a. What is the distribution of X? X N 68 12 N 68 1.7321 ✓) 0 08 b. What is the distribution of ? c. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.3 and 68.6. d. For the 48 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 67.3 and 68.6. e. For part d), is the assumption that the distribution is normal necessary? No Yes of بھی بھی
a. What is the distribution of X? X N 68 12 N 68 1.7321 ✓) 0 08 b. What is the distribution of ? c. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.3 and 68.6. d. For the 48 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 67.3 and 68.6. e. For part d), is the assumption that the distribution is normal necessary? No Yes of بھی بھی
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
Help with c and d
![**Distribution of Tire Mileages Before Replacement**
The following problem explores the statistical distribution of the mileage (in thousands) that a car's tire can function before needing replacement. Assume that the average number of miles before replacement is 68, with a standard deviation of 12. Given that 48 randomly selected tires are tested, we will round all answers to 4 decimal places and assume a normal distribution.
a. **Distribution of X**
Determine the distribution of the number of miles, X:
\( \bar{X} \sim N(\mu, \sigma^2) \)
Given:
\[
\mu = 68 \quad \sigma = 12
\]
b. **Distribution of the Sample Mean (x̄)**
Determine the distribution of the sample mean, x̄:
\( \bar{x} \sim N\left( \mu, \frac{\sigma}{\sqrt{N}} \right) \)
Given:
\[
\mu = 68 \quad \frac{12}{\sqrt{48}} = 1.7321
\]
c. **Probability Calculation**
If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.3 and 68.6:
\[
P(67.3 \leq X \leq 68.6)
\]
d. **Probability for the Sample Mean of 48 Tires**
For the 48 tires tested, find the probability that the average miles (in thousands) before needing replacement is between 67.3 and 68.6:
\[
P(67.3 \leq \bar{x} \leq 68.6)
\]
e. **Normal Distribution Assumption**
For part d), is the assumption that the distribution is normal necessary?
\[
\text{Yes}
\]
**Hint:**
- Check the relevant statistical theory and formulas.
- Utilize your textbook for detailed explanations and examples if needed.
Consider the above distribution and calculations when assessing the lifespan of tires and ensure that all assumptions and statistical methods are appropriately applied.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbc3b72c7-eb99-406f-83fe-82fc41b2a8b5%2F13ffe276-bc1d-416c-b3fa-d568434427a0%2Fu050y2e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Distribution of Tire Mileages Before Replacement**
The following problem explores the statistical distribution of the mileage (in thousands) that a car's tire can function before needing replacement. Assume that the average number of miles before replacement is 68, with a standard deviation of 12. Given that 48 randomly selected tires are tested, we will round all answers to 4 decimal places and assume a normal distribution.
a. **Distribution of X**
Determine the distribution of the number of miles, X:
\( \bar{X} \sim N(\mu, \sigma^2) \)
Given:
\[
\mu = 68 \quad \sigma = 12
\]
b. **Distribution of the Sample Mean (x̄)**
Determine the distribution of the sample mean, x̄:
\( \bar{x} \sim N\left( \mu, \frac{\sigma}{\sqrt{N}} \right) \)
Given:
\[
\mu = 68 \quad \frac{12}{\sqrt{48}} = 1.7321
\]
c. **Probability Calculation**
If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.3 and 68.6:
\[
P(67.3 \leq X \leq 68.6)
\]
d. **Probability for the Sample Mean of 48 Tires**
For the 48 tires tested, find the probability that the average miles (in thousands) before needing replacement is between 67.3 and 68.6:
\[
P(67.3 \leq \bar{x} \leq 68.6)
\]
e. **Normal Distribution Assumption**
For part d), is the assumption that the distribution is normal necessary?
\[
\text{Yes}
\]
**Hint:**
- Check the relevant statistical theory and formulas.
- Utilize your textbook for detailed explanations and examples if needed.
Consider the above distribution and calculations when assessing the lifespan of tires and ensure that all assumptions and statistical methods are appropriately applied.
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