) = 2500(0.6)(0.4) = 600 so σ = 600 = 24.49

P(X ≥ 1520) = 0.2070 using normcdf(1520, E99, 1500, 24.49)

a difference of 0.0061 or less than 0.6%

Histogram showing normal apx to binomial

- To simulate a binomial event, we must know:
- how random variable X and “success” is defined
- probability of success
- number of trials

- Calculator has a randbin(1,p,n) function that will generate results with “1”s as successes
- We can store the results in lists and do statistics on the lists as usual

Each entry in a table of random digits like Table B in our book has a probability of 0.1 of being a zero.

- Find probability of find exactly 4 zeros in a line 40 digits long.
- What is the probability that a group of five digits from the table will contain at least 1 zero?

P(X = 4) = 0.206 using binompdf(40, 0.1, 4)

P(X ≥ 1) = 1 – P(X=0)

= 1 – (0.9)5 = 0.4095 (before binomials)

= 1 – 0.59049 = 0.4095 using binompdf(40, 0.1, 4)

A university claims that 80% of its basketball players get their degree. An investigation examines the fates of a random sample of 20 players who entered the program over a period of several years. Of these players, 10 graduated and 10 are no longer in school. If the university's claim is true, what is the probability that exactly 10 out of 20 graduate? Can you conclude anything about the university's claim?

If the university's claim is true, what is the probability that exactly 10 out of 20 graduate?

Can you conclude anything about the university's claim?

P(Y) = 0.8

P(x) = nCx px(1-p)n-x

P(x=10 [p=0.8, n=20]) = 20C10 (0.8)10(1- 0.8)20-10

= 20C10 (0.8)10(0.2)10 = 0.00203

It is either a false claim or we got a very unusual sample!

10 – np 10 – 18 - 8 -8

Z = ----------- = ----------------- = -------- = --------------- = -4.47

np(1-p) 20(.8)(.2) 3.2 1.7889

- Summary
- Binomial experiments have 4 specific criteria that must be met
- Means and Variance for a Binomial
- E(X) = np and V(X) = np(1-p)

- Normal distribution (continuous) can approximate a Binomial (discrete)
- Calculator has a random binomial generator

- Homework
- pg