Respuesta :
Answer:
[tex]s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132[/tex]
And for the deviation we have:
[tex] s= \sqrt{0.00132}=0.0363[/tex]
And that value represent the best estimator for the population deviation since:
[tex] E(s) =\sigma[/tex]
Step-by-step explanation:
For this case we have the following data:
1.48,1.45,1.54,1.52,1.52
The first step for this cae is find the sample mean with the following formula:
[tex]\bar X =\frac{\sum_{i=1}^n X_}{n}[/tex]
And replacing we got:
[tex] \bar X= \frac{1.48+1.45+1.54+1.52+1.52}{5} = 1.502[/tex]
And now we can calculate the sample variance with the following formula:
[tex] s^2 =\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And replacing we got:
[tex]s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132[/tex]
And for the deviation we have:
[tex] s= \sqrt{0.00132}=0.0363[/tex]
And that value represent the best estimator for the population deviation since:
[tex] E(s) =\sigma[/tex]
Answer:
Point estimate for the population standard deviation = 0.036
Step-by-step explanation:
We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters;
X X - X bar [tex](X-Xbar)^{2}[/tex]
1.48 1.48 - 1.502 = -0.022 0.000484
1.45 1.45 - 1.502 = -0.052 0.002704
1.54 1.54 - 1.502 = 0.038 0.001444
1.52 1.52 - 1.502 = 0.018 0.000324
1.52 1.52 - 1.502 = 0.018 0.000324
Total = 0.00528
Firstly, mean of the data above, X bar = [tex]\frac{\sum X}{n}[/tex] = [tex]\frac{7.51}{5}[/tex] = 1.502
Standard deviation of data, S.D. = [tex]\sqrt{\frac{\sum (X-Xbar)^{2} }{n-1} }[/tex]
= [tex]\sqrt{\frac{0.00528}{5-1} }[/tex] = 0.036
Therefore, point estimate for the population standard deviation is 0.036 .
