Ch 5.3
Surveying 200 college students and recording their favorite TV show
No, because there are more than two possible outcomes
Determine whether the given procedure results in a binomial distribution. If it is not binomial, identify the requirements that are not satisfied.
There are certain requirements for a binomial probability experiment. A binomial probability experiment results from a procedure that meets all the following requirements. 1. The procedure has a fixed number of trials. 2. The trials must be independent. (The outcome of any individual trial doesn't affect the probabilities in the other trials.) 3. Each trial must have all outcomes classified in two categories (commonly referred to as success and failure.) 4. The probability of success remains the same in all trials.
Multiple-choice questions each have four possible answers (a, b, c, d), one of which is correct. Assume that you guess the answers to three such questions. a. Use the multiplication rule to find P(WCW), where C denotes a correct answer and W denotes a wrong answer. b. Beginning with WCW, make a complete list of the different possible arrangements of one correct answer and two wrong answers, then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made?
a. 3/4 x 1/4 x 3/4 = 0.140625 b. P(WCW)=0.140625 P(CWW)=0.140625 P(WWC)=0.140625 c. P(WCW) + P(CWW) + P(WWC)=0.421875
A certain TV show recently had a share of 80, meaning that among the TV sets in use, 80% were tuned to that show. Assume that an advertiser wants to verify that 80% share value by conducting its own survey, and a pilot survey begins with 10 households having TV sets in use at the time of the TV show broadcast. Complete parts (a) through (d) below. a. Find the probability that all of the households are tuned to the TV show. b. Find the probability that exactly 9 households are tuned to the TV show. c. Find the probability that at least 9 households are tuned to the TV show. d. If at least 9 households are tuned to the TV show, does it appear that the 80% share value is wrong? Why or why not?
a. p(10)= (0.80)^10 = 0.107 b. p(9)= 10!/(10-9)!9! x (0.8)^9 x (0.2)^(10-9)= 0.268 c. at least 9= 0.268 + 0.107= 0.375 d. No, because 9 households tuned to the TV show is not unusually high if the share is 80%. ****To determine if the share value may be wrong, recall that x successes among n trials is an unusually high number of successes if P(x or more)≤0.05; and x successes among n trials is an unusually low number of successes if P(x or fewer)≤0.05.
Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. n=7, x=3, p=0.35
q = 1 − p = 0.65 Now substitute the values of n, x, p, and q into the formula. P(x)=n!/(n-x)!x! * p^x*q^(n-x) 7!/(7-3)!3! x 0.35^3 x 0.65^(7-3) Subtract 7!/4!3! x 0.35^3 x 0.65^(7-3) Simplify 5040/(6)(24) x (0.0429)(0.1785) 5040/144 x (0.0429)(0.1785) 35 x (0.0429)(0.1785) =0.268 p(3)=0.268
Four cards are selected from a standard 52-card deck without replacement. The number of sevens selected is recorded.
No, because the trials of the experiment are not independent and the probability of success differs from trial to trial.
An experimental drug is administered to 40 randomly selected individuals, with the number of individuals responding favorably recorded.
Yes, because the experiment satisfies all the criteria for a binomial experiment.
