Help with Binomial Distribution please?
Let X be a random variable that follows the binomial distribution with parameters n=20 p=2/3 a)what is the probability that X takes the value 13? b) compute the E(X) and Var(X) of X. I think these are ok, 13 1/3, 4 4/9? c) Would this distribution be a good model for the population which you sampled in Q(1), why (not). ((Question 1 was a 20 person sample with yes/no answer with 13 yes answers)) I said no as sample too small?? d) Assuming this is the true model for the population sampled in Q(1). Before conducting the survey what is the probability of obtaining 13 yes answers? any help would be great fully appreciated, particularly a,d i think b,c are ok, but not sure! thanks in advance
Binomial Distribution HW Help?
Among employed women, 35% have never been married. Select 15 employed women at random. (a) The number in your sample who have never been married has a binomial distribution. What are n and p? n = ____ p = ____ (b) What is the probability that exactly 4 of the 15 women in your sample have never been married? (c) What is the probability that 2 or fewer women have never been married? Any help would be greatly appreciated. I am failing AP Statistics, and I can't seem to get the hang of it.
Binomial Distribution HW help, please?
We can use the binomial in this case. Technically it would better to use the hypergeometric, but that requires knowing the exact number of children, incarcerated parents and a few other values. We can use the Binomial when the sample is so small compared to the population that the removal of one object does not have a noticeable effect on on the success probability. Generally this mean you are sampling less than 5% of the population. Let X be the number of children with incarcerated parents. X has the binomial distribution with n = 160 trials and success probability p = 0.03 In general, if X has the binomial distribution with n trials and a success probability of p then P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x) for values of x = 0, 1, 2, ..., n P[X = x] = 0 for any other value of x. The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures. Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials. X ~ Binomial( n , p ) the mean of the binomial distribution is n * p = 4.8 the variance of the binomial distribution is n * p * (1 - p) = 4.656 the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 2.157777 the first 13 elements of the Probability Mass Function, PMF, f(X) = P(X = x) is: P( X = 0 ) = 0.00764676 ← none of the children sampled have incarcerated parents P( X = 1 ) = 0.03783964 ← only one child sampled has an incarcerated parent P( X = 2 ) = 0.0930387 ← exactly two children sample have an incarcerated parent P( X = 3 ) = 0.1515476 P( X = 4 ) = 0.1839663 P( X = 5 ) = 0.1775180 P( X = 6 ) = 0.1418314 P( X = 7 ) = 0.0965038 P( X = 8 ) = 0.0570815 P( X = 9 ) = 0.02981577 P( X = 10 ) = 0.01392427 P( X = 11 ) = 0.005872467 P( X = 12 ) = 0.002255148 c) P( X ≥ 3 ) = 0.8614749 P( X ≥ 3 ) = 1 - P(X < 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)
Binomial distribution?????
The normal approximation to the binomial is only okay to use if you have at least 10 expected successes and 10 expected failures. with only 14 flips of a fair coin you have 7 expected success and failures. thus it is not appropriate to use the normal approximation to the binomial. I'll do some work here to show you what is going on. Let X be the number of heads seen X ~ Binomial (n = 14, p = 1/2) In general, if X has the binomial distribution with n trials and a success probability of p then P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x) for values of x = 0, 1, 2, ..., n P[X = x] = 0 for any other value of x. Let Y ~ N( μ = np, σ² = np(1-p) ) = N(7, 1.75 ) I'll now put in the values for the CDF functions. When you do this don't forget about the continuity correction. P(X ≤ 0) = 0.0000610351 P(Y ≤ 0) = P( Z ≤ (0.5 - 7)/sqrt(1.75) ) = P(Z ≤ -4.913538) = 4.47e-7 P(X ≤ 5) = 0.2119751 P(Y ≤ 5) = P(Z ≤ (5.5 - 7)/sqrt(1.75)) = P( Z ≤ -1.1338 ) = 0.1284196 as you can see the estimates using the normal approximation to the binomial are not very good. You need a lot of trials before the binomial histogram will start to look like the normal density curve. To get 7 heads: P(X = 7) = 0.2094 P(Y = 7) = P( 6.5 < Y < 7.5) = P(-0.3779645 < Z < 0.3779645) = 0.294530
Binomial Distributions HW Help, PLEASE?
Suppose that James guesses on each question of a 49-item true-false quiz. Find the probability that James passes if each of the following is true. (a) A score of 25 or more correct is needed to pass. (b) A score of 30 or more correct is needed to pass. (c) A score of 31 or more correct is needed to pass. ***Any help is greatly appreciated, as I am dumb when it comes to math. Thank you so much
Statistic Help! Binomial Distribution?
The probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p is called a binomial distribution. You have yes/no experiments: vehicle needs service, yes or no? Each with a probability p (p=14%). They are independent. That is a reasonable assumption. If vehicle 1 needs service, that does not mean anything about vehicle 2 (unless they are all coming from a lemon production line. Oh well). Then the probability of having k vehicles out of n in need of service is: P(k,n) = nCk*p^k*(1-p)^(n-k) where nCk = n!/(k!(n-k)!) Question 1: exactly 3 vehicles require service: P(3,9) = 9C3*0.14^3*0.86^6 = 0.0133 = 1.33% exactly 3 vehicles don't require service: P(3,9) = 9C3*0.86^3*0.14^6 = 0.000057 = 0.0057% Question 2 The expected value of a binomial distribution is E(X)= np where n is the number of trials and p the probability of success in each trial. E(X) = 9*0.14 = 1.26 An average of 14% of your 9 vehicles will require service. That is 1.26 vehicles.
Statistics Help - Binomial Distribution?
Let X be the number of adults receiving less than five phone calls. X has the binomial distribution with n = 7 trials and success probability p = 0.43 In general, if X has the binomial distribution with n trials and a success probability of p then P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x) for values of x = 0, 1, 2, ..., n P[X = x] = 0 for any other value of x. The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures. Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials. X ~ Binomial( n = 7 , p = 0.43 ) the mean of the binomial distribution is n * p = 3.01 the variance of the binomial distribution is n * p * (1 - p) = 1.7157 the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.309847 The Probability Mass Function, PMF, f(X) = P(X = x) is: P( X = 0 ) = 0.01954897 P( X = 1 ) = 0.1032323 P( X = 2 ) = 0.233631 P( X = 3 ) = 0.2937466 P( X = 4 ) = 0.2215983 P( X = 5 ) = 0.1003024 P( X = 6 ) = 0.02522224 P( X = 7 ) = 0.002718186 The Cumulative Distribution Function, CDF, F(X) = P(X ≤ x) is: x ∑ P(X = t) = t = 0 P( X ≤ 0 ) = 0.01954897 P( X ≤ 1 ) = 0.1227813 P( X ≤ 2 ) = 0.3564123 P( X ≤ 3 ) = 0.6501589 P( X ≤ 4 ) = 0.8717572 P( X ≤ 5 ) = 0.9720596 P( X ≤ 6 ) = 0.9972818 P( X ≤ 7 ) = 1 1 - F(X) is: n ∑ P(X = t) = t = x P( X ≥ 0 ) = 1 P( X ≥ 1 ) = 0.980451 P( X ≥ 2 ) = 0.8772187 P( X ≥ 3 ) = 0.6435877 P( X ≥ 4 ) = 0.3498411 P( X ≥ 5 ) = 0.1282428 P( X ≥ 6 ) = 0.02794042 P( X ≥ 7 ) = 0.002718186