P. 403 – 404 #71 – 73, 75 – 78, 80, 82, 84 #72B: Binary? Yes – Success is a person is...

Day 1 Homeworkp. 403 – 404 #71 – 73, 75 – 78, 80, 82, 84

#72 B: Binary? Yes – Success is a person is left-handed.

I: Independent? Yes, since students are selected randomly, their handedness is independent.

N: fixed number of observations? Yes; n = 15

S: Same probability of success for each student p = 0.10

Yes, this is a binomial setting.

#74 (a) Yes, this is a binary setting.

B: Binary? Yes – Success is reaching a live person

I: Independent observations? It is reasonable to believe that each call is independent of the others.

N: fixed number of observations? Yes, n = 15

S: Same probability of success? Yes, p = 0.2

(b) This is NOT a binomial setting because there are not a fixed number of attempts.

Day 1 Homeworkp. 403 – 404 #71 – 73, 75 – 78, 80, 82, 84

#76 0.3151. There is about a 31.51% chance that exactly 1 of the 10 plants will die before producing any rhubarb

#78 0.0115. There is about a 1.15% chance that 3 or more of the plants will die before producing rhubarb. This would be surprising if it occurred.

#80 (a) 0.1285 (b) 0.0556. There is about a 6% chance of finding 4 or more lefties in a sample of 15. This would be moderately surprising but not completely unexpected.

#82 (a) μx = 2.4. You would expect to find an average of 2.4 people that the machine finds to be deceptive when testing 12 people actually telling the truth. (b) σx = 1.39. In actual practice, you would expect the number of “deceivers” to vary from 2.4 by an average of 1.39.

#84 (a) P(Y≥10) = 0.5584 (it is the same as P(X≤2) because if 2 or less are lying, then we are saying that 10 or more are telling the truth. (b) μy = 9.6, which is 12 - μx . The spread σx and σy are the same.

Warm-Up/Review

Tastes as good as the Real Thing?

Lesson Objecti ves

At the end of the lesson, students can:

• Sampling without replacement condition

• Determine when to use a normal approximation for a binomial distribution

• Graph binomial distributions on the calculator

Sampling Without Replacement Conditi on

Remember this example from last class?

Deal 10 cards from a shuffled deck of 52 cards. X = # of red cards.

• We said this was NOT a binary setting because the cards were not being replaced, hence making it not independent.

• However, in almost all real-world sampling, such as taking an SRS from a population of interest, is done without replacement.

• When the population is much larger than the sample, a count of successes in an SRS of size n has approximately the binomial distribution ….but what counts as “MUCH LARGER”?


When taking a SRS of size n from a population of size N, we can use a binomial distribution to model the count of successes in the sample AS LONG AS..

n ≤ 1/10 (N)

This is also known as the 10% rule – as long as the sample size n is less than or equal to 10% of the larger population N.

This does NOT mean that we need small samples! If we have a sample that is larger than 10% of the population, it just means that we should NOT USE the binomial distribution.


Example: Almost everyone has one – a drawer that holds miscellaneous batteries of all sizes. Suppose that your drawer contains 8 AAA batteries but only 6 of them are good. You need to choose 4 for your graphing calculator. If you randomly select 4 batteries, what is the probability that all 4 of the batteries you choose will work?

Problem: Explain why the answer isn’t:The actual probability is 0.2143.

3164.0)75.01()75.0(4

4)4( 04

XP

Solution: Since we are sampling without replacement, the selections of batteries aren’t independent. We can ignore this problem if the sample we are selecting is less than or equal to 10%. However, we are choosing 4 out of 8, which is 50%, so it is unreasonable to ignore the lack of independence.

The Normal Model to the Rescue!

Suppose the Tennessee Red Cross anticipates the need for at least 1850 units of O-negative blood this year. It estimates that it will collect blood from 32,000 donors. How great is the risk that the Tenn Red Cross will fall short of meeting its need?

We’ve just learned how to calculate such probabilities. We can use the Binomial model with n = 32,000 and p = 0.06. The probability of getting exactly 1850 units of O-negative blood from 32,000 donors is ()(). No calculator on earth can calculate that first term, as it has ≥ 100K digits! And that’s just the beginning! The problem said at least

1850, so we have to do it again for 1851, 1852 . . . all the way up to 32,000. NO THANKS!

The Normal Model to the Rescue!

When we’re dealing with a large number of trials like this, making direct calculations of the probabilities becomes tedious (or outright impossible). Here an old friend—the Normal model—comes to the rescue!

