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One vear consumers spent an average of $22 on a meal at a resturant. Assume that the amount spent on a resturant meal is normally distributed and that the standard deviation is $5. Complete parts (a) through (c) below a. What is the probability that a randomly selected person spent more than $24? P(X> $24) = 0.3446 (Round to four decimal places as needed.) b. What is the probability that a randomly selected person spent between $8 and $19? P($8

One vear consumers spent an average of $22 on a meal at a resturant. Assume that the amount spent on a resturant meal is normally distributed and that the standard deviation is $5. Complete parts (a) through (c) below a. What is the probability that a randomly selected person spent more than $24? P(X> $24) = 0.3446 (Round to four decimal places as needed.) b. What is the probability that a randomly selected person spent between $8 and $19? P($8

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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**Title: Understanding Probability and Normal Distribution in Restaurant Spending** **Overview:** This educational material explores the probability distribution of meal costs at a restaurant. Consumers spent an average of $22 on a meal. The spending is assumed to be normally distributed with a standard deviation of $5. We will examine various probabilities and intervals related to this data. **Content:** 1. **Probability of Spending More Than $24:** - **Question:** What is the probability that a randomly selected person spent more than $24? - **Solution:** \[ P(X > \$24) = 0.3446 \] (Rounded to four decimal places as needed) 2. **Probability of Spending Between $8 and $19:** - **Question:** What is the probability that a randomly selected person spent between $8 and $19? - **Solution:** \[ P(\$8 < X < \$19) = 0.2717 \] (Rounded to four decimal places as needed) 3. **Middle 95% Interval of Spending:** - **Question:** Between what two values, symmetrically distributed around the mean, will the middle 95% of the amounts of cash spent fall? - **Solution:** The middle 95% of the amounts of cash spent will fall between \(X = \$ \_\_\_\) and \(X = \$ \_\_\_\). (Round to the nearest cent as needed) **Conclusion:** These calculations demonstrate the application of the normal distribution to real-world spending scenarios, helping to predict the probability of certain spending behaviors within a given mean and standard deviation framework.
Transcribed Image Text:**Title: Understanding Probability and Normal Distribution in Restaurant Spending** **Overview:** This educational material explores the probability distribution of meal costs at a restaurant. Consumers spent an average of $22 on a meal. The spending is assumed to be normally distributed with a standard deviation of $5. We will examine various probabilities and intervals related to this data. **Content:** 1. **Probability of Spending More Than $24:** - **Question:** What is the probability that a randomly selected person spent more than $24? - **Solution:** \[ P(X > \$24) = 0.3446 \] (Rounded to four decimal places as needed) 2. **Probability of Spending Between $8 and $19:** - **Question:** What is the probability that a randomly selected person spent between $8 and $19? - **Solution:** \[ P(\$8 < X < \$19) = 0.2717 \] (Rounded to four decimal places as needed) 3. **Middle 95% Interval of Spending:** - **Question:** Between what two values, symmetrically distributed around the mean, will the middle 95% of the amounts of cash spent fall? - **Solution:** The middle 95% of the amounts of cash spent will fall between \(X = \$ \_\_\_\) and \(X = \$ \_\_\_\). (Round to the nearest cent as needed) **Conclusion:** These calculations demonstrate the application of the normal distribution to real-world spending scenarios, helping to predict the probability of certain spending behaviors within a given mean and standard deviation framework.
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