To determine the shape of the sampling distribution for [tex]\(\bar{x}_A-\bar{x}_C\)[/tex], we need to consider the properties of the underlying population distributions and the Central Limit Theorem (CLT).
### Step-by-Step Solution:
1. Population Distributions:
- Alex's times are normally distributed with a mean of 5.28 minutes and a standard deviation of 0.38 seconds.
- Chris's times are normally distributed with a mean of 5.45 seconds and a standard deviation of 0.2 seconds.
2. Conversions:
- Convert the standard deviations from seconds to minutes:
[tex]\[
\sigma_{\text{Alex}} = \frac{0.38}{60} \text{ minutes}
\][/tex]
[tex]\[
\sigma_{\text{Chris}} = \frac{0.2}{60} \text{ minutes}
\][/tex]
3. Sample Sizes:
- Sample size for Alex ([tex]\(n_A\)[/tex]): 10
- Sample size for Chris ([tex]\(n_C\)[/tex]): 15
4. Central Limit Theorem (CLT):
- According to the CLT, if the population distribution is normal, the sampling distribution of the sample mean will also be normal regardless of the sample size. Since both Alex's and Chris's times are normally distributed, the sampling distributions of [tex]\(\bar{x}_A\)[/tex] and [tex]\(\bar{x}_C\)[/tex] will also be normal.
5. Difference in Means:
- The difference in the mean times, [tex]\(\bar{x}_A - \bar{x}_C\)[/tex], involves the subtraction of two normally distributed variables:
[tex]\[
\bar{x}_A - \bar{x}_C
\][/tex]
- The difference of two normally distributed variables follows a normal distribution.
Conclusion:
Considering that both population distributions (Alex's and Chris's times) are normal, the difference in their sample means, [tex]\(\bar{x}_A - \bar{x}_C\)[/tex], will also follow a normal distribution. Thus, the shape of the sampling distribution for [tex]\(\bar{x}_A - \bar{x}_C\)[/tex] is normal.
Answer:
Normal, because both population distributions are Normal.
