A Manufacturer knows that their items have a normally distributed lifespan, with a mean of 5 years, and standard deviation of 0.5 years. If you randomly purchase one item, what is the probability that it will last longer than 6 years? (Give answer to 4 decimal places)
A Manufacturer knows that their items have a normally distributed lifespan, with a mean of 5 years, and standard deviation of 0.5 years. If you randomly purchase one item, what is the probability that it will last longer than 6 years? (Give answer to 4 decimal places)
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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![### Problem Statement
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5 years, and a standard deviation of 0.5 years.
If you randomly purchase one item, what is the probability that it will last longer than 6 years? (Give your answer to 4 decimal places.)
### Explanation and Solution
We are given:
- Mean (μ) = 5 years
- Standard deviation (σ) = 0.5 years
We need to find the probability that the lifespan of an item (X) is greater than 6 years (P(X > 6)).
#### Step-by-Step Solution
1. **Convert the problem into a standard normal distribution problem.**
We will transform the variable \( X \) into a standard normal variable \( Z \):
\[
Z = \frac{X - μ}{σ}
\]
Substituting the given values:
\[
Z = \frac{6 - 5}{0.5} = 2
\]
2. **Find the area to the right of Z = 2 in the standard normal distribution.**
The standard normal distribution table gives us the area to the left of a given Z value. For Z = 2, the area to the left is approximately 0.9772.
Therefore, the area to the right (P(Z > 2)) is:
\[
P(Z > 2) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228
\]
3. **Final Answer**
The probability that an item will last longer than 6 years is 0.0228.
### Conclusion
So, if you randomly purchase one item, the probability that it will last longer than 6 years is **0.0228** (to four decimal places).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5176b0c-078c-4393-8a73-67cd4f62d79f%2F21aebac4-20c9-499e-94b1-9c582c88a4bc%2F7gvxodl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 5 years, and a standard deviation of 0.5 years.
If you randomly purchase one item, what is the probability that it will last longer than 6 years? (Give your answer to 4 decimal places.)
### Explanation and Solution
We are given:
- Mean (μ) = 5 years
- Standard deviation (σ) = 0.5 years
We need to find the probability that the lifespan of an item (X) is greater than 6 years (P(X > 6)).
#### Step-by-Step Solution
1. **Convert the problem into a standard normal distribution problem.**
We will transform the variable \( X \) into a standard normal variable \( Z \):
\[
Z = \frac{X - μ}{σ}
\]
Substituting the given values:
\[
Z = \frac{6 - 5}{0.5} = 2
\]
2. **Find the area to the right of Z = 2 in the standard normal distribution.**
The standard normal distribution table gives us the area to the left of a given Z value. For Z = 2, the area to the left is approximately 0.9772.
Therefore, the area to the right (P(Z > 2)) is:
\[
P(Z > 2) = 1 - P(Z \leq 2) = 1 - 0.9772 = 0.0228
\]
3. **Final Answer**
The probability that an item will last longer than 6 years is 0.0228.
### Conclusion
So, if you randomly purchase one item, the probability that it will last longer than 6 years is **0.0228** (to four decimal places).
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