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Assume a famous coffee shop chain sells about 8 million cups of coffee per day with a standard deviation of 300,000 cups per day. Assuming the number of cups of coffee that they sell per day follows a normal distribution, what is the 90th percentile of the number of cups of coffee that they sell in a given day? 8,000,702 8,252,486 8,384,465 8,697,904

Assume a famous coffee shop chain sells about 8 million cups of coffee per day with a standard deviation of 300,000 cups per day. Assuming the number of cups of coffee that they sell per day follows a normal distribution, what is the 90th percentile of the number of cups of coffee that they sell in a given day? 8,000,702 8,252,486 8,384,465 8,697,904

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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### Question on Normal Distribution #### Problem Statement: Assume a famous coffee shop chain sells about 8 million cups of coffee per day with a standard deviation of 300,000 cups per day. Assuming the number of cups of coffee that they sell per day follows a normal distribution, what is the 90th percentile of the number of cups of coffee that they sell in a given day? #### Options: 1. **8,000,702** 2. **8,252,486** 3. **8,384,465** 4. **8,697,904** ### Explanation: To find the 90th percentile of a normally distributed variable, we use the formula: \[ X = \mu + Z \sigma \] Where: - \( \mu \) is the mean (8,000,000 cups), - \( \sigma \) is the standard deviation (300,000 cups), - \( Z \) is the standard score corresponding to the desired percentile (90th percentile). In statistical tables or using statistical software, the Z-value for the 90th percentile is approximately 1.28. Thus, the calculation would be: \[ X = 8,000,000 + (1.28 \times 300,000) \] \[ X = 8,000,000 + 384,000 \] \[ X = 8,384,000 \] So, the 90th percentile is approximately **8,384,000** cups. Therefore, the correct option is: ### Answer: **8,384,465** (Note that this accounts for rounding differences and slight variations) This concept can be applied to other scenarios in business, healthcare, and various fields where understanding the distribution of data is essential.
Transcribed Image Text:### Question on Normal Distribution #### Problem Statement: Assume a famous coffee shop chain sells about 8 million cups of coffee per day with a standard deviation of 300,000 cups per day. Assuming the number of cups of coffee that they sell per day follows a normal distribution, what is the 90th percentile of the number of cups of coffee that they sell in a given day? #### Options: 1. **8,000,702** 2. **8,252,486** 3. **8,384,465** 4. **8,697,904** ### Explanation: To find the 90th percentile of a normally distributed variable, we use the formula: \[ X = \mu + Z \sigma \] Where: - \( \mu \) is the mean (8,000,000 cups), - \( \sigma \) is the standard deviation (300,000 cups), - \( Z \) is the standard score corresponding to the desired percentile (90th percentile). In statistical tables or using statistical software, the Z-value for the 90th percentile is approximately 1.28. Thus, the calculation would be: \[ X = 8,000,000 + (1.28 \times 300,000) \] \[ X = 8,000,000 + 384,000 \] \[ X = 8,384,000 \] So, the 90th percentile is approximately **8,384,000** cups. Therefore, the correct option is: ### Answer: **8,384,465** (Note that this accounts for rounding differences and slight variations) This concept can be applied to other scenarios in business, healthcare, and various fields where understanding the distribution of data is essential.
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