Respuesta :
Answer:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And replacing we got:
[tex] \mu_{\bar X}= 32[/tex]
[tex] \sigma_{\bar X} = \frac{77}{\sqrt{23}} =10.06[/tex]
Step-by-step explanation:
Previous concepts
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".
Let X the random variable who represents the variable of interest, with the following properties:
[tex]\mu =32,\sigma =77[/tex]
We select a sample of n=23 nails.
From the central limit theorem we can approximate 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]
And replacing we got:
[tex] \mu_{\bar X}= 77[/tex]
[tex] \sigma_{\bar X} = \frac{77}{\sqrt{23}} =10.06[/tex]
