Question

g About 52% of American adults live in middle-income households. Suppose a sample of 20 American adults is selected. Let X be the number of adults from the sample that live in middle-income households. What is the probability distribution of X

Answer 1

P (x= 0) = 0.00384

P (x= 1) = 0.000009133

P (x= 2) = 0.00009399

P (x= 3) = 0.0061

P (x= 4) = 0.00281

P (x= 5) = 0.00975

P (x= 6) = 0.0264

P (x= 7) = 0.05772

P (x= 8) = 0.0202

P (x= 9) = 0.14550

P (x= 10) = 0.1733

P (x= 11) = 0.17077

P (x= 12) = 0.138751

P (x= 13) = 0.0925

P (x= 14) = 0.0501

P (x= 15) = 0.0217

P (x= 16) = 0.00735

P (x= 17) = 0.00187

P (x= 18) = 0.000259

P (x= 19) = 2 0.0000385

P (x= 20) =0.000000208

Explanation:

This is a binomila probability distribution as the success and failure are same and number of trials is fixed.

Here p =0.52 and q= 1-0.52= 0.48

n= 20 Now the probability distribution of X = 0,1,2,--------20

Applying the binomial distribution we can find the probability distribution of X taking the different values as

P (x= 0) = 20c0 (0.52)^0 (0.48)^20= 0.00384

P (x= 1) = 20c1 (0.52)^1 (0.48)^19= 0.000009133

P (x= 2) = 20c2 (0.52)^2 (0.48)^18= 0.00009399

P (x= 3) = 20c3 (0.52)^3 (0.48)^17= 0.0061

P (x= 4) = 20c4 (0.52)^4 (0.48)^16= 0.00281

P (x= 5) = 20c5 (0.52)^5 (0.48)^15= 0.00975

P (x= 6) = 20c6 (0.52)^6 (0.48)^14= 0.0264

P (x= 7) = 20c7 (0.52)^7 (0.48)^13= 0.05772

P (x= 8) = 20c8 (0.52)^8 (0.48)^12= 0.0202

P (x= 9) = 20c9 (0.52)^9 (0.48)^11= 0.14550

P (x= 10) = 20c10 (0.52)^10 (0.48)^10= 0.1733

P (x= 11) = 20c11 (0.52)^11 (0.48)^9= 0.17077

P (x= 12) = 20c12 (0.52)^12 (0.48)^8= 0.138751

P (x= 13) = 20c13 (0.52)^13 (0.48)^7= 0.0925

P (x= 14) = 20c14 (0.52)^14 (0.48)^6= 0.0501

P (x= 15) = 20c15 (0.52)^15 (0.48)^5= 0.0217

P (x= 16) = 20c16 (0.52)^16 (0.48)^4= 0.00735

P (x= 17) = 20c17 (0.52)^17 (0.48)^3= 0.00187

P (x= 18) = 20c18 (0.52)^18 (0.48)^2= 0.000259

P (x= 19) = 20c19 (0.52)^19 (0.48)^1= 0.0000385

P (x= 20) = 20c20 (0.52)^20 (0.48)^0= 0.000000208