Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me answer: - Statistics and Probability - JEE Main-6

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)

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

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: Closing ranks for AI, ML, data science at NITs highlight rising demand
anam-khan

JEE Main 2026 session 1 exam from January 21 to 30; registration open soon at jeemain.nta.nic.in
anam-khan

JEE Main 2026 Registration LIVE: NTA application form soon at jeemain.nta.nic.in; session 1 exam dates

Latest Question