Question 1198637: The average number of miles (in thousands) that a car's tire will function before needing replacement is 65 and the standard deviation is 14. Suppose that 11 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.
What is the distribution of X?X~ N(65Correct,14Correct)
What is the distribution of ¯x?¯x~ N(65Correct,4.2212Correct)
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 62.9 and 69.1. [-6.05 Incorrect]
For the 11 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 62.9 and 69.1.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! ### Step 1: Understanding the Problem
We are given:
- The population mean: \( \mu = 65 \) (in thousands of miles)
- The population standard deviation: \( \sigma = 14 \) (in thousands of miles)
- Sample size: \( n = 11 \)
- Assumption: The distribution is normal.
---
### **Part 1: Distribution of \( X \)**
The number of miles \( X \) for a randomly selected tire follows a normal distribution:
\[
X \sim N(\mu = 65, \sigma = 14)
\]
---
### **Part 2: Distribution of \( \bar{x} \)**
The sampling distribution of the sample mean \( \bar{x} \) also follows a normal distribution:
\[
\bar{x} \sim N\left(\mu = 65, \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\right)
\]
\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{14}{\sqrt{11}} \approx 4.2212
\]
Thus:
\[
\bar{x} \sim N(65, 4.2212)
\]
---
### **Part 3: Probability for a Single Tire (Range: 62.9 to 69.1)**
We calculate the probability \( P(62.9 \leq X \leq 69.1) \) using the standard normal distribution:
#### Step 1: Standardize the values
The z-score formula is:
\[
Z = \frac{X - \mu}{\sigma}
\]
For \( X = 62.9 \):
\[
Z = \frac{62.9 - 65}{14} = \frac{-2.1}{14} \approx -0.15
\]
For \( X = 69.1 \):
\[
Z = \frac{69.1 - 65}{14} = \frac{4.1}{14} \approx 0.29
\]
#### Step 2: Find probabilities from z-scores
Using standard normal tables or a calculator:
\[
P(Z \leq -0.15) \approx 0.4404, \quad P(Z \leq 0.29) \approx 0.6141
\]
#### Step 3: Compute the probability
\[
P(62.9 \leq X \leq 69.1) = P(Z \leq 0.29) - P(Z \leq -0.15)
\]
\[
P(62.9 \leq X \leq 69.1) = 0.6141 - 0.4404 = 0.1737
\]
---
### **Part 4: Probability for the Sample Mean (Range: 62.9 to 69.1)**
We calculate \( P(62.9 \leq \bar{x} \leq 69.1) \) using the sampling distribution:
#### Step 1: Standardize the values
The z-score formula for the sample mean is:
\[
Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}
\]
For \( \bar{x} = 62.9 \):
\[
Z = \frac{62.9 - 65}{4.2212} = \frac{-2.1}{4.2212} \approx -0.4973
\]
For \( \bar{x} = 69.1 \):
\[
Z = \frac{69.1 - 65}{4.2212} = \frac{4.1}{4.2212} \approx 0.9711
\]
#### Step 2: Find probabilities from z-scores
Using standard normal tables or a calculator:
\[
P(Z \leq -0.4973) \approx 0.3097, \quad P(Z \leq 0.9711) \approx 0.8330
\]
#### Step 3: Compute the probability
\[
P(62.9 \leq \bar{x} \leq 69.1) = P(Z \leq 0.9711) - P(Z \leq -0.4973)
\]
\[
P(62.9 \leq \bar{x} \leq 69.1) = 0.8330 - 0.3097 = 0.5233
\]
---
### Final Answers:
1. \( X \sim N(65, 14) \)
2. \( \bar{x} \sim N(65, 4.2212) \)
3. Probability for a single tire: \( P(62.9 \leq X \leq 69.1) = 0.1737 \)
4. Probability for the sample mean: \( P(62.9 \leq \bar{x} \leq 69.1) = 0.5233 \)
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