STAT Lab#5 Binomial Distribution & Midterm Review

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1 STAT Lab# Binomial Distribution & Midterm Review Binomial Distributions For X Bin(n, p), Assumptions: P (X = k) = n p k (1 p) n k k Only two possible outcomes The number of trials n must be fixed in advance The probability that the event occurs, p, must be the same from trial to trial The trials must be independent Practice Problems 1. An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for 30 years or more is 2/3. Calculate the probability that after 30 years: (a) All five people are still living. (b) Exactly two people are still living (c) at least three people are still living. The number of people still living has a Binomial(n =, p = 2/3) (a) P (X = ) = ( ) (2/3) (1/3) 0 = (2/3) (b) P (X = 2) = 2 (2/3) 2 (1/3) 3 =! 2!3! (2/3)2 (1/3) (c) P (X 3) = P (X = 3) + P (X = ) + P (X = ) = (2/3) 3 (1/3) 2 + (2/3) (1/3) 1 + (2/3) (1/3) 0 3 =! 3!2! (2/3)3 (1/3) 2 +!!1! (2/3) (1/3) 1 + (2/3) = 10(2/3) 3 (1/3) 2 + (2/3) (1/3) 1 + (2/3)

2 2. A pharmaceutical lab states that a drug causes negative side effects in 3 of every 100 patients. To confirm this affirmation, another laboratory chooses people at random who have consumed the drug. What is the probability of the following events? (a) None of the five patients experience side effects. (b) At least two had side effects. (c) It is highly plausible that Hispanic people experience side effects more often than Caucasian patients. Suppose of the people; three are Caucasian and two are Hispanic. Is this a problem for the previous two situations? Explain. The number of patients having side effects is Binomial(n =, p = 0.03) (a) P (X = 0) = ( 0) (0.03) 0 (0.97) = (0.97) (b) P (X 2) = 1 P (X = 0) P (X = 1) = 1 (0.03) 0 (0.97) (0.03) 1 (0.97) 0 1 = 1 (0.97)! 1!! (0.03)1 (0.97) = 1 (0.97) (0.03) 1 (0.97) (c) Yes, because the assumption of a constant probability for the binomial distribution will be violated, since Hispanic people will have a higher probability for a bad effect. 3. Let X = the number of 6- to 7-year-olds who suffer from diabetes in the sample of size 7. X is a Bin(7, 0.12) random variable. (a) What is the probability that two of the seven people have diabetes? (b) What is the probability that four of the seven people have diabetes? (a) P (X = 2) = 7 2 (0.12) 2 (0.87) in which ( (0.12) 2 (0.87) (b) P (X = ) = ( 7 3(0.12) (0.87) ) = 7! ) (0.12) (0.87) 3 in which 7 = 7! 2!! = !3! = = 21. So P (X = 2) = = 3. So P (X = ) =. Suppose you are interested in monitoring air pollution in LA over a one-week period. Let X be a random variable that represents the number of days out of seven on which the concentration of carbon monoxide surpasses a specified level. Do you believe X has a binomial distribution? Explain. No, because if one of the seven days has a concentration of CO above a certain level, then more than likely the probability of it still being above that level is higher for the next few days. Independence would be violated. 2

3 Normal Distributions Know how to calculate normal probabilities and normal percentiles is a vary important skill in STAT 220. Be sure you know how to compute them with a normal table and a calculator (but not R). 1. The length of the human pregnancy is not fixed. It is known that it varies according to a distribution which is roughly normal, with a mean of 266 days, and an SD of 16 days. What percent of pregnancies last more than 20 days (8 months)? Pregnancy length (days) N(266, 16) z score N(0,1) The z-score of 20 is So the answer is = P (X > 20) = P (Z > 1.62) = 1 P (Z < 1.62) How many days at least are the length of the longest % of pregnancies? 266 x=? 0 z=? % Pregnancy length (days) N(266, 16) z score N(0,1) Must find a z such that P (Z > z) = 0.0. That is, P (Z < z) = 0.9. we can find in the normal table that the z corresponds to an area of 0.9 is about 1.6. So the unknown x is 1.6 in z-score, we can find the value of x as Other Practice Problems x = z = = days. The following are some examples of computational problems that we may ask. Note that this list is not exclusive. The midterm exam also contains conceptual questions that require explanation rather than computation, which is not included here. Other good practice questions are the self-study problems in the first four homework assignments. 1. Find the mean, median and standard deviation (SD) for the list below. For full credit, you must show your work. 0,,, 7, 3 3

