Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the mean and variance of 12 observations be  and 4 respectively. Later on, it

Let the mean and variance of 12 observations be \frac{9}{2} and 4 respectively. Later on, it was observed that two
observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is \frac{m}{n} 

where m and n are coprime, then m + n is equal to

Option: 1

316


Option: 2

317


Option: 3

315


Option: 4

314


Answers (1)

best_answer

\begin{aligned} & \frac{\sum x}{12}=\frac{9}{2} \\ & \sum \mathrm{x}=54 \\ & \frac{\sum x^2}{12}-\left(\frac{9}{2}\right)^2=4 \\ & \sum \mathrm{x}^2=291 \\ & \sum \mathrm{x}_{\text {new }}=54-(9+10)+7+14=56 \\ & \sum \mathrm{x}_{\text {new }}^2=291-(81+100)+49+196=355 \\ & \sigma_{\text {new }}^2=\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\ & \sigma_{\text {new }}^2=\frac{281}{36}=\frac{m}{n} \\ & \mathrm{~m}+\mathrm{n}=317 \text { Option (2) } \end{aligned}

Posted by

rishi.raj

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 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question