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Tema [17]
3 years ago
9

A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point est

imate for the population variance. Round your answer to three decimal places. 1.15,1.24,1.15,1.27,1.13
Mathematics
2 answers:
xxTIMURxx [149]3 years ago
5 0

Answer:

Point estimate for the population variance = 3.92 * 10^{-3} .

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 - Xbar                                         (X-Xbar)^{2}

      1.13            1.13 - 1.188 = -0.058                                 3.364 * 10^{-3}

      1.15            1.15 - 1.188 = -0.038                                 1.444 * 10^{-3}    

      1.15            1.15 - 1.188 = -0.038                                 1.444 * 10^{-3}

      1.24           1.24 - 1.188 = 0.052                                 2.704 * 10^{-3}

      1.27           1.27 - 1.188 = 0.082                               <u>  6.724 * </u>10^{-3}<u>    </u>

                                                                    \sum (X-Xbar)^{2} <u>= 0.01568   </u>

Firstly, Mean of above data, Xbar = \frac{\sum X}{n} = \frac{1.15+1.24+1.15+1.27+1.13}{5} = 1.188

Point estimate of Population Variance = Sample variance

                                                               = \frac{\sum (X-Xbar)^{2}}{n-1} = \frac{0.01568}{4} = 3.92 * 10^{-3} .

Therefore, point estimate for the population variance = 3.92 * 10^{-3} .

       

MariettaO [177]3 years ago
3 0

Answer:

S² = 0.004

Step-by-step explanation:

Point estimate for the population variance is S²

S^{2}=\frac{\sum (x_{i}-\bar{x})^{2}}{n-1}

S² = Sample Variance

∑ = Sum

x_{i} = Term in data set

\bar{x} = Sample mean

n = Sample size

Sample mean (\bar{x})  =

\bar{x}=\frac{\sum x}{n}

   = \frac{1.15+1.24+1.15+1.27+1.13}{5}

   = \frac{5,94}{5}

  = 1.188

x_{i}     -     \bar{x}             (x_{i}-\bar{x})       (x_{i}-\bar{x})^{2}

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

<u>1.13  -   1.88  =       0.058        0.003364 </u>

<u>5.94                      0.268          0.01568   </u>

<u />(x_{i}-\bar{x})^{2}  =  0.0157

S² = \frac{0.0157}{5-1}   =  0.0039

S² = 0.004

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