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Ilia_Sergeevich [38]
3 years ago
10

The manager of a warehouse would like to know how many errors are made when a product’s serial number is read by a bar-code read

er. Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each.
Just to be sure, the manager has six more samples taken:


33, 45, 34, 17, 1, and 29 errors, per 1,000 scans each


How do the mean and standard deviation change, based on all 12 samples?
Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

The mean and standard deviation changed to 23.5 and 14.62 respectively, based on all 12 samples.

Step-by-step explanation:

We are given that the Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each.

Representing the data in tabular form;

           X                            X - \bar X                            (X - \bar X)^{2}

          36                     36 - 20.5 = 15.5                   240.25

          14                      14 - 20.5 = -6.5                     42.25  

          21                       21 - 20.5 = 0.5                      0.25

          39                      39 - 20.5 = 18.5                   342.25

           11                        11 - 20.5 = -9.5                     90.25

           2                         2 - 20.5 = -18.5               <u>    342.25    </u>

        Total                                                              <u>   1057.5      </u>

Now, the mean of these value is given by;

        Mean, \bar X  =  \frac{\sum X}{n}

                         =  \frac{36+14+21+39+11+2}{6}

                         =  \frac{123}{6}  =  20.5

Standard deviation formula for discrete distribution is given by;

           Standard deviation, \sigma  =  \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }

                                                 =  \sqrt{\frac{1057.5 }{6-1} } = 14.54

Now, the manager has six more samples taken:

33, 45, 34, 17, 1, and 29 errors, per 1,000 scans each

So, the modified table would be;

           X                            X - \bar X                            (X - \bar X)^{2}

          36                     36 - 23.5 = 12.5                   156.25

          14                      14 - 23.5 = -9.5                     90.25  

          21                       21 - 23.5 = -2.5                     6.25

          39                      39 - 23.5 = 15.5                   240.25

           11                        11 - 23.5 = -12.5                   156.25

           2                         2 - 23.5 = -21.5                   462.25    

          33                       33 - 23.5 = 9.5                     90.25

         45                        45 - 23.5 = 21.5                   462.25

         34                        34 - 23.5 = 10.5                   110.25

          17                         17 - 23.5 = -6.5                    42.25

           1                           1 - 23.5 = -22.5                   506.25

          29                        29 - 23.5 = 5.5               <u>     30.25      </u>

        Total                                                              <u>      2353      </u>

<u />

Now, the mean of these value is given by;

        Mean, \bar X  =  \frac{\sum X}{n}

                         =  \frac{36+14+21+39+11+2+33+45+34+17+1+29}{12}

                         =  \frac{282}{12}  =  23.5

Standard deviation formula for discrete distribution is given by;

           Standard deviation, \sigma  =  \sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }

                                                 =  \sqrt{\frac{2353 }{12-1} } = 14.62

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