Question

78% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 40 owned dogs are randomly selected, find the probability thata. Exactly 32 of them are spayed or neutered. b. At most 31 of them are spayed or neutered. c. At least 31 of them are spayed or neutered. d. Between 29 and 33 (including 29 and 33) of them are spayed or neutered.

Answer 1

$\begin{gathered} To\text{ answer this question we will use the binomial distribution that calculates the probability of success in n trials} \\ Px\text{ = \lparen nCx\rparen\lparen p}^x)(q^{n-x}) \\ When\text{ x = 32 and p =78\%, q = 22\%} \\ P(x=32)\text{ = \lparen40C32\rparen\lparen78\%}^{32})(22\%^8) \\ =\text{ 0.1487} \end{gathered}$$\begin{gathered} At\text{ most 31 of them are spayed:} \\ This\text{ is P\lparen x}\leq31)\text{ = 1-P\lparen x=40\rparen+P\lparen x=39\rparen + P\lparen x=38\rparen + P\lparen x=37\rparen...P\lparen x=32\rparen} \\ P(x=40)\text{ = 0} \\ P(x=39)\text{ =0} \\ P(x=38)=0.002 \\ P(x=37)\text{ =0.010} \\ P(x=36)\text{ =0.028} \\ P(x=35)\text{ =0.0567} \\ P\text{ \lparen x=34\rparen = 0.093} \\ P(x=33)\text{ = 0.128} \\ P(x=32)\text{ = 0.149} \\ Therefore\text{ P\lparen x}\leq31)\text{ = 0.5312} \\ \end{gathered}$$\begin{gathered} At\text{ least 31 of them are spayed:} \\ P(x\ge31) \\ P(x=31)\text{ = 0.149} \\ We\text{ will add P\lparen x=31,32,33....40\rparen} \\ =0.6179 \end{gathered}$$\begin{gathered} P(x=29)\text{ = 0.1} \\ P(x=30)\text{ = 0.13} \\ P(x=31)\text{ = 0.149} \\ P(x=32)\text{ =0.149} \\ P(x=33)\text{ = 0.128} \\ Adding\text{ these all together to get P\lparen29}\leq x\leq33) \\ =0.656 \end{gathered}$