Answer :
Let's tackle this question step-by-step:
a. Distribution of [tex]\( X \)[/tex]:
The random variable [tex]\( X \)[/tex] represents the miles (in thousands) before a car tire needs replacement. Since we are given that the average (mean) is 65 and the standard deviation is 18, we can say that [tex]\( X \)[/tex] follows a normal distribution with mean 65 and standard deviation 18. Therefore, the distribution of [tex]\( X \)[/tex] is:
[tex]\[ X \sim N(65, 18) \][/tex]
b. Distribution of [tex]\( \bar{x} \)[/tex]:
The distribution of the sample mean ([tex]\( \bar{x} \)[/tex]) from a normal distribution [tex]\( N(\mu, \sigma) \)[/tex] with sample size [tex]\( n \)[/tex] is given by:
[tex]\[ \bar{x} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) \][/tex]
Here, the mean ([tex]\(\mu\)[/tex]) is 65, the standard deviation ([tex]\(\sigma\)[/tex]) is 18, and the sample size ([tex]\(n\)[/tex]) is 48. We compute the standard error as [tex]\( \frac{18}{\sqrt{48}} \)[/tex].
So, the distribution of [tex]\( \bar{x} \)[/tex] is:
[tex]\[ \bar{x} \sim N\left(65, \frac{18}{\sqrt{48}}\right) \][/tex]
c. Probability for an individual tire:
To find the probability that a randomly selected tire lasts between 66.9 and 68.7 thousand miles, we need to calculate the z-scores for these values and use the standard normal distribution.
1. Calculate the z-scores:
[tex]\[
z_1 = \frac{66.9 - 65}{18}
\][/tex]
[tex]\[
z_2 = \frac{68.7 - 65}{18}
\][/tex]
2. Find the probability:
Use the z-scores to find the probabilities from the standard normal distribution table and calculate the probability that [tex]\( X \)[/tex] is between 66.9 and 68.7.
d. Probability for the average of 48 tires:
Similarly, find the probability for the average when 48 tires are tested to see if the average miles is between 66.9 and 68.7 thousand miles.
1. Calculate the z-scores using the adjusted standard error:
[tex]\[
z_1 = \frac{66.9 - 65}{\frac{18}{\sqrt{48}}}
\][/tex]
[tex]\[
z_2 = \frac{68.7 - 65}{\frac{18}{\sqrt{48}}}
\][/tex]
2. Find the probability:
Use these z-scores to calculate the corresponding probability from the standard normal distribution.
e. Is the assumption of a normal distribution necessary?
For part (d), the Central Limit Theorem ensures that the sampling distribution of the sample mean will be approximately normal, given that the sample size is large enough (in this case, 48), regardless of the shape of the original distribution. Therefore, the assumption that the distribution is normal is:
- No, the assumption is not necessary because of the Central Limit Theorem.
And that's how you solve each part of the question.
a. Distribution of [tex]\( X \)[/tex]:
The random variable [tex]\( X \)[/tex] represents the miles (in thousands) before a car tire needs replacement. Since we are given that the average (mean) is 65 and the standard deviation is 18, we can say that [tex]\( X \)[/tex] follows a normal distribution with mean 65 and standard deviation 18. Therefore, the distribution of [tex]\( X \)[/tex] is:
[tex]\[ X \sim N(65, 18) \][/tex]
b. Distribution of [tex]\( \bar{x} \)[/tex]:
The distribution of the sample mean ([tex]\( \bar{x} \)[/tex]) from a normal distribution [tex]\( N(\mu, \sigma) \)[/tex] with sample size [tex]\( n \)[/tex] is given by:
[tex]\[ \bar{x} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) \][/tex]
Here, the mean ([tex]\(\mu\)[/tex]) is 65, the standard deviation ([tex]\(\sigma\)[/tex]) is 18, and the sample size ([tex]\(n\)[/tex]) is 48. We compute the standard error as [tex]\( \frac{18}{\sqrt{48}} \)[/tex].
So, the distribution of [tex]\( \bar{x} \)[/tex] is:
[tex]\[ \bar{x} \sim N\left(65, \frac{18}{\sqrt{48}}\right) \][/tex]
c. Probability for an individual tire:
To find the probability that a randomly selected tire lasts between 66.9 and 68.7 thousand miles, we need to calculate the z-scores for these values and use the standard normal distribution.
1. Calculate the z-scores:
[tex]\[
z_1 = \frac{66.9 - 65}{18}
\][/tex]
[tex]\[
z_2 = \frac{68.7 - 65}{18}
\][/tex]
2. Find the probability:
Use the z-scores to find the probabilities from the standard normal distribution table and calculate the probability that [tex]\( X \)[/tex] is between 66.9 and 68.7.
d. Probability for the average of 48 tires:
Similarly, find the probability for the average when 48 tires are tested to see if the average miles is between 66.9 and 68.7 thousand miles.
1. Calculate the z-scores using the adjusted standard error:
[tex]\[
z_1 = \frac{66.9 - 65}{\frac{18}{\sqrt{48}}}
\][/tex]
[tex]\[
z_2 = \frac{68.7 - 65}{\frac{18}{\sqrt{48}}}
\][/tex]
2. Find the probability:
Use these z-scores to calculate the corresponding probability from the standard normal distribution.
e. Is the assumption of a normal distribution necessary?
For part (d), the Central Limit Theorem ensures that the sampling distribution of the sample mean will be approximately normal, given that the sample size is large enough (in this case, 48), regardless of the shape of the original distribution. Therefore, the assumption that the distribution is normal is:
- No, the assumption is not necessary because of the Central Limit Theorem.
And that's how you solve each part of the question.
