A sample of 5 buttons is randomly selected and the following diameters are measured in inches. Give a point estimate for the

[tex]s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

But we need to calculate the mean with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^n X_I}{n}[/tex]

And replacing we got:

[tex]\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072[/tex]

And for the sample variance we have:

[tex]s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003[/tex]

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance [tex]\sigma^2[/tex]

[tex]E(s^2) = \sigma^2[/tex]

Step-by-step explanation:

For this case we have the following data:

1.04,1.00,1.13,1.08,1.11

And in order to estimate the population variance we can use the sample variance formula:

[tex]s^2 = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]

But we need to calculate the mean with the following formula:

[tex]\bar X = \frac{\sum_{i=1}^n X_I}{n}[/tex]

And replacing we got:

[tex]\bar X = \frac{ 1.04+1.00+1.13+1.08+1.11}{5}= 1.072[/tex]

And for the sample variance we have:

[tex]s^2 = \frac{(1.04-1.072)^2 +(1.00-1.072)^2 +(1.13-1.072)^2 +(1.08-1.072)^2 +(1.11-1.072)^2}{5-1}= 0.00277\ approx 0.003[/tex]

And thi is the best estimator for the population variance since is an unbiased estimator od the population variance [tex]\sigma^2[/tex]

[tex]E(s^2) = \sigma^2[/tex]

We need the pictures of the graphs

[tex]x \times 6 = 108[/tex]