On average, Americans spent $940 dollars on holiday gifts in 2019, with a standard deviation of $210. If we define the random variable Y to be the sum of the dollars spent on holiday gifts by 70 randomly chosen Americans, find the probability that Y is greater than $68,000. 0.034 0.105 0.895 0.560
On average, Americans spent $940 dollars on holiday gifts in 2019, with a standard deviation of $210. If we define the random variable Y to be the sum of the dollars spent on holiday gifts by 70 randomly chosen Americans, find the probability that Y is greater than $68,000. 0.034 0.105 0.895 0.560
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
![### Probability Problem Related to Holiday Spending
---
**Problem Statement:**
On average, Americans spent $940 dollars on holiday gifts in 2019, with a standard deviation of $210. If we define the random variable \( Y \) to be the sum of the dollars spent on holiday gifts by 70 randomly chosen Americans, find the probability that \( Y \) is greater than $68,000.
---
**Answer Choices:**
- 0.034
- 0.105
- 0.895
- 0.560
---
To solve this problem, we need to understand the distribution of the sum of dollars spent on holiday gifts by 70 people. Since each person’s spending is independent and normally distributed, the sum \( Y \) will also be normally distributed.
**Steps to solve:**
1. **Find \( \mu_Y \) (Mean of \( Y \))**
\[
\mu_Y = n \times \mu = 70 \times 940 = 65800
\]
2. **Find \( \sigma_Y \) (Standard Deviation of \( Y \))**
\[
\sigma_Y = \sqrt{n} \times \sigma = \sqrt{70} \times 210 \approx 1753.632
\]
3. **Calculate the Z-score for \( Y = 68000 \)**
\[
Z = \frac{68000 - 65800}{1753.632} \approx 1.254
\]
4. **Find the probability**
Using Z-tables or a standard normal distribution calculator, find the probability that \( Z \) is greater than 1.254.
\[
P(Z > 1.254) = 1 - P(Z < 1.254) \approx 1 - 0.895 = 0.105
\]
Hence, the probability that \( Y \) is greater than $68,000 is approximately 0.105.
**Correct Answer:**
- 0.105](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7e444e2-a441-4a28-bca8-26e04764874a%2F4646c4f9-f11b-459f-8e2b-944a323e7a42%2Fl9lh1u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability Problem Related to Holiday Spending
---
**Problem Statement:**
On average, Americans spent $940 dollars on holiday gifts in 2019, with a standard deviation of $210. If we define the random variable \( Y \) to be the sum of the dollars spent on holiday gifts by 70 randomly chosen Americans, find the probability that \( Y \) is greater than $68,000.
---
**Answer Choices:**
- 0.034
- 0.105
- 0.895
- 0.560
---
To solve this problem, we need to understand the distribution of the sum of dollars spent on holiday gifts by 70 people. Since each person’s spending is independent and normally distributed, the sum \( Y \) will also be normally distributed.
**Steps to solve:**
1. **Find \( \mu_Y \) (Mean of \( Y \))**
\[
\mu_Y = n \times \mu = 70 \times 940 = 65800
\]
2. **Find \( \sigma_Y \) (Standard Deviation of \( Y \))**
\[
\sigma_Y = \sqrt{n} \times \sigma = \sqrt{70} \times 210 \approx 1753.632
\]
3. **Calculate the Z-score for \( Y = 68000 \)**
\[
Z = \frac{68000 - 65800}{1753.632} \approx 1.254
\]
4. **Find the probability**
Using Z-tables or a standard normal distribution calculator, find the probability that \( Z \) is greater than 1.254.
\[
P(Z > 1.254) = 1 - P(Z < 1.254) \approx 1 - 0.895 = 0.105
\]
Hence, the probability that \( Y \) is greater than $68,000 is approximately 0.105.
**Correct Answer:**
- 0.105
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