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The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 61 and a standard deviation of 5. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 61 and 71? Do not enter the percent symbol. ans = %

The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 61 and a standard deviation of 5. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 61 and 71? Do not enter the percent symbol. ans = %

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Lightbulb Replacement Requests at State University - Data Analysis Using Statistical Methods** The maintenance department at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped (normal distribution) with a mean of 61 and a standard deviation of 5. **Problem Statement:** Using the 68-95-99.7 rule (Empirical Rule), what is the approximate percentage of lightbulb replacement requests numbering between 61 and 71? **Empirical Rule Explanation:** The 68-95-99.7 rule states that for a normal distribution: - About 68% of data falls within one standard deviation of the mean. - About 95% of data falls within two standard deviations of the mean. - About 99.7% of data falls within three standard deviations of the mean. **Calculations:** 1. **Identify the Range:** - Mean (μ) = 61 - Standard Deviation (σ) = 5 - Range in question: 61 to 71 2. **Determine Standard Deviation Range:** - Lower bound: Mean (61) + 0*σ = 61 - Upper bound: Mean (61) + 2*σ = 61 + 10 = 71 3. **Apply the Empirical Rule:** - According to the rule, 95% of the data falls within two standard deviations from the mean. - This means the percentage of lightbulb replacement requests numbered between 61 (mean) and 71 (mean + 2σ) is approximately 47.5%. (Since 95% is for both tails, we consider half for between mean and two standard deviations) **Conclusion:** Therefore, the approximate percentage of lightbulb replacement requests between 61 and 71 is 47.5%. **Interactive Help Videos:** - [Video 1] - [Video 2] **Submission Section:** To answer, do not enter the percent symbol. **Answer:** \[ \text{ans} = \underline{\hspace{5cm}} \% \] **Submit Answer:** [Submit Question]
Transcribed Image Text:**Lightbulb Replacement Requests at State University - Data Analysis Using Statistical Methods** The maintenance department at the main campus of a large state university receives daily requests to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped (normal distribution) with a mean of 61 and a standard deviation of 5. **Problem Statement:** Using the 68-95-99.7 rule (Empirical Rule), what is the approximate percentage of lightbulb replacement requests numbering between 61 and 71? **Empirical Rule Explanation:** The 68-95-99.7 rule states that for a normal distribution: - About 68% of data falls within one standard deviation of the mean. - About 95% of data falls within two standard deviations of the mean. - About 99.7% of data falls within three standard deviations of the mean. **Calculations:** 1. **Identify the Range:** - Mean (μ) = 61 - Standard Deviation (σ) = 5 - Range in question: 61 to 71 2. **Determine Standard Deviation Range:** - Lower bound: Mean (61) + 0*σ = 61 - Upper bound: Mean (61) + 2*σ = 61 + 10 = 71 3. **Apply the Empirical Rule:** - According to the rule, 95% of the data falls within two standard deviations from the mean. - This means the percentage of lightbulb replacement requests numbered between 61 (mean) and 71 (mean + 2σ) is approximately 47.5%. (Since 95% is for both tails, we consider half for between mean and two standard deviations) **Conclusion:** Therefore, the approximate percentage of lightbulb replacement requests between 61 and 71 is 47.5%. **Interactive Help Videos:** - [Video 1] - [Video 2] **Submission Section:** To answer, do not enter the percent symbol. **Answer:** \[ \text{ans} = \underline{\hspace{5cm}} \% \] **Submit Answer:** [Submit Question]
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