The Binomial model has mean np = (32,000)(0.06) =1920 and standard deviation = = 42.48. We could try approximating its distribution with a Normal model, using the same mean and st.dev. Remarkably enough, that turns out to be a very good approximation. With that, we can find the probability P(X < 1850) by finding the z-score and using Table A! P(z < -1.65) = 0.05.

There seems to be about a 5% chance that the Red Cross chapter will run short of O-negative blood.

Normal Approximati ons to Binomial Distr ibuti ons

As the number of observations ( n ) gets larger, the binomial distribution gets close to being a Normal distribution !

As n increases, our binomial formulas become LESS accurate and the Normal approximations become MORE accurate.


RULE OF THUMB: (remember this!)

Success/Failure Condition:We will use the Normal approximation to the binomial distribution when n and p satisfy:

1. np ≥ 10 2. n(1 – p) ≥ 10 (# successes ≥10) (# failures≥10)

Before you compute any probabilities, check to see if these conditions are satisfied. If they are, then use the Normal approximation; if not, use the binomial formula or calculator.


Suppose that X has the binomial distribution with n trials and probability of success p. When n is large (i.e. the Rule of Thumb is satisfied), then the distribution of X is approximately Normal with: Mean = μx = np Standard Deviation = σx = √np(1 – p)

N (np, √np(1 – p) ) The accuracy of the Normal approximation improves as the sample size (n) increases. It is most accurate when p is close to 1/2 and is least accurate when p is close to 0 or 1..

Normal Approx. of Binomial Distr.Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” The population that the poll wants to draw conclusions about is all US residents aged 18 and over.

Suppose that exactly 60% of all adult US residents would say “agree” if asked the same question. Let X = the number in the sample who agree.

(a) Show that X is approximately a binomial random variable.

(b) Check the conditions for using a Normal approximation to this setting.

(c) Use a Normal distribution to estimate the probability that 1520 or more of the sample agree.


Suppose that exactly 60% of all adult US residents would say “agree” if asked the same question. Let X = the number in the sample who agree.(a) Show that X is approximately a binomial random variable.Binary? Yes Success = people who agreeIndependent? Yes We can assume that each adult’s response does not affect another’s response. Since we are sampling w/o replacement, check the 10% rule, too! It is reasonable to believe that there are at least 25,000 US adults in the population to draw the sample.Number? Yes 2500 adultsSuccess? Yes 60%


Suppose that exactly 60% of all adult US residents would say “agree” if asked the same question. Let X = the number in the sample who agree.

(b) Check the conditions for using a Normal approximation to this setting.1. np ≥ 10 2. n(1 – p) ≥ 10

(2500)(.60) ≥ 10? (2500)(1 - .60) ≥ 101500 ≥ 10 ☺ 1000 ≥ 10 ☺

Normal Approx. of Binomial Distr.(c) Use a Normal distribution to estimate the probability that 1520 or more of the sample agree.Since this is a binomial setting that follows the rule of thumb for Normal approximation, we can use the Normal curve to find the P(X≥1520) with a mean of μ=np = 2500(.60) = 1500 and a standard deviation of √np(1 – p)= √2500(.60)(1-.60) = 24.49We want to find P(X≥1520), so to standardize the variable of X=1520, we will find the z-score = (1520 –1500)/24.49. This will calculate z = .8167. Using Table A, we find that P(X<1520) = 0.7939.So P(X ≥1520) = 1 – 0.7939 = .2061.

So there is about a 21% chance that the sample will agree that although they like to buy new clothes, but shopping is often frustrating and time-consuming.


(We answered this in the last problem, but if we were to do a full write up, this is what you would include):

State: State what you are trying to find.Plan: First, check the BINS to decide if it’s a binomial distribution!! Then state why you can use a Normal approximation to the binomial.Do: Find the mean and standard deviation. Draw a Normal curve and standardize the variable. Use Table A, to find the probability.Conclude: Write your conclusion with the probability.

Homework

Read Textbook p. 393 – 397

Do exercises p. 404 – 405 #85, 87, 88, 91, and 92

Check answers to odd problems

Happy Thanksgiving!I am very thankful for your continued efforts and

good attitudes!

Lesson Objecti ves

At the end of the lesson, students can:

• Sampling without replacement condition

• Determine when to use a normal approximation for a binomial distribution

• Graph binomial distributions on the calculator

P. 403 – 404 #71 – 73, 75 – 78, 80, 82, 84 #72B: Binary? Yes – Success is a person is...

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