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The daily temperatures for the winter months in Virginia are Normally distributed with a mean of [tex]\(59^{\circ} F\)[/tex] and a standard deviation of [tex]\(10^{\circ} F\)[/tex]. The daily temperatures for the winter months in California are Normally distributed with a mean of [tex]\(64^{\circ} F\)[/tex] and a standard deviation of [tex]\(12^{\circ} F\)[/tex]. Random samples of 10 temperatures are taken from the winter months for both Virginia and California. The mean temperature is recorded for both samples. Let [tex]\(\bar{x}_V - \bar{x}_C\)[/tex] represent the difference in the mean temperatures for the winter months in Virginia and California. Which of the following represents the standard deviation of the sampling distribution for [tex]\(\bar{x}_V - \bar{x}_C\)[/tex]?

A. 1.6
B. 2.2
C. 4.9
D. 7.0


Sagot :

To determine the standard deviation of the sampling distribution for [tex]\(\bar{x}_v - \bar{x}_c\)[/tex], we must consider the properties of the means and standard deviations for each sample. Here's how you can approach this problem step-by-step:

1. Identify the parameters:
- For Virginia:
- Mean [tex]\( \mu_V = 59 \)[/tex]
- Standard deviation [tex]\( \sigma_V = 10 \)[/tex]
- Sample size [tex]\( n_V = 10 \)[/tex]
- For California:
- Mean [tex]\( \mu_C = 64 \)[/tex]
- Standard deviation [tex]\( \sigma_C = 12 \)[/tex]
- Sample size [tex]\( n_C = 10 \)[/tex]

2. Understand the formula:

The standard deviation of the sampling distribution of the difference between two means, [tex]\(\bar{x}_v - \bar{x}_c\)[/tex], is found using the formula:
[tex]\[ \sigma_{\bar{x}_v - \bar{x}_c} = \sqrt{\left(\frac{\sigma_V^2}{n_V}\right) + \left(\frac{\sigma_C^2}{n_C}\right)} \][/tex]

3. Substitute the values:
- Compute the variance for each state:
- For Virginia: [tex]\(\sigma_V^2 = (10)^2 = 100\)[/tex]
- For California: [tex]\(\sigma_C^2 = (12)^2 = 144\)[/tex]

- Now, calculate the standard error for each sample:
- For Virginia: [tex]\(\frac{\sigma_V^2}{n_V} = \frac{100}{10} = 10\)[/tex]
- For California: [tex]\(\frac{\sigma_C^2}{n_C} = \frac{144}{10} = 14.4\)[/tex]

4. Sum the variances and take the square root:
[tex]\[ \sigma_{\bar{x}_v - \bar{x}_c} = \sqrt{10 + 14.4} = \sqrt{24.4} \][/tex]

5. Calculate the result:
[tex]\[ \sigma_{\bar{x}_v - \bar{x}_c} = \sqrt{24.4} \approx 4.9396356140913875 \][/tex]

Hence, the standard deviation of the sampling distribution for [tex]\(\bar{x}_v - \bar{x}_c\)[/tex] is approximately [tex]\(4.9\)[/tex].

Considering the options provided:

1. 6
2. 2
3. 4.9
4. 7.0

The correct answer is:
[tex]\[ \boxed{4.9} \][/tex]