70% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 31 owned dogs are randomly selected, find the probability thata. Exactly 20 of them are spayed or neutered.&nbsp;b. At most 24 of them are spayed or neutered.&nbsp;c. At least 22 of them are spayed or neutered.&nbsp;d. Between 16 and 23 (including 16 and 23) of them are spayed or neutered.&nbsp;Round all answers to 4 decimal places.

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70% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 31 owned dogs are randomly selected, find the probability thata. Exactly 20 of them are spayed or neutered.&amp;nbsp;b. At most 24 of them are spayed or neutered.&amp;nbsp;c. At least 22 of them are spayed or neutered.&amp;nbsp;d. Between 16 and 23 (including 16 and 23) of them are spayed or neutered.&amp;nbsp;Round all answers to 4 decimal places.

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Given that 70% of owned dogs in the United States are spayed or neutered. Let X be the number of spayed or neutered dogs out of 31 selected dogs.We know that X follows a binomial distribution with parameters n=31 and p=0.7.a. We need to find the probability that exactly 20 of the 31 dogs are spayed or neutered, i.e., P(X=20).Using the binomial probability formula, we have:P(X=20)=(3120)(0.7)20(0.3)11≈0.0912Therefore, the probability that exactly 20 of the 31 dogs are spayed or neutered is approximately 0.0912.b. We need to find the probability that at most 24 of the 31 dogs are spayed or neutered, i.e., P(X≤24).Using the cumulative distribution function of the binomial distribution, we have:P(X≤24)=∑x=024(31x)(0.7)x(0.3)31−x≈0.9605Therefore, the probability that at most 24 of the 31 dogs are spayed or neutered is approximately 0.9605.c. We need to find the probability that at least 22 of the 31 dogs are spayed or neutered, i.e., P(X≥22).Using the cumulative distribution function of the binomial distribution, we have:P(X≥22)=∑x=2231(31x)(0.7)x(0.3)31−x≈0.6586Therefore, the probability that at least 22 of the 31 dogs are spayed or neutered is approximately 0.6586.d. We need to find the probability that between 16 and 23 (including 16 and 23) of the 31 dogs are spayed or neutered, i.e., P(16≤X≤23).Using the cumulative distribution function of the binomial distribution, we have:P(16≤X≤23)=∑x=1623(31x)(0.7)x(0.3)31−x≈0.9355Therefore, the probability that between 16 and 23 (including 16 and 23) of the 31 dogs are spayed or neutered is approximately 0.9355.

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