Respuesta :
Answer:
[tex] \mu_{\bar X} =\mu = 1.8[/tex]
And for the standard deviation we have:
[tex] \sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{0.4}{\sqrt{50}}=0.0566[/tex]
And then the distribution for the sample mean is:
[tex] \bar X \sim N(\mu= 1.8 grams, \sigma_{\bar X}= 0.0566)[/tex]
Step-by-step explanation:
Assuming this question "Describe the shape, center, and spread of the sampling distribution of [tex]\bar x[/tex]"
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".
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".
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]
For this case we satisfy the condition that the sample size is large because n>30 so then we have:
[tex] \mu_{\bar X} =\mu = 1.8[/tex]
And for the standard deviation we have:
[tex] \sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{0.4}{\sqrt{50}}=0.0566[/tex]
And then the distribution for the sample mean is:
[tex] \bar X \sim N(\mu= 1.8 grams, \sigma_{\bar X}= 0.0566)[/tex]
