University:
Missouri State University
Course:
MTH 343 | Foundations of Probability and Statistics for Teachers
Academic year:

2021
Views:

137

Pages:

3
Author:

skarvurdld6

Unit 5 Binomial Chapter 18- Day 2: practice binomial and geometric, normal approximation of binomial Here's your warm up packet... 1414 Warm ups: MC #3, 4, 6, 7, 8 A LTATA C What are the characteristics of a binomial distribution? 1. set number of trials (outcomes) 2. prob of success the same for each trial 3. each trial is independent 4. two possible outcomes: success or failure B(n, p) Geom(p) mu = np mu = 1 / p sigma = sqrt(npq) sigma = (sqrt(q)) / p AP Formula Sheet IIX has a binomial distribution with parameters and p P(X = k) = binomial(alpha,k) * p ^ k * (1 - p) ^ (k - k) A_{1} = lambda * psi ^ p mean 1/017.11 9001 denurs (89)(11)-11383 underline tupd (120, 11, 2) - 20 ,23=P(y=2) [[20], [8]] * (0) ^ 1 * (0.84) ^ 18 7(X ^ (2k)) * st - lig * (k!)/(6 * k!) * (10/64, 11, 1) = 0.3025 S(10, 11) 150(11) sqrt(100(- 5)(0.5 deg)) Hint ne observation is 16-5-2(783)-8.83 Jetations away Normal Approximation to the Binomial WS (a) Verify that a binomial situation applied fival number 2500 p = omega_{0}*Omega B(2500, 0.6) 4 independence agree success 3. ng 210 (b) Verify that you can use the normal distribution to approximate the binomial distribution P10(2500) LPP пр102200(4)150010 2500(14)-1000 210 (c) Find the mean and standard deviation of the number of people who would say agree. M(1500, 24.495) mu = Lambda*p = 1500 delta = sqrt(npq) = sqrt 2500(.w)(4) Chapter 18-Sampling Distributions (Day 1] Normal Approximation so the Binomial What happens when n gets so large the binomial distribution becomes too difficult? When the following are true. 1. Population is at least 10 times the sample 2. np >= 10 10 Suvass 3. q(1 - p) >= 10 10 failures nq >= 10 .....a can be used to approximate a binomial probability!!! The mean and standard deviation are... mu_{x} = np sigma_{x} = sqrt(np(1 - p)) (d) Find the probability that at least 1520 people agree? N(1500, 24495) P(X Rightarrow 152P ) Z = (1520 - 1550)/(244 ^ 0 * 85) = 0.816 P(z >= 0.816) 207 (e) Find the probability at most 1430 people agree? P(x <= 1430) z = (1430 - 1505)/244485 z = - 2.46 P(z <= - 2.86) (Find the omhshility that hatuwen 1430 and 1520 people agree? P( 1430 <*41520)=.7908 1-(207-0021)