Respuesta :
Answer:
Point estimate for the population variance = 3.92 * [tex]10^{-3}[/tex] .
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 - [tex]Xbar[/tex] [tex](X-Xbar)^{2}[/tex]
1.13 1.13 - 1.188 = -0.058 3.364 * [tex]10^{-3}[/tex]
1.15 1.15 - 1.188 = -0.038 1.444 * [tex]10^{-3}[/tex]
1.15 1.15 - 1.188 = -0.038 1.444 * [tex]10^{-3}[/tex]
1.24 1.24 - 1.188 = 0.052 2.704 * [tex]10^{-3}[/tex]
1.27 1.27 - 1.188 = 0.082 6.724 * [tex]10^{-3}[/tex]
[tex]\sum (X-Xbar)^{2}[/tex] = 0.01568
Firstly, Mean of above data, [tex]Xbar[/tex] = [tex]\frac{\sum X}{n}[/tex] = [tex]\frac{1.15+1.24+1.15+1.27+1.13}{5}[/tex] = 1.188
Point estimate of Population Variance = Sample variance
= [tex]\frac{\sum (X-Xbar)^{2}}{n-1}[/tex] = [tex]\frac{0.01568}{4}[/tex] = 3.92 * [tex]10^{-3}[/tex] .
Therefore, point estimate for the population variance = 3.92 * [tex]10^{-3}[/tex] .
Answer:
S² = 0.004
Step-by-step explanation:
Point estimate for the population variance is S²
[tex]S^{2}=\frac{\sum (x_{i}-\bar{x})^{2}}{n-1}[/tex]
S² = Sample Variance
∑ = Sum
[tex]x_{i}[/tex] = Term in data set
[tex]\bar{x}[/tex] = Sample mean
n = Sample size
Sample mean ([tex]\bar{x}[/tex]) =
[tex]\bar{x}=\frac{\sum x}{n}[/tex]
= [tex]\frac{1.15+1.24+1.15+1.27+1.13}{5}[/tex]
= [tex]\frac{5,94}{5}[/tex]
= 1.188
[tex]x_{i}[/tex] - [tex]\bar{x}[/tex] [tex](x_{i}-\bar{x})[/tex] [tex](x_{i}-\bar{x})^{2}[/tex]
1.15 - 1.88 = 0.038 0.001444
1.24 - 1.88 = 0.052 0.002704
1.15 - 1.88 = 0.038 0.001444
1.27 - 1.88 = 0.082 0.006724
1.13 - 1.88 = 0.058 0.003364
5.94 0.268 0.01568
[tex](x_{i}-\bar{x})^{2}[/tex] = 0.0157
S² = [tex]\frac{0.0157}{5-1}[/tex] = 0.0039
S² = 0.004
