Respuesta :
Answer:
[tex]n \geq 36[/tex]
Step-by-step explanation:
For this case we know the mean and the deviation:
[tex] \mu = 1.12 kg , \sigma= 0.0009[/tex]
The mean is given by this:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
If we find the expected value for the sample mean we have:
[tex] E(\bar X) = E(\frac{\sum_{i=1}^n X_i}{n}) =\frac{1}{n} \sum_{i=1}^n E(X_i)[/tex]
Since each observation in the sample [tex] X_1, X_2,...., X_{25}[/tex] have an expectation [tex] E(X_i) = \mu , i =1,2,...,25[/tex] so then we have that:
[tex] E(\bar X) = \frac{1}{n} n\mu = \mu[/tex]
Now for the variance of the sample mean we have this:
[tex]Var (\bar X) = Var(\frac{\sum_{i=1}^n X_i}{n})= \frac{1}{n^2} \sum_{i=1}^n Var(X_i)[/tex]
And again each observation have a variance [tex] \sigma^2_i = 0.0009 , i =1,2,...,25[/tex] then we have:
[tex]Var (\bar X) =\frac{1}{n^2} n(\sigma^2) =\frac{\sigma^2}{n}[/tex]
And then the standard deviation would be:
[tex] Sd(\bar X) =\sqrt{\frac{\sigma^2}{n}}= \frac{\sigma}{\sqrt{n}}[/tex]
And we want that the standard deviation for the sample no more than 0.005 so we have this condition:
[tex]\frac{\sigma}{\sqrt{n}} \leq 0.005[/tex]
And since we know the value of [tex] \sigma= \sqrt{0.0009}=0.03[/tex] we can solve for the value of n like this:
[tex] \frac{0.03}{0.005} \leq \sqrt{n}[/tex]
[tex] 6 \leq \sqrt{n}[/tex]
And if we square both sides we got:
[tex] n\geq 36[/tex]
