Use The Following Information To Answer The Question: A Pesc

Use The Following Information To Answer The Question A Pescatarian Is

Use the following information to answer the question. A pescatarian is a person who eats fish and seafood but no other animal. An event planner finds that approximately 2.75% of the people in the area are pescatarian. The event expects 250 guests, which can be considered a simple random sample from a population of about 150,000. The question asks for the average number and the range (give or take) of how many guests would be expected to be pescatarian, rounded to the nearest whole person.

Paper For Above instruction

The problem involves estimating the expected number of pescatarians among the guests at a large event, based on the prevalence of pescatarian diet preferences in the local population. Given that 2.75% of the local population are pescatarians, and that the event guests are a simple random sample, we can apply basic probability principles to determine the expected value and the variability around this estimate.

First, we identify the key pieces of information:

- Proportion of pescatarians in the population, p = 2.75% = 0.0275

- Number of guests, n = 250

The expected number of pescatarians among the guests is calculated by multiplying the total number of guests by the population proportion:

\[ \text{Expected number} = n \times p = 250 \times 0.0275 = 6.875 \]

Since we cannot have a fraction of a person, we round this to the nearest whole number, which gives approximately 7 pescatarians.

Next, to understand the variability and to construct an approximate range, we consider the standard deviation for the number of pescatarians in a simple random sample:

\[ \sigma = \sqrt{n \times p \times (1 - p)} \]

\[ \sigma = \sqrt{250 \times 0.0275 \times (1 - 0.0275)} \]

\[ \sigma = \sqrt{250 \times 0.0275 \times 0.9725} \]

\[ \sigma = \sqrt{250 \times 0.02677875} \]

\[ \sigma = \sqrt{6.6947} \]

\[ \sigma \approx 2.588 \]

This standard deviation indicates the typical fluctuation around the mean. For an approximate 95% range, which involves about two standard deviations on either side of the mean, we estimate:

\[ \text{Range} \approx \text{Expected} \pm 2 \times \sigma \]

\[ \approx 7 \pm 2 \times 2.588 \]

\[ \approx 7 \pm 5.176 \]

Rounding to whole persons, the expected number of pescatarians is about 7, with a plausible range from approximately 2 to 12 guests.

Therefore, based on the local population's pescatarian proportion, we expect around 7 guests to be pescatarian, give or take about 5 persons, which encompasses typical sampling variability.

References

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