A manufacturer knows that their items have a normally distributed lifespan, with a mean of 4.7 years, and standard deviation of 1.2 years. If you randomly purchase 19 items, what is the probability that their life will be longer than 5 years? (Give answer to 4 decimal places)
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 4.7 years, and standard deviation of 1.2 years. If you randomly purchase 19 items, what is the probability that their life will be longer than 5 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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**Probability and Statistics Exercise**
**Exercise Description:**
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 4.7 years, and standard deviation of 1.2 years.
- If you randomly purchase 19 items, what is the probability that their life will be longer than 5 years?
*(Give answer to 4 decimal places)*
**Instructions:**
1. To solve this problem, use the properties of normal distribution.
2. Begin by standardizing the distribution.
3. Use the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where:
- \(X\) = 5 years
- \(\mu\) (mean) = 4.7 years
- \(\sigma\) (standard deviation) = 1.2 years
4. Find the Z-score for \(X = 5\) years.
5. Look up the corresponding probability for the calculated Z-score using the Z-table.
6. Subtract this probability from 1 to find the probability of the items lasting longer than 5 years.
**Example Calculation:**
1. Calculate the Z-score for \(X = 5\):
\[
Z = \frac{5 - 4.7}{1.2} = \frac{0.3}{1.2} = 0.25
\]
2. Use the Z-table to find the probability corresponding to \(Z = 0.25\).
3. Find the cumulative area under the curve to the left of \(Z = 0.25\) (let's assume it is 0.5987).
4. Subtract this value from 1 to get the probability of items lasting longer than 5 years:
\[
P(X > 5) = 1 - P(X \leq 5) = 1 - 0.5987 = 0.4013
\]
Thus, the probability that the lifespan of the items will be longer than 5 years is approximately 0.4013.
---
**Note:**
Make sure you explain each step in detail for students to follow the logic clearly. Encourage them to use statistical tables or software calculators to verify probabilities.
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Transcribed Image Text:---
**Probability and Statistics Exercise**
**Exercise Description:**
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 4.7 years, and standard deviation of 1.2 years.
- If you randomly purchase 19 items, what is the probability that their life will be longer than 5 years?
*(Give answer to 4 decimal places)*
**Instructions:**
1. To solve this problem, use the properties of normal distribution.
2. Begin by standardizing the distribution.
3. Use the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Where:
- \(X\) = 5 years
- \(\mu\) (mean) = 4.7 years
- \(\sigma\) (standard deviation) = 1.2 years
4. Find the Z-score for \(X = 5\) years.
5. Look up the corresponding probability for the calculated Z-score using the Z-table.
6. Subtract this probability from 1 to find the probability of the items lasting longer than 5 years.
**Example Calculation:**
1. Calculate the Z-score for \(X = 5\):
\[
Z = \frac{5 - 4.7}{1.2} = \frac{0.3}{1.2} = 0.25
\]
2. Use the Z-table to find the probability corresponding to \(Z = 0.25\).
3. Find the cumulative area under the curve to the left of \(Z = 0.25\) (let's assume it is 0.5987).
4. Subtract this value from 1 to get the probability of items lasting longer than 5 years:
\[
P(X > 5) = 1 - P(X \leq 5) = 1 - 0.5987 = 0.4013
\]
Thus, the probability that the lifespan of the items will be longer than 5 years is approximately 0.4013.
---
**Note:**
Make sure you explain each step in detail for students to follow the logic clearly. Encourage them to use statistical tables or software calculators to verify probabilities.
---
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