Question

A sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters. Give a point estimate for the population standard deviation. Round your answer to three decimal places. 1.48,1.45,1.54,1.52,1.52

Answer 1

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                           $(X-Xbar)^{2}$    

    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 = $\frac{\sum X}{n}$  = $\frac{7.51}{5}$ = 1.502

Standard deviation of data, S.D. = $\sqrt{\frac{\sum (X-Xbar)^{2} }{n-1} }$

                                                     = $\sqrt{\frac{0.00528}{5-1} }$ = 0.036

Therefore, point estimate for the population standard deviation is 0.036 .

Answer 2

Answer:

$s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132$

And for the deviation we have:

$s= \sqrt{0.00132}=0.0363$

And that value represent the best estimator for the population deviation since:

$E(s) =\sigma$

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:

$\bar X =\frac{\sum_{i=1}^n X_}{n}$

And replacing we got:

$\bar X= \frac{1.48+1.45+1.54+1.52+1.52}{5} = 1.502$

And now we can calculate the sample variance with the following formula:

$s^2 =\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

And replacing we got:

$s^2 = \frac{(1.48-1.502)^2 +(1.45-1.502)^2 +(1.54-1.502)^2 +(1.52-1.502)^2 +(1.52-1.502)^2}{5-1} =0.00132$

And for the deviation we have:

$s= \sqrt{0.00132}=0.0363$

And that value represent the best estimator for the population deviation since:

$E(s) =\sigma$