Answer :
Sure, let's tackle this question step by step.
a. Distribution of [tex]\( X \)[/tex]:
The problem tells us that the average number of miles (in thousands) that a car's tire will function before needing replacement is 65, with a standard deviation of 18 miles. Since we're assuming a normal distribution, we can express this as:
[tex]\[ X \sim N(65, 18) \][/tex]
b. Distribution of [tex]\(\bar{x}\)[/tex]:
When we take a sample, in this case, 48 randomly selected tires, the distribution of the sample mean [tex]\(\bar{x}\)[/tex] is also normal. The mean of this distribution is the same as the population mean, 65. However, the standard deviation is the population standard deviation divided by the square root of the sample size [tex]\(n\)[/tex]. Therefore, the standard deviation for [tex]\(\bar{x}\)[/tex] is:
[tex]\[ \frac{18}{\sqrt{48}} \approx 2.5981 \][/tex]
So, the distribution of the sample mean is:
[tex]\[ \bar{x} \sim N(65, 2.5981) \][/tex]
c. Probability for an Individual Tire:
We want to find the probability that a randomly selected tire will need replacement between 66.9 and 68.7 thousand miles. To do this, we calculate the z-scores for 66.9 and 68.7, using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
For 66.9, [tex]\( z = \frac{66.9 - 65}{18} \)[/tex]
For 68.7, [tex]\( z = \frac{68.7 - 65}{18} \)[/tex]
The probability is the difference between the cumulative probabilities from these z-scores:
The probability that a tire will need replacement between 66.9 and 68.7 thousand miles is approximately 0.0394.
d. Probability for the Average of 48 Tires:
Now, we find the probability that the average mileage for 48 tires lies between 66.9 and 68.7. We use the standard deviation of the sample mean ([tex]\(2.5981\)[/tex]) to calculate the z-scores:
For 66.9, the z-score is calculated using:
[tex]\[ z = \frac{66.9 - 65}{2.5981} \][/tex]
For 68.7, the z-score is:
[tex]\[ z = \frac{68.7 - 65}{2.5981} \][/tex]
The probability that the average mileage will be between 66.9 and 68.7 for 48 tires is approximately 0.1551.
e. Is the Normality Assumption Necessary?
No, the normality assumption is not necessary. This is because, according to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal regardless of the population distribution when the sample size is large (typically [tex]\(n \geq 30\)[/tex]), which is the case here with a sample size of 48.
a. Distribution of [tex]\( X \)[/tex]:
The problem tells us that the average number of miles (in thousands) that a car's tire will function before needing replacement is 65, with a standard deviation of 18 miles. Since we're assuming a normal distribution, we can express this as:
[tex]\[ X \sim N(65, 18) \][/tex]
b. Distribution of [tex]\(\bar{x}\)[/tex]:
When we take a sample, in this case, 48 randomly selected tires, the distribution of the sample mean [tex]\(\bar{x}\)[/tex] is also normal. The mean of this distribution is the same as the population mean, 65. However, the standard deviation is the population standard deviation divided by the square root of the sample size [tex]\(n\)[/tex]. Therefore, the standard deviation for [tex]\(\bar{x}\)[/tex] is:
[tex]\[ \frac{18}{\sqrt{48}} \approx 2.5981 \][/tex]
So, the distribution of the sample mean is:
[tex]\[ \bar{x} \sim N(65, 2.5981) \][/tex]
c. Probability for an Individual Tire:
We want to find the probability that a randomly selected tire will need replacement between 66.9 and 68.7 thousand miles. To do this, we calculate the z-scores for 66.9 and 68.7, using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
For 66.9, [tex]\( z = \frac{66.9 - 65}{18} \)[/tex]
For 68.7, [tex]\( z = \frac{68.7 - 65}{18} \)[/tex]
The probability is the difference between the cumulative probabilities from these z-scores:
The probability that a tire will need replacement between 66.9 and 68.7 thousand miles is approximately 0.0394.
d. Probability for the Average of 48 Tires:
Now, we find the probability that the average mileage for 48 tires lies between 66.9 and 68.7. We use the standard deviation of the sample mean ([tex]\(2.5981\)[/tex]) to calculate the z-scores:
For 66.9, the z-score is calculated using:
[tex]\[ z = \frac{66.9 - 65}{2.5981} \][/tex]
For 68.7, the z-score is:
[tex]\[ z = \frac{68.7 - 65}{2.5981} \][/tex]
The probability that the average mileage will be between 66.9 and 68.7 for 48 tires is approximately 0.1551.
e. Is the Normality Assumption Necessary?
No, the normality assumption is not necessary. This is because, according to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal regardless of the population distribution when the sample size is large (typically [tex]\(n \geq 30\)[/tex]), which is the case here with a sample size of 48.
