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# Engineering

If both the mean and the standard deviation of 50 observations 

x_1,x_2,x_3,.................,x_{50} are equal to 16 ,  then the mean

of (x_1-4)^{2},(x_2-4)^{2},............,(x_{50}-4)^{2} is : 

  • Option 1)

    400

  • Option 2)

    380

  • Option 3)

    525

  • Option 4)

    480

 

Answers (1)
V Vakul

50 observations 

x_1,x_2,x_3,.................,x_{50} 

mean,\bar{x}=\frac{\sum x_i}{50}=16.....................(1)

variance,\sigma ^{2}=\frac{\sum x_i^{2}}{50}-(\bar{x})^{2}=16^{2}

=>\frac{\sum x_i^{2}}{50}=16^{2}+(\bar{x})^{2}=16^{2}+(16)^{2}=512......................(2)

So, mean value of (x_1-4)^{2},(x_2-4)^{2},.........................,(x_{50}-4)^{2}

\\=> \sum\frac{(x_i-4)^{2}}{50}=\frac{\sum x_i^2-8\sum x_i+16\times 50}{50}\\\frac{\sum x_i^2}{50}-\frac{8\sum x_i}{50}+16\\512-8\times16+16\\=400


Option 1)

400

Option 2)

380

Option 3)

525

Option 4)

480

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