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1.A pet food company wants to know the number of pets owned by adults ages 21 to 70. The frequency table shows the data from a simple random sample of the targeted population.

Total pets owned

1

2

3

4

5

6

7

Number of adults

248

567

1402

728

419

456

203

What is the estimated mean for the population? Explain your answer and show your work.

Guest Feb 19, 2019

#1**+2 **

\(\text{The expected number of pets given some sampling of }N \text{ people is }\\ \bar{p}=\dfrac 1 N \sum \limits_{k=1}^N ~p_k,~\text{ where }p_k \text{ is the number of pets the }kth \text{ person owns}\\ \text{The frequency chart simplifies this a bit by by allowing us to do this instead }\\ \bar{p}=\dfrac{(1)(248)+(2)(567)+(3)(1402)+(4)(728)+(5)(419) + (6)(456)+(7)(203)}{248+567+1402+728+419+456+203}\)

I leave it to you to chug out the numbers.

Rom Feb 19, 2019

#1**+2 **

Best Answer

\(\text{The expected number of pets given some sampling of }N \text{ people is }\\ \bar{p}=\dfrac 1 N \sum \limits_{k=1}^N ~p_k,~\text{ where }p_k \text{ is the number of pets the }kth \text{ person owns}\\ \text{The frequency chart simplifies this a bit by by allowing us to do this instead }\\ \bar{p}=\dfrac{(1)(248)+(2)(567)+(3)(1402)+(4)(728)+(5)(419) + (6)(456)+(7)(203)}{248+567+1402+728+419+456+203}\)

I leave it to you to chug out the numbers.

Rom Feb 19, 2019