The daily sales at a convenience store produce a normal distribution with a mean of $1,990 and a standar deviation of $179. Determine the probability that the sales on a given day at this store are more than $2,024. O 0.5753 0.8235 00.3192 004247
The daily sales at a convenience store produce a normal distribution with a mean of $1,990 and a standar deviation of $179. Determine the probability that the sales on a given day at this store are more than $2,024. O 0.5753 0.8235 00.3192 004247
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
17).
![### Probability in Normal Distribution
**Problem Statement**: The daily sales at a convenience store produce a normal distribution with a mean of $1,990 and a standard deviation of $179. Determine the probability that the sales on a given day at this store are more than $2,024.
#### Multiple Choice Options:
- ○ 0.5753
- ○ 0.8235
- ○ 0.3192
- ○ 0.4247
#### Explanation:
In this problem, we need to calculate the probability that daily sales exceed $2,024. To solve this, we use the properties of the normal distribution.
**Steps:**
1. **Calculate the Z-score**: The Z-score tells us how many standard deviations away $2,024 is from the mean.
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \(X\) is the value we are interested in ($2,024)
- \(\mu\) is the mean ($1,990)
- \(\sigma\) is the standard deviation ($179)
2. **Plug in the values**:
\[
Z = \frac{(2,024 - 1,990)}{179} = \frac{34}{179} \approx 0.190
\]
3. **Use the Z-score to find the probability**:
With a Z-score of approximately 0.190, we refer to Z-tables or use software tools to find the cumulative probability for Z = 0.190.
- The cumulative probability for Z = 0.190 is approximately 0.5753.
4. **Find the desired probability**:
Since we want the probability of sales being more than $2,024, we subtract the cumulative probability from 1:
\[
P(X > 2,024) = 1 - P(Z \leq 0.190) = 1 - 0.5753 = 0.4247
\]
Therefore, the probability that the sales on a given day are more than $2,024 is **0.4247**.
#### Correct Answer:
- ○ 0.5753
- ○ 0.8235
- ○ 0.3192
- ● 0.4247 (Selected)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc94cb5d1-997a-49d0-b0ef-9ec02bf0a187%2Fd07fd3e3-2578-4829-88b1-766d3d2d5dd1%2Fh94cwkd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Probability in Normal Distribution
**Problem Statement**: The daily sales at a convenience store produce a normal distribution with a mean of $1,990 and a standard deviation of $179. Determine the probability that the sales on a given day at this store are more than $2,024.
#### Multiple Choice Options:
- ○ 0.5753
- ○ 0.8235
- ○ 0.3192
- ○ 0.4247
#### Explanation:
In this problem, we need to calculate the probability that daily sales exceed $2,024. To solve this, we use the properties of the normal distribution.
**Steps:**
1. **Calculate the Z-score**: The Z-score tells us how many standard deviations away $2,024 is from the mean.
\[
Z = \frac{(X - \mu)}{\sigma}
\]
Where:
- \(X\) is the value we are interested in ($2,024)
- \(\mu\) is the mean ($1,990)
- \(\sigma\) is the standard deviation ($179)
2. **Plug in the values**:
\[
Z = \frac{(2,024 - 1,990)}{179} = \frac{34}{179} \approx 0.190
\]
3. **Use the Z-score to find the probability**:
With a Z-score of approximately 0.190, we refer to Z-tables or use software tools to find the cumulative probability for Z = 0.190.
- The cumulative probability for Z = 0.190 is approximately 0.5753.
4. **Find the desired probability**:
Since we want the probability of sales being more than $2,024, we subtract the cumulative probability from 1:
\[
P(X > 2,024) = 1 - P(Z \leq 0.190) = 1 - 0.5753 = 0.4247
\]
Therefore, the probability that the sales on a given day are more than $2,024 is **0.4247**.
#### Correct Answer:
- ○ 0.5753
- ○ 0.8235
- ○ 0.3192
- ● 0.4247 (Selected)
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