Question

the weights of the largemouth bass in one location are skewed right with a mean of 5.13 pounds and a standard deviation of 4.07 pounds.

create a sampling distribution for the mean weight of a sample of size \\(n = 43\\) of largemouth bass in this location.

(a) the mean of the sampling distribution is and the standard deviation is
round to 2 decimal places.

(b) which graph most closely resembles a sampling distribution of sample means?

Explanation:

Calculate the mean of the sampling distribution

Using the Sampling Distribution of the Mean knowledge point

$$ \mu_{\bar{x}} = \mu = 5.13 $$

Calculate the standard deviation of the sampling distribution

Using the Standard Error of the Mean knowledge point

$$ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.07}{\sqrt{43}} \approx \frac{4.07}{6.5574} \approx 0.6207 \approx 0.62 $$

Determine the shape of the sampling distribution

Using the Central Limit Theorem knowledge point
Since the sample size \(n = 43\) is sufficiently large (\(n \ge 30\)), the sampling distribution of the sample mean \(\bar{x}\) is approximately normally distributed (symmetric and bell-shaped), centered at \(\mu_{\bar{x}} = 5.13\). The first graph shows a highly right-skewed distribution of individual weights, whereas the second graph is symmetric, bell-shaped, and centered near 5. Thus, the second graph represents the sampling distribution.

Answer:

Question a

The mean of the sampling distribution is <blank>5.13</blank> and the standard deviation is <blank>0.62</blank>

Question b

• (A) The first graph (skewed right, representing the population or a single sample)
• (B) The second graph (symmetric and bell-shaped, representing the sampling distribution of the mean) (Correct answer)