D E F pls Let us assume that a turkey is classified by its weight as low, medium

Question

D E F pls

Let us assume that a turkey is classified by its weight as low, medium, or high. We can also assume that 3% of turkeys are classified as low, 88% as medium, and 9% as high. A sample of 20 turkeys is selected for weighing. Let X, Y, and Z represent the number of turkeys that are independently classified as low, medium, and high weight, respectively. a. What are the name and values of the parameters of the joint probability distribution of the joint distribution of X, Y, and Z? O multivariate normal; Ax = 0.03, pur – 0.88, Hz = 0.09, ox – oy — 07 – 1. O multinomial; PX = 0.88, Py = 0.09, Pz = 0.03. O multinomial; px = 0.03, Py = 0.88, Pz = 0.09. O binomial; n = 20,p= 0.88 Submit Answer Answer submitted: Your final submission will be graded after the due date. Trles 1/10 Previous Trles b. What are the name and values of the parameters of the marginal probability distribution of X? • binomial; n = 20,p= 0.03 O binomial; n = 20, p= 0.88 O multinomial; PX = 0.03, y = 0.88, Pz = 0.09. O normal; pax = 0.03, X = 1. |Submit Answer Answer Submitted: Your final submission will be graded after the due date. Tries 1/10 Previous Tries Round all answers to 4 decimal points. c. What is the expected number of medium weight turkeys? 17.6000 Computer's answer now shown above. You are correct. Previous Tries Your receipt no. Is 163-61610

Explanation / Answer

d)

P=n!/(n1!)(n2!)(n3!)*p1^n1*p2^n2*p3^n3

P=20!/(1!)(18!)(2!)*(0.03^1)*(0.88^18)*(0.09^2)

P=0.0046

e)

P=n!/(n1!)(n2!)(n3!)*p1^n1*p2^n2*p3^n3

P=20!/(0!)(18!)(2!)*(0.03^0)*(0.88^18)*(0.09^2)+P=20!/(1!)(18!)(2!)*(0.03^1)*(0.88^18)*(0.09^2)

P=0.1541+0.0046=0.1587

f)

p=0.03, q=0.97, n=20, x<=1

n!/x!(n-x)!*p^x*q^n-x

20!/0!20!*0.03^0*0.97^20+20!/1!19!*0.03^1*0.97^19

0.88

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