Question
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Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go") so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that \( 48 \% \) of its customers order their food to go. Suppose that this proportion is correct and that a random sample of 50 individual customers is taken. Answer the following. (If necessary, consult a list of formulas.) (a) Estimate the number of customers in the sample who order their food to go by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response. (b) Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.

Ask by Ramos Schmidt. in the United States
Jan 27,2025

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Answer

(a) The mean number of customers ordering "to go" is 24. (b) The standard deviation is 3.533.

Solution

Let's analyze the problem step by step. **Given:** - **Probability of ordering to go (\( p \))**: \( 48\% = 0.48 \) - **Sample size (\( n \))**: 50 customers ### (a) Estimating the Number of Customers Ordering "To Go" The number of customers who order "to go" follows a **binomial distribution** \( X \sim \text{Binomial}(n, p) \). - **Expectation (Mean) of \( X \)**: \[ E[X] = n \times p = 50 \times 0.48 = 24 \] **Answer for (a):** The mean number of customers who order "to go" is **24**. ### (b) Quantifying the Uncertainty (Standard Deviation) The **standard deviation (\( \sigma \))** of a binomial distribution is calculated using the formula: \[ \sigma = \sqrt{n \times p \times (1 - p)} \] - **Calculations:** \[ \sigma = \sqrt{50 \times 0.48 \times (1 - 0.48)} = \sqrt{50 \times 0.48 \times 0.52} \] \[ \sigma = \sqrt{50 \times 0.2496} = \sqrt{12.48} \approx 3.533 \] **Answer for (b):** The standard deviation is **3.533**. --- **Summary of Answers:** (a) The mean number of customers ordering "to go" is 24. (b) The standard deviation is 3.533.

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The Deep Dive

To tackle these questions, let’s dive right in! For (a), we need to estimate the number of customers in the sample who order their food to go. Since \( 48\% \) of customers order to go, we can calculate the expected number of customers in a sample of 50. The mean can be calculated as: \[ \text{Mean} = n \times p = 50 \times 0.48 = 24 \] So, we can expect about 24 customers in the sample to order their food to go. For (b), to quantify the uncertainty of this estimate, we need to calculate the standard deviation of the distribution. The standard deviation for a binomial distribution can be calculated using the formula: \[ \text{Standard Deviation} = \sqrt{n \times p \times (1 - p)} \] Plugging in the values we have: \[ \text{Standard Deviation} = \sqrt{50 \times 0.48 \times (1 - 0.48)} = \sqrt{50 \times 0.48 \times 0.52} \approx \sqrt{12.48} \approx 3.535 \] Thus, the standard deviation rounds to approximately \( 3.535 \). In summary, approximately 24 customers in the sample are expected to order food to go, with an uncertainty measured by a standard deviation of \( 3.535 \).

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