4 The sorted list is, 3, 0,, 7, Median = 0, the middle number ( ) + ( 3) Mean = = = 1 ( 1) 2 + ( 3 1) 2 + (0 1) 2 + ( 1) 2 + (7 1) 2 SD = = = A card is selected at random from a deck of 2 poker cards. Let A be the event that the selected card is a King, and let B be the event that the selected card is a diamond. (a) Are events A and B disjoint? (b) Find P (A) and P (B). (c) Find P (A or B). (a) No, because (A and B) =the event that the selected card is the King of diamonds, which is non-empty. (b) P (A) = /2, P (B) = 13/2. (c) Since A and B are not disjoint, P (A or B) = P (A) + P (B) P (A and B) = /2 + 13/2 1/2 = 16/2 = / According to Current Population Reports, published by the U.S. Census Bureau, 1.6% of U.S. adults are female, 10.% of U.S. adults are divorced, and 6.0% of U.S. adults are divorced females. For a U.S. adult selected at random, let F = event the person is female, and D = event the person is divorced. (a) Obtain P (F ), P (D), and P (F and D). (b) Determine the proportion of U.S. adults are either female or divorced. (c) Determine the proportion of U.S. females that are divorced. (d) Find the probability that a randomly selected adult is male. (a) P (F ) = 0.16, P (D) = 0.10, P (F and D) = (b) By the General Addition Rule: P (F or D) = P (F ) + P (D) P (F and D) = = 0.6 which means 6% of U.S. adults are either female or divorced. (c) P (D F ) = P (F and D)/P (F ) = 0.06/ (d) P ({male}) = P ({female} c ) = P (F c ) = 1 P (F ) = = 0.8.

5 . According to the Arizona Chapter of the American Lung Association, 7.0% of the population has lung disease. Of those having lung disease, 90.0% are smokers; of those not having lung disease, 2.3% are smokers. (a) What percentage of people in the population are smokers? (b) Determine the probability that a randomly selected smoker has lung disease. (a) = (b) = Suppose that the turnout rate in Chicago area for the 2016 general election is %. If we randomly select eligible voters in the Chicago area, what is the probability that (a) all of them will vote? (b) none of them will vote? (c) at least one of them will vote (a) (0.) (b) (0.) (c) 1 (0.) = A person is said to be overweight if his or her body mass index (BMI) is between 2 and 29, inclusive; a person is said to be obese if his or her BMI is 30 or greater. From the document Utah Behavioral Risk Factor Surveillance System (BRFSS) Local Health District Report, issued by the Utah Department of Health, we obtained the following table. The first two columns of the table provide an age distribution for adults living in Utah. The third column gives the percentage of adults in each age group who are either obese or overweight. % obese or Age (yr) % adults overweight & over (a) What percentage of Utah adults are overweight or obese? (b) Of those Utah adults who are between 3 and 9 years old, inclusive, what percentage are overweight or obese? (c) Of those Utah adults who are overweight or obese, what percentage are between 3 and 9 years old, inclusive? (a) = 2.1%. (b) 7.9% (c) = 31.7% Suppose that a four-sided die has faces with 1, 2, 3, spots respectively. All four faces are equally likely to come up. (a) Find the expected value and the standard deviation of the number of spots on the face that come up in a roll

6 (b) Rolling this die twice, find the expected value and the standard deviation of the sum of spots obtained in the two roll. (a) Let X be the number of spots obtained. It s obvious that X are equally likely to be any of 1, 2, 3, or, so the probability model for X is value of X probability 1/ 1/ 1/ 1/ The expected value, variance, and SD for this distribution is E(X) = = 2 = 2. V (X) = (1 2.) (2 2.)2 1 + (3 2.)2 1 + ( 2.)2 1 = SD(X) = V (X) = (b) Let X 1, X 2 be respectively the spots obtained in the first and the second roll. Both X 1, X 2 have the same distribution as X in part (a). Since the rolls are independent, E(X 1 + X 2 +) = E(X 1 ) + E(X 2 ) = = V (X 1 + X 2 ) = V (X 1 ) + V (X 2 ) = + = 2 SD(X 1 + X 2 ) = V (X 1 + X 2 ) =

BIOS 4120: Introduction to Biostatistics Breheny. Lab #7. I. Binomial Distribution. RCode: dbinom(x, size, prob) binom.test(x, n, p = 0.

BIOS 4120: Introduction to Biostatistics Breheny Lab #7 I. Binomial Distribution P(X = k) = ( n k )pk (1 p) n k RCode: dbinom(x, size, prob) binom.test(x, n, p = 0.5) P(X < K) = P(X = 0) + P(X = 1) + +