Respuesta :
Answer:
Mean of sampling distribution = 25 inches
Standard deviation of sampling distribution = 4 inches
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 25 inches
Standard Deviation, σ = 12 inches
Sample size, n = 9
We are given that the distribution of length of the widgets is a bell shaped distribution that is a normal distribution.
a) Mean of the sampling distribution
The best approximator for the mean of the sampling distribution is the population mean itself.
Thus, we can write:
[tex]\bar{x} = \mu = 25\text{ inches}[/tex]
b) Standard deviation of the sampling distribution
[tex]s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{12}{\sqrt{9}} = 4\text{ inches}[/tex]
Using the Central Limit Theorem, it is found that:
- The mean of the distribution would be of 25 inches.
- The standard deviation of the distribution would be of 4 inches.
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For the population:
- Mean of 25 inches, thus [tex]\mu = 25[/tex]
- Standard deviation of 12 inches, thus [tex]\sigma = 12[/tex].
Samples of size 9, thus [tex]n = 9[/tex].
By the Central Limit Theorem:
- The mean of the sampling distribution is [tex]\mu = 25[/tex].
- The standard deviation of the sampling distribution is [tex]s = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{9}} = 4[/tex].
A similar problem is given at https://brainly.com/question/15519778
