Respuesta :
Answer:
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex]\mu_{\bar X}= \mu = 119.6[/tex]
And now for the deviation we have this:
[tex] SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3[/tex]
So then the correct answer for this caee would be:
c. 1.30 ounces.
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex]\mu_{\bar X}= \mu = 119.6[/tex]
And now for the deviation we have this:
[tex] SE_{\bar X} = \frac{6.5}{\sqrt{25}}=1.3[/tex]
So then the correct answer for this caee would be:
c. 1.30 ounces.
