Question

In: Statistics and Probability

In 2017, the city of Burrusville contained 26,898 households. According to city records, 16,275 of these...

In 2017, the city of Burrusville contained 26,898 households. According to city records, 16,275 of these households had a dog. Use this information to answer the following questions.

a) What is p, the population proportion? Show your calculation and round your solution to three decimal places. (2 points)

b) Suppose a simple random sample of 140 homes were chosen at random, and 91 of these homes had a dog. Verify that the sampling distribution is approximately normal. (3 points)

c) What is the probability that a random sample of 140 homes from Burrusville would contain 91 or fewer with a dog? (4 points)

Expert Solution

Number of household in Burrusville = 26898

households had a dog in Burrusville = 16275

a) What is p, the population proportion? Show your calculation and round your solution to three decimal places.

p = households had a dog in Burrusville / Number of household in Burrusville

p = 16275 / 26898

p = 0.605

b) Suppose a simple random sample of 140 homes were chosen at random, and 91 of these homes had a dog. Verify that the sampling distribution is approximately normal.

The sample size is greater than 30 hence it is normal

Sample proportion, p = 91/140 = 0.65

c) What is the probability that a random sample of 140 homes from Burrusville would contain 91 or fewer with a dog?

Let x be the number of dogs

orchestra answered 1 year ago

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