One truck from Lakeland Trucking, Inc. can carry a load of [tex]$694.4$[/tex] lb. Records show that the weights of boxes it carries have a mean of [tex]$95$[/tex] lb and a standard deviation of [tex]$21$[/tex] lb. For samples of size [tex]$49$[/tex], find the mean and standard deviation of [tex]$\bar{x}$[/tex].

A. [tex]$\mu_{\bar{x}}=95; \sigma_{\bar{x}}=3$[/tex]
B. [tex]$\mu_{\bar{x}}=95; \sigma_{\bar{x}}=21$[/tex]
C. [tex]$\mu_{\bar{x}}=85; \sigma_{\bar{x}}=3$[/tex]
D. [tex]$\mu_{\bar{x}}=7.937; \sigma_{\bar{x}}=4$[/tex]



Answer :

Alright, let's solve this step-by-step.

We are given the following information:
1. The mean weight ([tex]\(\mu\)[/tex]) of boxes is [tex]\(95 \text{ lbs}\)[/tex].
2. The standard deviation ([tex]\(\sigma\)[/tex]) of the weights of the boxes is [tex]\(21 \text{ lbs}\)[/tex].
3. The sample size ([tex]\(n\)[/tex]) is [tex]\(49\)[/tex].

We are asked to find the mean ([tex]\(\mu_{\bar{x}}\)[/tex]) and the standard deviation ([tex]\(\sigma_{\bar{x}}\)[/tex]) of the sample mean ([tex]\(\bar{x}\)[/tex]) for samples of size 49.

### Step 1: Calculate the Mean of the Sample Mean ([tex]\(\mu_{\bar{x}}\)[/tex])
The mean of the sample mean ([tex]\(\mu_{\bar{x}}\)[/tex]) is the same as the mean of the population (the mean weight of boxes).
[tex]\[ \mu_{\bar{x}} = \mu = 95 \text{ lbs} \][/tex]

### Step 2: Calculate the Standard Deviation of the Sample Mean ([tex]\(\sigma_{\bar{x}}\)[/tex])
The standard deviation of the sample mean ([tex]\(\sigma_{\bar{x}}\)[/tex]) is given by the formula:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
where:
- [tex]\(\sigma\)[/tex] is the population standard deviation,
- [tex]\(n\)[/tex] is the sample size.

Plugging in the values:
[tex]\[ \sigma_{\bar{x}} = \frac{21}{\sqrt{49}} = \frac{21}{7} = 3 \text{ lbs} \][/tex]

So, we have found:
[tex]\[ \mu_{\bar{x}} = 95 \text{ lbs} \][/tex]
[tex]\[ \sigma_{\bar{x}} = 3 \text{ lbs} \][/tex]

### Step 3: Compare with the Multiple-Choice Answers
Let's compare the calculated mean and standard deviation with the given options:

A. [tex]\(H_4 = 21 ; \theta_x = 4\)[/tex] — This is not correct.
B. [tex]\(B = 95 ; 0 = 21\)[/tex] — This gives the population standard deviation, not the standard deviation of the sample mean.
C. [tex]\(H _{ j }=85 ; 0_x=3\)[/tex] — The mean is incorrect.
D. [tex]\(H_{ x }=7.937 ; \sigma_{ x }=4\)[/tex] — Both the mean and standard deviation are incorrect.

None of the provided options correctly list [tex]\( \mu_{\bar{x}} = 95 \)[/tex] and [tex]\( \sigma_{\bar{x}} = 3 \)[/tex].

So, based on the calculations for the mean and standard deviation of the sample mean, the correct values are:

[tex]\[ \mu_{\bar{x}} = 95 \][/tex]
[tex]\[ \sigma_{\bar{x}} = 3 \][/tex]

It appears there was no correct match in the given multiple-choice options. If there was a need to identify the appropriate mean and standard deviation, the closest valid computation from our detailed steps would be as correctly deduced above.