Answer :
To determine the shape of the sampling distribution for the difference in mean times between Alex and Chris ([tex]\(\bar{x}_A - \bar{x}_C\)[/tex]), let's carefully analyze the distributions of their times:
1. Alex's Times:
- Mean ([tex]\(\mu_A\)[/tex]): 5.28 minutes
- Standard Deviation ([tex]\(\sigma_A\)[/tex]): 0.38 seconds
- Distribution: Normally distributed
2. Chris's Times:
- Mean ([tex]\(\mu_C\)[/tex]): 5.45 minutes
- Standard Deviation ([tex]\(\sigma_C\)[/tex]): 0.2 seconds
- Distribution: Normally distributed
Given these details, we have the following key points to consider:
### Distribution Properties:
- Both Alex’s and Chris’s running times are normally distributed, which means each of their time distributions follows a bell curve.
- The Central Limit Theorem (CLT) states that for any population with a finite mean and variance, as the sample size grows, the distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution.
Since we are dealing with differences of means:
- The difference between two normally distributed variables is itself normally distributed. That is because if [tex]\(X \sim N(\mu_x, \sigma_x)\)[/tex] and [tex]\(Y \sim N(\mu_y, \sigma_y)\)[/tex], then [tex]\(X - Y \sim N(\mu_x - \mu_y, \sqrt{\sigma_x^2 + \sigma_y^2})\)[/tex].
### Analysis of the Sampling Distribution:
- [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] represents the difference in the sample means of Alex’s and Chris’s running times.
- Because both of these sample means come from normally distributed populations, the distribution of [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] will also be normally distributed.
### Possible Answers:
1. Normal, because both population distributions are Normal.
- This is correct because, as stated, the difference of sample means from normally distributed populations will itself be normally distributed.
2. Uniform, because both sample sizes are less than 30.
- This is incorrect. The shape of the sampling distribution is determined by the original distribution of the data (Normal in this case), not by the sample size alone.
3. Skewed right, because the difference in times cannot be negative.
- This is incorrect as the times can be both positive and negative. Regardless, the populations are normally distributed, not skewed.
4. Skewed left, because the sample sizes are less than 30 and the sampling variability is unknown.
- This is incorrect because the population distributions are Normal, therefore their sample means’ difference will also be Normally distributed regardless of the sample size being less than 30.
### Conclusion:
The sampling distribution of [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] is Normal, because both population distributions are Normal.
Therefore, the correct answer is: Normal, because both population distributions are Normal.
1. Alex's Times:
- Mean ([tex]\(\mu_A\)[/tex]): 5.28 minutes
- Standard Deviation ([tex]\(\sigma_A\)[/tex]): 0.38 seconds
- Distribution: Normally distributed
2. Chris's Times:
- Mean ([tex]\(\mu_C\)[/tex]): 5.45 minutes
- Standard Deviation ([tex]\(\sigma_C\)[/tex]): 0.2 seconds
- Distribution: Normally distributed
Given these details, we have the following key points to consider:
### Distribution Properties:
- Both Alex’s and Chris’s running times are normally distributed, which means each of their time distributions follows a bell curve.
- The Central Limit Theorem (CLT) states that for any population with a finite mean and variance, as the sample size grows, the distribution of the sample mean approaches a normal distribution, regardless of the shape of the population distribution.
Since we are dealing with differences of means:
- The difference between two normally distributed variables is itself normally distributed. That is because if [tex]\(X \sim N(\mu_x, \sigma_x)\)[/tex] and [tex]\(Y \sim N(\mu_y, \sigma_y)\)[/tex], then [tex]\(X - Y \sim N(\mu_x - \mu_y, \sqrt{\sigma_x^2 + \sigma_y^2})\)[/tex].
### Analysis of the Sampling Distribution:
- [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] represents the difference in the sample means of Alex’s and Chris’s running times.
- Because both of these sample means come from normally distributed populations, the distribution of [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] will also be normally distributed.
### Possible Answers:
1. Normal, because both population distributions are Normal.
- This is correct because, as stated, the difference of sample means from normally distributed populations will itself be normally distributed.
2. Uniform, because both sample sizes are less than 30.
- This is incorrect. The shape of the sampling distribution is determined by the original distribution of the data (Normal in this case), not by the sample size alone.
3. Skewed right, because the difference in times cannot be negative.
- This is incorrect as the times can be both positive and negative. Regardless, the populations are normally distributed, not skewed.
4. Skewed left, because the sample sizes are less than 30 and the sampling variability is unknown.
- This is incorrect because the population distributions are Normal, therefore their sample means’ difference will also be Normally distributed regardless of the sample size being less than 30.
### Conclusion:
The sampling distribution of [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] is Normal, because both population distributions are Normal.
Therefore, the correct answer is: Normal, because both population distributions are Normal.
